Re: [R] Finding convex hull? [Broadcast]
From: Dong-hyun Oh
Dear UseRs,
I would like to know which function is the most efficient in finding
convex hull of points in 3(or 2)-dimensional case?
Functions for finding convex hull is the following:
convex.hull (tripack), chull (grDevices), in.chull (sgeostat),
convhulln (geometry), convexhull.xy (spatstat), calcConvexHull
(PBSmapping).
I also would like to know if there is a function that can be
used for
finding convex hull in multi-dimensional case, that is more than 3-
dimension.
If you had look a bit more carefully, you should have seen that
convhulln (geometry) will handle more than 3 dimensions.
Andy
Thank you in advance.
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Re: [R] Q: loess-like function that allows more predictors?
locfit() in the locfit package can do that.
Andy
From: D. R. Evans
I have a feeling that this may be a stupid question, but here
goes anyway:
is there a function that I can use to replace loess but which allows a
larger number of predictors?
(I have a situation in which it would be very convenient to use 5
predictors, which violates the constraint in loess that the number of
predictors be in the range from 1 to 4.)
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Re: [R] Monotonic interpolation
Not if Mr. excalibur really want interpolating (as oppose to
smooting) splines. Other than linear, I'm not even sure if
it can be done (though I'm no expert on this).
One possibility is to use the cobs package and play with
the amount of smoothing...
Andy
From: Bert Gunter
RSiteSearch(monotone, restr=func) will give you several
packages and
functions for monotone smoothing, including the isoreg()
function in the
standard stats package. You can determine if any of these
does what you
want.
Bert Gunter
Genetech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of excalibur
Sent: Thursday, September 06, 2007 8:04 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] Monotonic interpolation
Le jeu. 6 sept. à 09:45, excalibur a écrit :
Hello everybody, has anyone got a function for smooth monotonic
interpolation
(splines ...) of a univariate function (like a distribution
function for
example) ?
approxfun() might be what your looking for.
Is the result of approxfun() inevitably monotonic ?
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Re: [R] Recursive concatenation
Or something like:
R do.call(paste, c(expand.grid(LETTERS[1:3], 1:3), sep=))
[1] A1 B1 C1 A2 B2 C2 A3 B3 C3
(The ordering is bit different, but that shouldn't matter.)
Andy
From: Dimitris Rizopoulos
try this:
paste(rep(LETTERS[1:3], each = 3), 1:3, sep = )
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
Quoting Dennis Fisher [EMAIL PROTECTED]:
Colleagues,
I want to create the following array:
A1, A2, A3, B1, B2, B3, C1, C2, C3
I recall that there is a trick using c or paste permitting me to
form all combinations of c(A, B, C) and 1:3. But, I can't
recall the trick.
Dennis
Dennis Fisher MD
P (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-415-564-2220
www.PLessThan.com
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Re: [R] Synchronzing workspaces
See the example in ?save on how to set defaults via options().
Andy
From: Gabor Grothendieck
You could try saving prior to quitting in the future if you
want to try
those arguments.
On 9/3/07, Paul August [EMAIL PROTECTED] wrote:
Thanks for sharing your experience. In my case, the
involved machines are Windows Vista, XP and 2000. Not sure
whether it contributes to my problem or not. I will look into
this further.
I just noticed the two arguments ascii and compress for
save. However, my .RData file was created by q() with yes.
The manual says that q() is equivalent to save(list =
ls(all=TRUE), file = .RData). There seems to be no way to
set ascii or compression of save through q function, unless
the q function is replaced explicitly with save(list =
ls(all=TRUE), file = .RData, ascii = T).
Paul.
- Original Message
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Paul August [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Thursday, August 30, 2007 11:24:31 PM
Subject: Re: [R] Synchronzing workspaces
I haven't had similar experience but note that save has ascii=
and compress= arguments. You could check if varying those
parameter values makes a difference.
On 8/30/07, Paul August [EMAIL PROTECTED] wrote:
I used to work on several computers and to use a flash
drive to synchronize the workspace on each machine before
starting to work on it. I found that .RData always caused
some trouble: Often it is corrupted even though there is no
error in copying process. Does anybody have the similar experience?
Paul.
- Original Message
From: Barry Rowlingson [EMAIL PROTECTED]
To: Eric Turkheimer [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Wednesday, August 22, 2007 9:43:57 AM
Subject: Re: [R] Synchronzing workspaces
Eric Turkheimer wrote:
How do people go about synchronizing multiple
workspaces on different
workstations? I tend to wind up with projects spread
around the various
machines I work on. I find that placing the
directories on a server and
reading them remotely tends to slow things down.
If R were to store all its workspace data objects in
individual files
instead of one big .RData file, then you could use a
revision control
system like SVN. Check out the data, work on it, check
it in, then on
another machine just update to get the changes.
However SVN doesn't work too well for binary files -
conflicts being
hard to resolve without someone backing down - so maybe
its not such a
good idea anyway...
On unix boxes and derivatives, you can keep things in
sync efficiently
with the 'rsync' command. I think there are GUI addons
for it, and
Windows ports.
Barry
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Re: [R] Variable Importance - Random Forest
I'm slowly clearing my back-log of r-help messages...
Please see reply inline below.
Andy
From: Mathe, Ewy (NIH/NCI) [F]
Hello,
I am trying to explore the use of random forests for
classification and
am certain about the interpretation of the importance measurements.
When having the option importance = T in the randomForest call, the
resulting 'importance' element matrix has four columns with the
following headings:
0 - mean raw importance score of variable x for class 0 (where
importance is the difference between the permutated data error and the
original test set error)
1 - mean raw importance score of variable x for class 1
MeanDecreaseAccuracy : average lowering of the margin across all cases
(where margin is the proportion of votes for the true class - the
maximum proportion of votes for the other classes)
MeanDecreaseGini : summation of the gini decreases over all
trees in the
forest
Are these definitions correct? Why is the raw importance score
calculated for each class? Could one just average the raw importance
scores for class 0 and 1 to get a composite importance score?
The permutation-based importance measures are based on OOB data. For
each tree in the forest, the difference in error rates on the OOB data
with and without permuting the variable of interest is computed. Call
this d[i] for the i-th tree. The overall importance measure is
mean(d[i]) / se(d[i]), where se(d[i]) is sd(d[i])/sqrt(ntree) (the
standard error). The numbers in the 0 and 1 columns are the
analogs computed separately for the 0 class and 1 class separately.
These are useful, e.g., when balanced sampling is used.
Now, when having the option importance = F in the randomForest call,
the 'importance' element is now a vector. What values are those?
That's the MeanDecreaseGini, because they come at nearly zero additional
computation, so we might as well keep them.
Thank you in advance for any input you may have.
Best,
Ewy
Ewy Mathe, Ph. D.
Laboratory of Human Carcinogenesis
National Cancer Institute, NIH
37 Convent Drive
Building 37, Room 3068
Bethesda, MD 20892-4255
Tel: 301-496-5835
Fax: 301-496-0497
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Re: [R] categorical variable coefficients in QSAR [Broadcast]
No one seemed to have picked up on this, so I'll take a stab:
You need to read para and meta into R as factors, and if you want the
coefficients to match the way you showed, you also need to take care that the
factor levels are in the same order as you showed in the coefficient table.
I cut-and-pasted the three columns of data into R separately, like so:
[copy para data to the clipboard]
R para - factor(scan(clipboard, what=))
Read 22 items
[copy meta data to the clipboard]
R meta - factor(scan(clipboard, what=))
Read 22 items
[copy biological activity to the clipboard]
R y - scan(clipboard)
Read 22 items
[copy the column heading of the coefficient table to the clipboard]
R lvl - scan(clipboard, what=)
Read 6 items
R para - factor(as.character(para), levels=lvl)
R meta - factor(as.character(meta), levels=lvl)
R qsar - lm(y ~ para + meta)
R qsar
Call:
lm(formula = y ~ para + meta)
Coefficients:
(Intercept)paraF paraCl paraBrparaI paraMe
7.8213 0.3400 0.7675 1.0200 1.4287 1.2560
metaF metaCl metaBrmetaI metaMe
-0.3013 0.2068 0.4340 0.5787 0.4540
These coefficients match the ones you showed quite closely.
If you don't reorder the levels of the factors, then by default R orders them
alphabetically, so that Br becomes the reference and all coefficients are
differences from Br.
HTH,
Andy
From: [EMAIL PROTECTED]
Dear list:
I am interested in the following sort of problem, as is found
frequently
in the field of QSAR. I have biological activity as a
function of chemical
structure, with structure defined in a categorical manner in that the
SUBSTITUENT is the levels of the POSITION factor. For
example, data from
Kubinyi (http://www.kubinyi.de/dd-12.pdf) for this type of analysis is
presented as follows:
factor para:
H
F
Cl
Br
I
Me
H
H
H
H
H
F
F
F
Cl
Cl
Cl
Br
Br
Br
Me
Me
factor meta:
H
H
H
H
H
H
F
Cl
Br
I
Me
Cl
Br
Me
Cl
Br
Me
Cl
Br
Me
Me
Br
observed biological activity:
7.46
8.16
8.68
8.89
9.25
9.30
7.52
8.16
8.30
8.40
8.46
8.19
8.57
8.82
8.89
8.92
8.96
9.00
9.35
9.22
9.30
9.52
I then think the following analysis should be appropriate
meta-factor(scan(file=meta,what=character))
para-factor(scan(file=para,what=character))
ba-scan(file=ba)
rslt-lm(ba~meta+para-1)
What I wish to obtain is a coefficient for each substituent at each
position, as does Kubinyi:
H F Cl Br I Me
meta 0.00 -0.30 0.21 0.43 0.58 0.45
para 0.00 0.34 0.77 1.02 1.43 1.26
However, I do not get a coefficient for the Br substituent at the para
position. I would like to know if there is an error in this
formulation.
The technique is quite well established in the field of medicinal
chemistry and it is traditional that the binary incidence
matrix is formed
by hand as an intermediate step in the analysis, instead of the much
simpler formulation that I am considering here.
Thank you for whatever insight you may give.
Prof. Roy Little
Dept. Chem.
Universidad de los Andes
Mérida, Venezuela
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Re: [R] randomForest help
What software are you using, exactly? I'm the maintainer of the
randomForest package, yet I do not know which manual you are quoting.
If you are using the randomForest package, the model object can be saved
to a file by save(Rfobject, file=myRFobject.rda). If you need that to
be in ascii, use ascii=TRUE in save(). You can get it back into R by
using load() or attach().
To run data down the model, use predict(Rfobject, datatopredict) (see
?predict.randomForest).
What exactly do you want to print to a csv file, the prediction? See
?write or ?write.table.
Andy
From: Jennifer Dawn Watts
Hello! As a new R user, I'm sure this will be a silly
question for the
rest of you. I've been able to successfully run a forest but yet to
figure out proper command lines for the following:
1. saving the forest. The guide just says isavef=1. I'm unsure how
expand on this to create the command.
2. Running new data down the mode. Again, the guide just states irunf
3. Print to file. I need to be able to export this data to a cvs
file, to then incorporate into an Arc shapefile. The manual
just says
ntestout.
Again, I feel like these should be easy steps that I just
can't relate
to as a beginner. Any advice would be greatly appreciated.
Thanks,
Jenny
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Re: [R] ecological meaning of randomForest vegetation classification? [Broadcast]
Hi Christoph,
I'm not exactly sure what you're looking for, but I'll take a stab
anyway.
The trees in a random forest is not designed to be interpreted as one
would
with an ordinary tree. There are several things you may try to see if
they help you any. One is the distribution of votes. It looks like you
are
classifying each data point into one of many possible classes. RF with
give
you the fraction of trees in the forest that classified the observation
as
a particular class (and the class with the highest fraction of votes is
the
predicted class). Another is the partial dependence plot: You can
use
plot(importance(rf.object)) to see which variables are the most
important,
and then use partialPlot() to examine their marginal effects. These
offer
some clue of what the RF black box is doing, and hopefully will make
some
sense to you.
Best,
Andy
From: Christoph Muller
Hi, everyone,
I haven't found anything similar in the forum, so here's my
problem (I'm no
expert in R nor statistics):
I have a data set of 59.000 cases with 9 variables each (fractional
coverage of 9 different plant types, such as deciduous broad-leaved
temperate trees or evergreen tropical trees etc.), which was
generated by a
vegetation model.
In order to evaluate the quality of the vegetation model's
output, I want
to compare it to a land-cover data set which has 23 different
land-cover
types (such as needle leaved evergreen forest, dense
broad-leaved forest,
barren, etc.).
A statistician advised me to use the randomForest package in
R and using a
sub-set to generate the random Forest, I get a very good
prediction for the
rest.
However, I need to evaluate how meaningful this
classification is in an
ecological sense (boreal trees should not play a role in the
definition of
tropical land-cover types, for example), otherwise I cannot judge the
quality of the vegetation model's output.
Unfortunately, randomForest gives me about 15.000 splits of
which about
5000 are end branches (rough guess), so it's very hard and
time-consuming
to check each single branch of one of the final trees for its
ecological
meaning.
Is there any utility to summarize the characteristics of each
of the 23
prediction classes? Such as land-cover class 1 has less than
5% of plant
types 1-5, 20-50% of plant type 7 and at least 30% of plant type 8.
Or is there a more suitable method to classify my data?
Thanks a lot in advance!
Christoph
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Re: [R] (Most efficient) way to make random sequences of random sequences
Similarly:
s - c(replicate(N, sample(3)))
Andy
From: roger koenker
One way:
N - 10
s - c(apply(matrix(rep(1:3,N),3,N),2,sample))
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Aug 21, 2007, at 3:49 PM, Emmanuel Levy wrote:
Hi,
I was wondering the what would be the (most efficient) way
to generate
a sequence
of sequences, i mean:
if I have 1,2 and 3.
I'd like to generate a sequence of length N*3 (N ~
1,000,000 or more)
Where random permutations of the sequence 1,2,3 follow each other.
i.e 1,2,3,1,3,2,3,2,1
/!\ The thing is that there should never be twice the same number of
in the same sub-sequence, meaning that this is different from
generating a vector with the numbers 1,2 and 3 randomly distributed.
Any suggestion very welcome! Thanks,
Emmanuel
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Re: [R] RFclustering - is it available in R?
Basically the random forest algorithm can generate a proximity
matrix of the data, and it's up to you how you would want to
proceed from there. You can feed that into clustering
algorithms that accept a similarity matrix, or turn it into a
distance matrix for clustering algorithms that need a distance
matrix (e.g., hclust()). You may or may not want to do
ordination as the UCLA folks suggest.
I think this is one of the great things about working in R:
you have the freedom to choose how you want to proceed from
some intermediate result, and not locked in to something some
one decide to hardwire into the software.
Andy
From: Gavin Simpson
On Wed, 2007-08-15 at 09:44 -0700, David Katz wrote:
Several searches turned up nothing. Perhaps I will try to
implement it if
nobody else has. Thanks.
You can do this with Andy Liaw's randomForest package can do this and
the first hit on a Google search (on term RFclustering) was this:
http://www.genetics.ucla.edu/labs/horvath/RFclustering/RFclust
ering.htm
which shows how one might go about this with some helper functions.
G
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Re: [R] Loading JMP Files
JMP can write CSV, and that's probably a safer choice than XPT.
Andy
From: Diana C. Dolan
Hi,
I know how to use SPSS and JMP, and have quite a few
JMP files I would like to use in R. I converted them
to .xpt files, downloaded the 'foreign' library then
tried this command:
read.xport(D:\\Databases\nameoffile.xpt)
to which I get:
Error in lookup.xport(file) : unable to open file
I have read FAQ lists and Google searched and cannot
figure out what I'm doing wrong that my files won't
open. I tried saving to the C drive, but no luck
there. I also have no luck getting R to read my SPSS
files with read.spss
My file names do have spaces and dashes, and I do have
variables/variable names longer than 8 characters.
Please help! I am very new to R and do not understand
all the package reference manuals...I can not seem to
find a simple, basic guide to how to command R and use
basic functions without a bunch of jargon (eg 'usage'
and 'arguments'). It would help to at least be able
to load my files to practice on.
Any help would be appreciated!
Thanks,
Diana
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Re: [R] rfImpute
I seem to recall that rfImpute() can sometimes come up with NAs at some
point in the iterations. Could you please send me a (small) set of
data/code that reproduces the problem?
Andy
From: Eric Turkheimer
I am having trouble with the rfImpute function in the
randomForest package.
Here is a sample...
clunk.roughfix-na.roughfix(clunk)
clunk.impute-rfImpute(CONVERT~.,data=clunk)
ntree OOB 1 2
300: 26.80% 3.83% 85.37%
ntree OOB 1 2
300: 18.56% 5.74% 51.22%
Error in randomForest.default(xf, y, ntree = ntree, ...,
do.trace = ntree,
:
NA not permitted in predictors
So roughFix works, but rfImpute doesn't
Thanks,
Eric
ent3c *at* virginia.edu
--
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Department of Psychology
University of Virginia
PO Box 400400
Charlottesville, VA 22904-4400
434-982-4732
434-982-4766 (FAX)
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Re: [R] Sourcing commands but delaying their execution
Here's one possibility:
The file garbage.R has
x - rnorm(100)
print(summary(x))
You can do:
cmds - parse(file=garbage.R, n=NA)
and when you want to execute those commands, do
eval(cmds)
Andy
From: Dennis Fisher
Colleagues:
I have encountered the following situation:
SERIES OF COMMANDS
source(File1)
MORE COMMANDS
source(File2)
Optimally, I would like File1 and File2 to be merged into a single
file (FileMerged). However, if I wrote the following:
SERIES OF COMMANDS
source(FileMerged)
MORE COMMANDS
I encounter an error: the File2 portion of FileMerged contains
commands that cannot be executed properly until MORE COMMANDS are
executed. Similarly, sourcing FileMerged after MORE COMMANDS does
not work because MORE COMMANDS requires the information from the
File1 portion of FileMerged.
I am looking for a means to source FileMerged but not execute
some of
the commands immediately. Functionally this would look like:
SERIES OF COMMANDS
source(FileMerged)# but withhold execution of
some of the commands
MORE COMMANDS
COMMAND TO EXECUTE THE WITHHELD COMMANDS
Does R offer some option to accomplish this?
Dennis
Dennis Fisher MD
P (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-415-564-2220
www.PLessThan.com
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Re: [R] getting the name of variables passed to a function
Here's one possibility:
R f - function(...) { call - match.call(); sapply(as.list(call[-1]),
deparse) }
R f(x, y)
[1] x y
R f(x=x, y=y)
x y
x y
You basically need to know how to manipulate call objects. The relevant
section in the R Language Definition should help.
Andy
From: Horace Tso
Folks,
I've entered into an R programming territory I'm not very
familiar with, thus this probably very elementary question
concerning the mechanic of a function call.
I want to know from within a function the name of the
variables I pass down. The function makes use of the ... to
allow for multiple unknown arguments,
myfun = function(...) { do something }
In the body I put,
{
nm - names(list(...))
nm
}
When the function is called with two vectors x, and y
myfun(x, y)
It returns NULL. However, when the call made is,
myfun(x=x, y=y)
The result is
[1] x y
Question : how do i get the names of the unknown variables
without explicitly saying x=x...
Thanks in advance.
Horace
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Re: [R] randomForest importance problem with combine [Broadcast]
I've been fixing some problems in the combine() function, but that's
only for regression data. Looks like you are doing classification, and
I don't see the problem:
R library(randomForest)
randomForest 4.5-19
Type rfNews() to see new features/changes/bug fixes.
R set.seed(1)
R rflist - replicate(50, randomForest(iris[-5], iris[[5]], ntree=50,
importance=TRUE), simplify=FALSE)
R rfall - do.call(combine, rflist)
R importance(rfall)
setosa versicolor virginica MeanDecreaseAccuracy
Sepal.Length 0.4457861 0.53883425 0.55806570.4120840
Sepal.Width 0.3266790 0.07652383 0.36202400.2128450
Petal.Length 1.1950989 1.42014628 1.32204710.7989841
Petal.Width 1.1986973 1.40855969 1.36406200.7951053
MeanDecreaseGini
Sepal.Length 9.578580
Sepal.Width 2.301172
Petal.Length42.935832
Petal.Width 44.409058
R importance(rflist[[1]])
setosa versicolor virginica MeanDecreaseAccuracy
Sepal.Length 0.401714 0.71583422 0.49464200.4166555
Sepal.Width 0.00 -0.03155946 0.68292870.2317111
Petal.Length 1.290430 1.47915219 1.34567700.8219003
Petal.Width 1.110142 1.44996777 1.35847990.7881210
MeanDecreaseGini
Sepal.Length 6.168439
Sepal.Width 2.240723
Petal.Length48.821726
Petal.Width 42.059112
Please provide a reproducible example.
Andy
From: Joseph Retzer
My apologies, subject corrected.
I'm building a RF 50 trees at a time due to memory limitations (I have
roughly .5 million observations and around 20 variables). I thought I
could combine some or all of my forests later and look at global
importance.
If I have say 2 forests : tree1 and tree2, they have similar Gini and
Raw importances and, additionally, are similar to one another. After
combining (using the combine command) the trees into one however, the
combined tree Raw importances have changed in rank order
rather dramtically
(e.g. the top most important becomes least important. It is not
however a completely reversed ordering). In addtion, the
scale of both the
Raw and Gini importances is orders of magnitude smaller for
the combined
tree.
Note that the combined tree Gini importance looks roughly similar to
the individual tree Gini (and Raw) importance, at least in
terms of rank
ordering.
I'm using the non-formula randomForest specification along with
norm.votes=FALSE to facilitate large sample estimation and tree
combining.
I'm using R 2.5.0 on a windows XP machine with 2 gig RAM. I'm also
using randomForest 4.5-18.
Any advice is appreciated,
Many thanks,
Joe
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Re: [R] Viewing a data object
I believe JGR has an object browser. See the screenshots at the bottom
of http://rosuda.org/JGR/.
Andy
From: Stephen Tucker
Hi Horace,
I have also thought that it may be useful but I don't know of
any Object
Explorer available for R.
However, (you may alread know this but)
(1) you can view your list of objects in R with objects(),
(2) view objects in a spreadsheet-like table (if they are
matrices or data
frames) with invisible(edit(objectName)) [which isn't easy on
the fingers].
fix(objectName) is also a shorter option but it has the side effect of
possibly changing your object when you close the viewing
data. For instance,
this can happen if you mistakenly type something into a cell;
it can also
change your column classes when you don't - for example:
options(stringsAsFactors=TRUE)
x - data.frame(letters[1:5],1:5)
sapply(x,class)
letters.1.5. X1.5
factorinteger
fix(x) # no user-changes made
sapply(x,class)
letters.1.5. X1.5
factornumeric
(3) I believe Deepayan Sarkar contributed the tab-completion
capability at
the command line. So unless you have a lot of objects beginning with
'AuroraStoch...' you should be able to type a few letters and let the
auto-completion handle the rest.
Best regards,
ST
--- Horace Tso [EMAIL PROTECTED] wrote:
Dear list,
First apologize that this is trivial and just betrays my
slothfulness at
the keyboard. I'm sick of having to type a long name just
to get a glimpse
of something. For example, if my data frame is named
'AuroraStochasticRunsJune1.df and I want to see what the
middle looks
like, I have to type
AuroraStochasticRunsJune1.df[ 400:500, ]
And often I'm not even sure rows 400 to 500 are what I want
to see. I
might have to type the same line many times.
Is there sort of a R-equivalence of the Object Explorer,
like in Splus,
where I could mouse-click an object in a list and a window
pops up? Short
of that, is there any trick of saving a couple of
keystrokes here and
there?
Thanks for tolerating this kind of annoying questions.
H.
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Re: [R] R for bioinformatics
Just to complete this thread: A colleague sent me the following
regarding the book.
Following up on this post from a few months back...
The author has recently posted a public-domain version of this
book on CRAN under Documentation - Contributed -
Statistics Using R with Biological Examples by Kim Seefeld and Ernst
Linder (PDF).
Unfortunately not all the mirror sites have it yet.
Andy
From: Benoit Ballester
Marc Schwartz wrote:
On Thu, 2007-02-01 at 21:32 +0100, Peter Dalgaard wrote:
Marc Schwartz wrote:
On Thu, 2007-02-01 at 10:45 -0800, Seth Falcon wrote:
Benoit Ballester [EMAIL PROTECTED] writes:
Hi,
I was wondering if someone could tell me more about
this book, (if it's
a good or bad one).
I can't find it, as it seems that O'Reilly doesn't
publish any more.
I've never seen a copy so I can't comment about its quality (has
anyone seen a copy?).
You might want to take a look at _Bioinformatics and
Computational
Biology Solutions Using R and Bioconductor_.
http://www.bioconductor.org/pub/docs/mogr/
I'll stand (or sit) to be corrected on this as I cannot
find the source,
but I have a recollection from seeing something quite
some time ago that
the book may have never been published.
It's been a while since the status was something along the
lines that
the authors may or may not complete it. Subject matter
moving faster
than pen, I suspect
Peter, that wording does seem familiar, just cannot recall
where I saw
it. Perhaps on the O'Reilly web site, where it is no longer listed.
For confirmation, I called O'Reilly's customer service in
Cambridge, MA.
They confirm that the book was indeed cancelled and never published.
No reasons were given.
Thanks for those replies.
I did also contacted the O'reilly offices in UK, and they told me the
same thing. The book was never published. I just wanted to
compare the
R for bioinformatics with the Bioinformatics and Computational
Biology Solutions Using R and Bioconductor, and see which
one suit me
more - But guess I don't have the choice now :-)
Ben
--
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Ensembl Team
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Re: [R] normality tests [Broadcast]
From: [EMAIL PROTECTED]
On 25/05/07, Frank E Harrell Jr [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] wrote:
Hi all,
apologies for seeking advice on a general stats question. I ve run
normality tests using 8 different methods:
- Lilliefors
- Shapiro-Wilk
- Robust Jarque Bera
- Jarque Bera
- Anderson-Darling
- Pearson chi-square
- Cramer-von Mises
- Shapiro-Francia
All show that the null hypothesis that the data come from a normal
distro cannot be rejected. Great. However, I don't think
it looks nice
to report the values of 8 different tests on a report. One note is
that my sample size is really tiny (less than 20
independent cases).
Without wanting to start a flame war, are there any
advices of which
one/ones would be more appropriate and should be reported
(along with
a Q-Q plot). Thank you.
Regards,
Wow - I have so many concerns with that approach that it's
hard to know
where to begin. But first of all, why care about
normality? Why not
use distribution-free methods?
You should examine the power of the tests for n=20. You'll probably
find it's not good enough to reach a reliable conclusion.
And wouldn't it be even worse if I used non-parametric tests?
I believe what Frank meant was that it's probably better to use a
distribution-free procedure to do the real test of interest (if there is
one) instead of testing for normality, and then use a test that assumes
normality.
I guess the question is, what exactly do you want to do with the outcome
of the normality tests? If those are going to be used as basis for
deciding which test(s) to do next, then I concur with Frank's
reservation.
Generally speaking, I do not find goodness-of-fit for distributions very
useful, mostly for the reason that failure to reject the null is no
evidence in favor of the null. It's difficult for me to imagine why
there's insufficient evidence to show that the data did not come from a
normal distribution would be interesting.
Andy
Frank
--
Frank E Harrell Jr Professor and Chair School
of Medicine
Department of Biostatistics
Vanderbilt University
--
yianni
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Re: [R] in unix opening data object created under win
What are the versions of R on the two platform? Is the version on Unix
at least as new as the one on Windows?
Andy
From: [EMAIL PROTECTED]
Hi All
I am saving a dataframe in my MS-Win R with save().
Then I copy it onto my personal AFS space.
Then I start R and run it with emacs and load() the data.
It loads only 2 lines: head() shows only two lines nrow() als
say it has only 2
lines, I get error message, when trying to use this data
object, saying that
some row numbers are missing.
If anyone had similar situation, I appreciate letting me know.
Best Toby
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Re: [R] using lm() with variable formula [Broadcast]
One way to do it is by giving a data frame with the right variables to
lm() as the first argument each time. If lm() is given a data frame as
the first argument, it will treat the first variable as the LHS and the
rest as the RHS of the formula.
As examples, you can do:
lm(myData[c(height, weight, BP, Cals)])
(The drawback to this is that the formula in the fitted model object
looks a bit strange...)
Andy
From: Chris Elsaesser
New to R; please excuse me if this is a dumb question. I
tried to RTFM;
didn't help.
I want to do a series of regressions over the columns in a data.frame,
systematically varying the response variable and the the
terms; and not
necessarily including all the non-response columns. In my case, the
columns are time series. I don't know if that makes a difference; it
does mean I have to call lag() to offset non-response terms. I can not
assume a specific number of columns in the data.frame; might
be 3, might
be 20.
My central problem is that the formula given to lm() is different each
time. For example, say a data.frame had columns with the following
headings: height, weight, BP (blood pressure), and Cals
(calorie intake
per time frame). In that case, I'd need something like the following:
lm(height ~ weight + BP + Cals)
lm(height ~ weight + BP)
lm(height ~ weight + Cals)
lm(height ~ BP + Cals)
lm(weight ~ height + BP)
lm(weight ~ height + Cals)
etc.
In general, I'll have to read the header to get the argument labels.
Do I have to write several functions, each taking a different
number of
arguments? I'd like to construct a string or list representing the
varialbes in the formula and apply lm(), so to say [I'm mainly a Lisp
programmer where that part would be very simple. Anyone have
a Lisp API
for R? :-}]
Thanks,
chris
Chris Elsaesser, PhD
Principal Scientist, Machine Learning
SPADAC Inc.
7921 Jones Branch Dr. Suite 600
McLean, VA 22102
703.371.7301 (m)
703.637.9421 (o)
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Re: [R] more woes trying to convert a data.frame to a numerical matrix
I think this might be a bit more straight forward: R mat - do.call(cbind, scan(clipboard, what=list(NULL, 0, 0, 0), sep=,, skip=2)) Read 3 records R mat [,1] [,2] [,3] [1,]123 [2,]456 [3,]789 Andy From: Andrew Yee Thanks again to everyone for all your help. I think I've figured out the solution to my dilemma. Instead of using data.matrix or sapply, this works for me: sample.data-read.csv(sample.csv) sample.matrix.raw-as.matrix(sample.data[-1,-1]) sample.matrix - matrix(as.numeric(sample.matrix.raw), nrow=attributes(sample.matrix.raw)$dim[1], ncol=attributes( sample.matrix.raw)$dim[2]) With the above code, I get the desired matrix of: 1 2 3 4 5 6 7 8 9 (I'd like to be able to import the whole csv and then subset the relevant header and data sections (rather than creating a separate csv for the header and csv for the data) Of course, the above code seems kind of clunky, and welcome any suggestions for improvement. Thanks, Andrew On 5/16/07, Andrew Yee [EMAIL PROTECTED] wrote: Thanks for the suggestion. However, I've tried sapply and data.matrix. The problem is that it while it returns a numeric matrix, it gives back: 1 1 1 2 2 2 3 3 3 instead of 1 2 3 4 5 6 7 8 9 The latter matrix is the desired result Thanks, Andrew On 5/16/07, Marc Schwartz [EMAIL PROTECTED] wrote: On Wed, 2007-05-16 at 08:40 -0400, Andrew Yee wrote: Thanks for the suggestion and the explanation for why I was running into these troubles. I've tried: as.numeric(as.matrix(sample.data[-1, -1])) However, this creates another vector rather than a matrix. Right. That's because I'm an idiot and need more caffeine... :-) Is there a straight forward way to convert this directly into a numeric matrix rather than a vector? Yeah, Dimitris' approach below of using data.matrix(). You could also use: mat - sapply(sample.data[-1, -1], as.numeric) rownames(mat) - rownames(sample.data[-1, -1]) mat x y z 2 1 1 1 3 2 2 2 4 3 3 3 Though, this is essentially what data.matrix() does internally. Additionally, I've also considered: data.matrix(sample.data[-1,-1] but bizarrely, it returns: x y z 2 1 1 1 3 2 2 2 4 3 3 3 That is a numeric matrix: str(data.matrix(sample.data[-1, -1])) int [1:3, 1:3] 1 2 3 1 2 3 1 2 3 - attr(*, dimnames)=List of 2 ..$ : chr [1:3] 2 3 4 ..$ : chr [1:3] x y z HTH, Marc Thanks, Andrew On 5/16/07, Marc Schwartz [EMAIL PROTECTED] wrote: On Wed, 2007-05-16 at 08:10 -0400, Andrew Yee wrote: I have the following csv file: name,x,y,z category,delta,gamma,epsilon a,1,2,3 b,4,5,6 c,7,8,9 I'd like to create a numeric matrix of just the numbers in this csv dataset. I've tried the following program: sample.data - read.csv(sample.csv) numerical.data - as.matrix (sample.data[-1,-1]) However, print(numerical.data ) returns what appears to be a matrix of characters: x y z 2 1 2 3 3 4 5 6 4 7 8 9 How do I force it to be numbers rather than characters? Thanks, Andrew The problem is that you have two rows which contain alpha entries. The first row is treated as the header, but the second row is treated as actual data, thus overriding the numeric values in the subsequent rows. You could use: as.numeric(as.matrix(sample.data [-1, -1])) to coerce the matrix to numeric, or if you don't need the alpha entries, you could modify the read.csv() call to something like: read.csv(sample.csv, header = FALSE, skip = 2, row.names = 1, col.names = c(name, x, y, z) This will skip the first two rows, set the first column to the row names and give you a data frame with numeric columns, which in most cases can be treated as a numeric matrix and/or you could explicitly coerce it to one. HTH, Marc Schwartz [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and
Re: [R] Testing for existence inside a function [Broadcast]
Not sure which one you want, but the following should cover it:
R f - function(x) c(x=missing(x), y=exists(y))
R f(1)
x y
FALSE FALSE
R f()
x y
TRUE FALSE
R y - 1
R f()
xy
TRUE TRUE
R f(1)
x y
FALSE TRUE
Andy
From: Talbot Katz
Hi.
I'm having trouble testing for existence of an object inside
a function.
Suppose I have a function:
f-function(x){
...
}
and I call it with argument y:
f(y)
I'd like to check inside the function whether argument y
exists. Is this
possible, or do I have to either check outside the function
or pass the name
of the argument as a separate argument?
If I do exists(x) or exists(eval(x)) inside the function and
y does not
exist, it generates an error message. If I do exists(x) it
says that x
exists even if y does not. If I had a separate argument to
hold the text
string y then I could check that. But is it possible to check the
existence of the argument inside the function without passing
its name as a
separate argument?
Thanks!
-- TMK --
212-460-5430 home
917-656-5351 cell
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Testing for existence inside a function
Just need a bit more work:
R f - function(x) exists(deparse(substitute(x)))
R f(y)
[1] FALSE
R y - 1
R f(y)
[1] TRUE
R f(z)
[1] FALSE
Andy
From: Talbot Katz
Hi, Andy.
Thank you for the quick response! Unfortunately, none of
these are exactly
what I'm looking for. I'm looking for the following:
Suppose object y
exists and object z does not exist. If I pass y as the value of the
argument to my function, I want to be able to verify, inside
my function,
the existence of y; similarly, if I pass z as the value of
the argument, I
want to be able to see, inside the function, that z doesn't exist.
The missing function just checks whether the argument is
missing; in my
case, the argument is not missing, but the object may not
exist. And the
way you use the exists function inside the user-defined
function doesn't
test the argument to the user-defined function, it's just
hard-coded for the
object y. So I'm sorry if I wasn't clear before, and I hope
this is clear
now. Perhaps what I'm attempting to do is unavailable
because it's a bad
programming paradigm. But even an explanation if that's the
case would be
appreciated.
-- TMK --
212-460-5430 home
917-656-5351 cell
From: Liaw, Andy [EMAIL PROTECTED]
To: Talbot Katz [EMAIL PROTECTED],r-help@stat.math.ethz.ch
Subject: RE: [R] Testing for existence inside a function [Broadcast]
Date: Tue, 15 May 2007 11:03:12 -0400
Not sure which one you want, but the following should cover it:
R f - function(x) c(x=missing(x), y=exists(y))
R f(1)
x y
FALSE FALSE
R f()
x y
TRUE FALSE
R y - 1
R f()
xy
TRUE TRUE
R f(1)
x y
FALSE TRUE
Andy
From: Talbot Katz
Hi.
I'm having trouble testing for existence of an object inside
a function.
Suppose I have a function:
f-function(x){
...
}
and I call it with argument y:
f(y)
I'd like to check inside the function whether argument y
exists. Is this
possible, or do I have to either check outside the function
or pass the name
of the argument as a separate argument?
If I do exists(x) or exists(eval(x)) inside the function and
y does not
exist, it generates an error message. If I do exists(x) it
says that x
exists even if y does not. If I had a separate argument to
hold the text
string y then I could check that. But is it possible
to check the
existence of the argument inside the function without passing
its name as a
separate argument?
Thanks!
-- TMK --
212-460-5430 home
917-656-5351 cell
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
-
-
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Whitehouse Station,
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outside the United States as Merck Frosst, Merck Sharp Dohme or MSD
and in Japan, as Banyu - direct contact information for affiliates is
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Testing for existence inside a function
Another thing to watch out for is that an argument to a function can be
an expression (or even literal constants), instead of just the name of
an object. exists() wouldn't really do the right thing. I'm not sure
how to properly do the exhaustive check.
Andy
From: Gabor Grothendieck
Try this modification:
chk - function(x) exists(deparse(substitute(x)),
parent.env(environment()))
ab - 1
chk(ab)
[1] TRUE
exists(x)
[1] FALSE
chk(x)
[1] FALSE
On 5/15/07, Talbot Katz [EMAIL PROTECTED] wrote:
Hi.
Thanks once more for the swift response. This solution
works pretty well.
The only small glitch is if I pass the function an argument
with the same
name as the function argument. That is, suppose x is the
argument name in
my user-defined function, and that object x does not
exist. If I call the
function f(x), i.e., using the non-existent object x as the
argument value,
then the function says that x exists.
Here is my example log:
chkex5 - function(objn){
+ c(exob=exists(deparse(substitute(objn
+ }
exists(objn)
[1] FALSE
chkex5(objn)
exob
TRUE
But I suppose I can live with this. Thanks again!
-- TMK --
212-460-5430home
917-656-5351cell
From: Liaw, Andy [EMAIL PROTECTED]
To: Talbot Katz [EMAIL PROTECTED],r-help@stat.math.ethz.ch
Subject: RE: [R] Testing for existence inside a function
Date: Tue, 15 May 2007 11:41:17 -0400
Just need a bit more work:
R f - function(x) exists(deparse(substitute(x)))
R f(y)
[1] FALSE
R y - 1
R f(y)
[1] TRUE
R f(z)
[1] FALSE
Andy
From: Talbot Katz
Hi, Andy.
Thank you for the quick response! Unfortunately, none of
these are exactly
what I'm looking for. I'm looking for the following:
Suppose object y
exists and object z does not exist. If I pass y as the
value of the
argument to my function, I want to be able to verify, inside
my function,
the existence of y; similarly, if I pass z as the value of
the argument, I
want to be able to see, inside the function, that z
doesn't exist.
The missing function just checks whether the argument is
missing; in my
case, the argument is not missing, but the object may not
exist. And the
way you use the exists function inside the user-defined
function doesn't
test the argument to the user-defined function, it's just
hard-coded for the
object y. So I'm sorry if I wasn't clear before, and I hope
this is clear
now. Perhaps what I'm attempting to do is unavailable
because it's a bad
programming paradigm. But even an explanation if that's the
case would be
appreciated.
-- TMK --
212-460-5430home
917-656-5351cell
From: Liaw, Andy [EMAIL PROTECTED]
To: Talbot Katz [EMAIL PROTECTED],r-help@stat.math.ethz.ch
Subject: RE: [R] Testing for existence inside a
function [Broadcast]
Date: Tue, 15 May 2007 11:03:12 -0400
Not sure which one you want, but the following should cover it:
R f - function(x) c(x=missing(x), y=exists(y))
R f(1)
x y
FALSE FALSE
R f()
x y
TRUE FALSE
R y - 1
R f()
xy
TRUE TRUE
R f(1)
x y
FALSE TRUE
Andy
From: Talbot Katz
Hi.
I'm having trouble testing for existence of an object inside
a function.
Suppose I have a function:
f-function(x){
...
}
and I call it with argument y:
f(y)
I'd like to check inside the function whether argument y
exists. Is this
possible, or do I have to either check outside the function
or pass the name
of the argument as a separate argument?
If I do exists(x) or exists(eval(x)) inside the
function and
y does not
exist, it generates an error message. If I do
exists(x) it
says that x
exists even if y does not. If I had a separate argument to
hold the text
string y then I could check that. But is it possible
to check the
existence of the argument inside the function
without passing
its name as a
separate argument?
Thanks!
-- TMK --
212-460-5430home
917-656-5351cell
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
reproducible code.
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-
Notice: This e-mail message, together with any
attachments, contains
information of Merck Co., Inc. (One Merck
Re: [R] Optimized File Reading with R
If it's a matrix, use scan(). If the columns are not all the same type,
use the colClasses argument to read.table() to specify their types,
instead of leaving it to R to guess. That will speed things up quite a
lot.
Andy
From: Lorenzo Isella
Dear All,
Hope I am not bumping into a FAQ, but so far my online search
has been fruitless
I need to read some data file using R. I am using the (I think)
standard command:
data_150-read.table(y_complete06000, header=FALSE)
where y_complete06000 is a 6000 by 40 table of numbers.
I am puzzled at the fact that R is taking several minutes to
read this file.
First I thought it may have been due to its shape, but even
re-expressing and saving the matrix as a 1D array does not help.
It is not a small file, but not even huge (it amounts to about 5Mb of
text file).
Is there anything I can do to speed up the file reading?
Many thanks
Lorenzo
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Re: [R] geeting name of an object to which a variable refers?
Something like this?
R f - function(x) deparse(substitute(x))
R a - 1:3
R f(a)
[1] a
Andy
From: new ruser
#Sorry for the convoluted subject line.
#I have:
a=c(1,2,3)
x=a #example of user supplied input
#Is there any function that will tell me the name of the
object x refers to, referring only to x itself?
#i.e. the answer I want is a
#I want:
#fun(x) == 'a'
#(I don't think this is possible, but figured I'd ask.)
-
Got a little couch potato?
Check out fun summer activities for kids.
[[alternative HTML version deleted]]
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Re: [R] Allocating shelf space
I don't know if there's an R solution, but this sounds to me like some
variation of the knapsack problem...
http://en.wikipedia.org/wiki/Knapsack_problem
Andy
From: [EMAIL PROTECTED]
Hi Folks,
This is not an R question as such, though it may well have
an R answer. (And, in any case, this community probably
knows more about most things than most others ... indeed,
has probably pondered this very question).
I: Given a catalogue of hundreds of books, where each
entry has author and title (or equivalent ID), and also
Ia) The dimensions (thickness, height, depth) of the book
Ib) A sort of classification of its subject/type/genre
II: Given also a specification of available and possibly
potential bookshelf space (numbers of book-cases, the width,
height and shelf-spacing of each, and the dimensions of any
free wall-space where further book-cases may be placed),
where some book-cases have fixed shelves and some have shelves
with (discretely) adjustable position, and additional book-cases
can be designed to measure (probably with adjustable shelves).
Question: Is there a resource to approach the solution of the
problem of optimising the placement of adjustable shelves,
the design of additional bookcases, and the placement of the
books in the resulting shelf-space so as to
A: Make the efficient use of space
B: Minimise the spatial disclocation of related books
(it is acceptable to separate large books from small books
on the same subject, for the sake of efficient packing).
Awaiting comments and suggestions with interest!
With thanks,
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 09-May-07 Time: 18:23:53
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Re: [R] summing values according to a factor
Howdy!
I guess what you want to do is compare Q1/T1 among the sections? If you
want to compute the sum of Q1/T1 by Section, you can do something like:
sum.by.section - with(mydata, tapply(Q1/T1, section, sum))
Substitute sum with anything you want to compute.
Cheers,
Andy
From: Salvatore Enrico Indiogine
Greetings!
I have exam scores of students in several sections. The data
looks like:
StuNum Section Q1 T1
111 502 45 123
112 502 23123
113 503 58123
114 504 63 123
115 504 83 123
..
where Q1 is the score for question 1 and T1 is the maximum possible
score for question 1
I need to check whether the section has an effect on the scores. I
thought about using chisq.test and calculate the sums of scores per
section.
I think that I have to use apply() but I am lost here.
Thanks in advance,
Enrico
--
Enrico Indiogine
Mathematics Education
Texas AM University
[EMAIL PROTECTED]
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Re: [R] R package development in windows
I guess it depends on what you want to be able to do with such a private
package; e.g., does it not need to have any documentation (i.e., the Rd
files)? If all you want is to be able to access the objects, you can
just save() all those objects (mostly functions, I presume) in a .rda
file, and whenever you need them. just attach() the .rda file.
Andy
From: Lucke, Joseph F
Might there be an (semi-)automated procedure to create a minimal,
personal package, for my eyes only, that I can load with a
libray(MyStuff) command? This would be preferable to having to
source() the files. Is there already such a procedure?
Joe
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Doran, Harold
Sent: Thursday, May 03, 2007 2:33 PM
To: Duncan Murdoch
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] [SPAM] - Re: R package development in windows -
BayesianFilter detected spam
Thanks, Duncan. I'll look into that. Is there an
authoritative document
that codifies the new package development procedures for 2.5.0
(windows-specific), or is that Writing R Extensions? In this thread
alone I've received multiple emails pointing to multiple web
sites with
instructions for windows. Inasmuch as its appreciated, I'm a bit
confused as to which I should consider authoritative.
I do hope I can resolve this and appreciate the help I've received.
However, I feel a bit compelled to note how very difficult
this process
is.
Harold
-Original Message-
From: Duncan Murdoch [mailto:[EMAIL PROTECTED]
Sent: Thursday, May 03, 2007 3:24 PM
To: Doran, Harold
Cc: Gabor Grothendieck; r-help@stat.math.ethz.ch
Subject: [SPAM] - Re: [R] R package development in windows
- Bayesian
Filter detected spam
On 5/3/2007 3:04 PM, Doran, Harold wrote:
Thanks Gabor, Sundar, and Tony. Indeed, Rtools was
missing from the
path. With that resolved, and another 10 minute windows
restart, I get
the following below. The log suggests that hhc is not
installed. It
is, and, according to the directions I am following, I have
placed it
in the c:\cygwin directory.
I think the problem is that you are following a real mix of
instructions, and they don't make sense.
It would be nice if folks would submit patches to the R
Admin manual
(or to the Rtools web site) rather than putting together web sites
with advice that is bad from day one, and quickly gets
worse when it
is not updated.
BTW, package.skeleton() doesn't seem to create the correct
DESCRIPTION
template. I had to add the DEPENDS line. Without this, I
get another
error.
C:\Program Files\R\R-2.4.1\binRcmd build --force --binary g:\foo
R 2.4.1 is no longer current; the package building
instructions in R
2.5.0 have been simplified a bit. You might want to try those.
Duncan Murdoch
* checking for file 'g:\foo/DESCRIPTION' ... OK
* preparing 'g:\foo':
* checking DESCRIPTION meta-information ... OK
* removing junk files
* checking for LF line-endings in source files
* checking for empty or unneeded directories
* building binary distribution
WARNING
* some HTML links may not be found
installing R.css in c:/TEMP/Rinst40061099
Using auto-selected zip options ''
latex: not found
latex: not found
latex: not found
-- Making package foo
latex: not found
adding build stamp to DESCRIPTION
latex: not found
latex: not found
latex: not found
installing R files
latex: not found
installing data files
latex: not found
installing man source files
installing indices
latex: not found
not zipping data
installing help
Warning: \alias{foo} already in foo-package.Rd -- skipping
the one in
foo.Rd Building/Updating help pages for package 'foo'
Formats: text html latex example chm
foo-package texthtmllatex
example chm
foo texthtmllatex
example chm
mydatatexthtmllatex
example chm
hhc: not found
cp: cannot stat `c:/TEMP/Rbuild40048815/foo/chm/foo.chm':
No such file
or direct ory
make[1]: *** [chm-foo] Error 1
make: *** [pkg-foo] Error 2
*** Installation of foo failed ***
Removing 'c:/TEMP/Rinst40061099/foo'
ERROR
* installation failed
C:\Program Files\R\R-2.4.1\bin
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Thursday, May 03, 2007 2:50 PM
To: Doran, Harold
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] R package development in windows
It can find sh.exe so you haven't installed Rtools.
There are several HowTo's listed in the links section here that
include pointers to R manuals and other step by step
instructions:
Re: [R] R question [Broadcast]
Bill,
A couple more points:
1. Please use an informative subject line. I'd deleted the original
post w/o reading if I didn't catch Marc's reply.
2. Are you sure you have bivariate response? To me bivariate means
two variables, and randomForest surely does not handle that (at least
for now).
Andy
From: Marc Schwartz
On Fri, 2007-05-04 at 12:05 -0500, Bill Vorias wrote:
I had a question about Random Forests. I have a text file with 10
dichotomous variables and a bivariate response vector. I
read this file
into R as a data frame, and then used the command
randomForest(Response ~.,
dataset, etc.. where Response is the column header of
the response
variable and dataset is the name of the data frame. I
get an error that
says Response not found. I was looking at the Iris data
example in the R
help files, and it seems like this is exactly what they
did. Do you have
any suggestions? Thanks.
R you sure that you have correctly specified the column and data frame
names in the call to randomForest()?
Be sure to check for typos, including capitalization.
You can use:
ls()
to check for the current objects in your working environment
and you can
then use:
str(YourDataFrame)
or
names(YourDataFrame)
to display information about the detailed structure and/or
column names,
respectively, in the data frame that you created from the
imported data.
HTH,
Marc Schwartz
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Re: [R] how to concatinate the elements of some text vectors cat() or print() ?
Is paste() what you're looking for?
Andy
From: John Kane
I have some comment text taken from a SAS data file.
It is stored in two vectors and is difficult to read.
I would like to simply concatentate the individual
entries and end up with a character vector that give
me one line of text per comment.
I cannot see how to do this, yet it must be very easy.
I have played around with cat() and print with no
success. Would someone kindly point out where I
am going wrong?
Thanks
Simple Example:
aa - LETTERS[1:5]
bb - letters[1:5]
cat(aa[1], bb[1])# works for individuals
cat(aa,bb)#(concatinates entire vectors)
# Using sink I might get it to work if I could figure
out how to escape a
# new line command. encodeString does not seem
appropriate here.
harry - c(rep(NA,5))
for (i in 1:5 ) {
cat (aa[i],bb[i])
}
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[R] thousand separator (was RE: weight)
I've run into this occasionally. My current solution is simply to read it into Excel, re-format the offending column(s) by unchecking the thousand separator box, and write it back out. Not exactly ideal to say the least. If anyone can provide a better solution in R, I'm all ears... Andy From: Natalie O'Toole Hi, These are the variables in my file. I think the variable i'm having problems with is WTPP which is of the Factor type. Does anyone know how to fix this, please? Thanks, Nat data.frame': 290 obs. of 5 variables: $ PROV : num 48 48 48 48 48 48 48 48 48 48 ... $ REGION: num 4 4 4 4 4 4 4 4 4 4 ... $ GRADE : num 7 7 7 7 7 7 7 7 7 7 ... $ Y_Q10A: num 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 ... $ WTPP : Factor w/ 1884 levels 1,106.8250,1,336.5138,..: 1544 67 1568 40 221 1702 1702 1434 310 310 ... __ --- Douglas Bates [EMAIL PROTECTED] wrote: On 4/28/07, John Kane [EMAIL PROTECTED] wrote: IIRC you have a yes/no smoking variable scored 1/2 ? It is possibly being read in as a factor not as an integer. try class(df$smoking.variable) to see . Good point. In general I would recommend using str(df) to check on the class or storage type of all variables in a data frame if you are getting unexpected results when manipulating it. That function is carefully written to provide a maximum of information in a minimum of space. Yes but I'm an relative newbie at R and didn't realise that str() would do that. I always thought it was some kind of string function. Thanks, it makes life much easier. --- Natalie O'Toole [EMAIL PROTECTED] wrote: Hi, I'm getting an error message: Error in df[, 1:4] * df[, 5] : non-numeric argument to binary operator In addition: Warning message: Incompatible methods (Ops.data.frame, Ops.factor) for * here is my code: ##reading in the file happyguys-read.table(c:/test4.dat, header=TRUE, row.names=1) ##subset the file based on Select If test-subset(happyguys, PROV==48 GRADE == 7 Y_Q10A 9) ##sorting the file mydata-test mydataSorted-mydata[ order(mydata$Y_Q10A), ] print(mydataSorted) ##assigning a different name to file happyguys-mydataSorted ##trying to weight my data data.frame-happyguys df-data.frame df1-df[, 1:4] * df[, 5] ##getting error message here?? Error in df[, 1:4] * df[, 5] : non-numeric argument to binary operator In addition: Warning message: Incompatible methods (Ops.data.frame, Ops.factor) for * Does anyone know what this error message means? I've been reviewing R code all day getting more familiar with it Thanks, Nat -- -- This communication is intended for the use of the recipient to which it is addressed, and may contain confidential, personal, and or privileged information. Please contact the sender immediately if you are not the intended recipient of this communication, and do not copy, distribute, or take action relying on it. Any communication received in error, or subsequent reply, should be deleted or destroyed. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Be smarter than spam. See how smart SpamGuard is at giving junk email the boot with the All-new Yahoo! Mail at http://mrd.mail.yahoo.com/try_beta?.intl=ca -- -- This communication is intended for the use of the recipient to which it is addressed, and may contain confidential, personal, and or privileged information. Please contact the sender immediately if you are not the intended recipient of this communication, and do not copy, distribute, or take action relying on it. Any communication received in error, or subsequent reply, should be deleted or destroyed. [[alternative HTML version deleted]] __
Re: [R] thousand separator (was RE: weight)
Still, though, it would be nice to have the data read in correctly in the first place, instead of having to do this kind of post-processing afterwards... Andy From: Bert Gunter Nothing! My mistake! gsub -- not sub -- is what you want to get 'em all. -- Bert Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marc Schwartz Sent: Monday, April 30, 2007 10:18 AM To: Bert Gunter Cc: r-help@stat.math.ethz.ch Subject: Re: [R] thousand separator (was RE: weight) Bert, What am I missing? print(as.numeric(gsub(,, , 1,123,456.789)), 10) [1] 1123456.789 FWIW, this is using: R version 2.5.0 Patched (2007-04-27 r41355) Marc On Mon, 2007-04-30 at 10:13 -0700, Bert Gunter wrote: Except this doesn't work for 1,123,456.789 Marc. I hesitate to suggest it, but gregexpr() will do it, as it captures the position of **every** match to ,. This could be then used to process the vector via some sort of loop/apply statement. But I think there **must** be a more elegant way using regular expressions alone, so I, too, await a clever reply. -- Bert Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marc Schwartz Sent: Monday, April 30, 2007 10:02 AM To: Liaw, Andy Cc: r-help@stat.math.ethz.ch Subject: Re: [R] thousand separator (was RE: weight) One possibility would be to use something like the following post-import: WTPP [1] 1,106.8250 1,336.5138 str(WTPP) Factor w/ 2 levels 1,106.8250,1,336.5138: 1 2 as.numeric(gsub(,, , WTPP)) [1] 1106.825 1336.514 Essentially strip the ',' characters from the factors and then coerce the resultant character vector to numeric. HTH, Marc Schwartz On Mon, 2007-04-30 at 12:26 -0400, Liaw, Andy wrote: I've run into this occasionally. My current solution is simply to read it into Excel, re-format the offending column(s) by unchecking the thousand separator box, and write it back out. Not exactly ideal to say the least. If anyone can provide a better solution in R, I'm all ears... Andy From: Natalie O'Toole Hi, These are the variables in my file. I think the variable i'm having problems with is WTPP which is of the Factor type. Does anyone know how to fix this, please? Thanks, Nat data.frame': 290 obs. of 5 variables: $ PROV : num 48 48 48 48 48 48 48 48 48 48 ... $ REGION: num 4 4 4 4 4 4 4 4 4 4 ... $ GRADE : num 7 7 7 7 7 7 7 7 7 7 ... $ Y_Q10A: num 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 ... $ WTPP : Factor w/ 1884 levels 1,106.8250,1,336.5138,..: 1544 67 1568 40 221 1702 1702 1434 310 310 ... __ --- Douglas Bates [EMAIL PROTECTED] wrote: On 4/28/07, John Kane [EMAIL PROTECTED] wrote: IIRC you have a yes/no smoking variable scored 1/2 ? It is possibly being read in as a factor not as an integer. try class(df$smoking.variable) to see . Good point. In general I would recommend using str(df) to check on the class or storage type of all variables in a data frame if you are getting unexpected results when manipulating it. That function is carefully written to provide a maximum of information in a minimum of space. Yes but I'm an relative newbie at R and didn't realise that str() would do that. I always thought it was some kind of string function. Thanks, it makes life much easier. --- Natalie O'Toole [EMAIL PROTECTED] wrote: Hi, I'm getting an error message: Error in df[, 1:4] * df[, 5] : non-numeric argument to binary operator In addition: Warning message: Incompatible methods (Ops.data.frame, Ops.factor) for * here is my code: ##reading in the file happyguys-read.table(c:/test4.dat, header=TRUE, row.names=1) ##subset the file based on Select If test-subset(happyguys, PROV==48 GRADE == 7 Y_Q10A 9) ##sorting the file mydata-test mydataSorted-mydata[ order(mydata$Y_Q10A), ] print(mydataSorted) ##assigning a different name to file happyguys-mydataSorted ##trying to weight my data data.frame-happyguys df-data.frame df1-df[, 1:4] * df[, 5] ##getting error message here?? Error in df[, 1:4] * df[, 5] : non-numeric argument to binary operator In addition: Warning message
Re: [R] thousand separator (was RE: weight)
Looks very neat, Gabor! I just cannot fathom why anyone who want to write numerics with those separators in a flat file. That's usually not for human consumption, and computers don't need those separators! Andy From: Gabor Grothendieck That could be accomplished using a custom class like this: library(methods) setClass(num.with.junk) setAs(character, num.with.junk, function(from) as.numeric(gsub(,, , from))) ### test ### Input - A B 1,000 1 2,000 2 3,000 3 DF - read.table(textConnection(Input), header = TRUE, colClasses = c(num.with.junk, numeric)) str(DF) On 4/30/07, Liaw, Andy [EMAIL PROTECTED] wrote: Still, though, it would be nice to have the data read in correctly in the first place, instead of having to do this kind of post-processing afterwards... Andy From: Bert Gunter Nothing! My mistake! gsub -- not sub -- is what you want to get 'em all. -- Bert Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marc Schwartz Sent: Monday, April 30, 2007 10:18 AM To: Bert Gunter Cc: r-help@stat.math.ethz.ch Subject: Re: [R] thousand separator (was RE: weight) Bert, What am I missing? print(as.numeric(gsub(,, , 1,123,456.789)), 10) [1] 1123456.789 FWIW, this is using: R version 2.5.0 Patched (2007-04-27 r41355) Marc On Mon, 2007-04-30 at 10:13 -0700, Bert Gunter wrote: Except this doesn't work for 1,123,456.789 Marc. I hesitate to suggest it, but gregexpr() will do it, as it captures the position of **every** match to ,. This could be then used to process the vector via some sort of loop/apply statement. But I think there **must** be a more elegant way using regular expressions alone, so I, too, await a clever reply. -- Bert Bert Gunter Genentech Nonclinical Statistics -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marc Schwartz Sent: Monday, April 30, 2007 10:02 AM To: Liaw, Andy Cc: r-help@stat.math.ethz.ch Subject: Re: [R] thousand separator (was RE: weight) One possibility would be to use something like the following post-import: WTPP [1] 1,106.8250 1,336.5138 str(WTPP) Factor w/ 2 levels 1,106.8250,1,336.5138: 1 2 as.numeric(gsub(,, , WTPP)) [1] 1106.825 1336.514 Essentially strip the ',' characters from the factors and then coerce the resultant character vector to numeric. HTH, Marc Schwartz On Mon, 2007-04-30 at 12:26 -0400, Liaw, Andy wrote: I've run into this occasionally. My current solution is simply to read it into Excel, re-format the offending column(s) by unchecking the thousand separator box, and write it back out. Not exactly ideal to say the least. If anyone can provide a better solution in R, I'm all ears... Andy From: Natalie O'Toole Hi, These are the variables in my file. I think the variable i'm having problems with is WTPP which is of the Factor type. Does anyone know how to fix this, please? Thanks, Nat data.frame': 290 obs. of 5 variables: $ PROV : num 48 48 48 48 48 48 48 48 48 48 ... $ REGION: num 4 4 4 4 4 4 4 4 4 4 ... $ GRADE : num 7 7 7 7 7 7 7 7 7 7 ... $ Y_Q10A: num 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 ... $ WTPP : Factor w/ 1884 levels 1,106.8250,1,336.5138,..: 1544 67 1568 40 221 1702 1702 1434 310 310 ... __ --- Douglas Bates [EMAIL PROTECTED] wrote: On 4/28/07, John Kane [EMAIL PROTECTED] wrote: IIRC you have a yes/no smoking variable scored 1/2 ? It is possibly being read in as a factor not as an integer. try class(df$smoking.variable) to see . Good point. In general I would recommend using str(df) to check on the class or storage type of all variables in a data frame if you are getting unexpected results when manipulating it. That function is carefully written to provide a maximum of information in a minimum of space. Yes but I'm an relative newbie at R and didn't realise that str() would do that. I always thought it was some kind of string function. Thanks, it makes life much easier. --- Natalie O'Toole [EMAIL PROTECTED] wrote: Hi, I'm getting an error message: Error in df[, 1:4] * df[, 5] : non-numeric argument to binary operator
Re: [R] NA and NaN randomForest
Hi Clayton,
If you use the formula interface, then it should do what you want:
R library(randomForest)
randomForest 4.5-18
Type rfNews() to see new features/changes/bug fixes.
R iris1 - iris[-(1:5),]
R iris2 - iris[1:5,]
R iris2[1, 3] - NA
R iris2[3, 1] - NA
R iris.rf - randomForest(Species ~ ., iris1)
R predict(iris.rf, iris2[-5])
[1] NA setosa NA setosa setosa
Levels: setosa versicolor virginica
The problem, of course, is that the formula interface is not suitable
for data with large number of variables. I'll look into doing the same
thing in the default method.
Andy
From: [EMAIL PROTECTED]
Dear R-help,
This is about randomForest's handling of NA and NaNs in test set data.
Currently, if the test set data contains an NA or NaN then
predict.randomForest will skip that row in the output.
I would like to change that behavior to outputting an NA.
Can this be done with flags to randomForest?
If not can some sort of wrapper be built to put the NAs back in?
thanks,
Clayton
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Re: [R] Getting Confused [Broadcast]
If you are serious in getting useful help, please do try to follow
suggestions in the Posting Guide. You have not told us anything about
your OS, the versions of R you tried to install, and exactly what you
typed to build/install them.
Many Linux distro by default do not install the Fortran part of GCC, so
don't be surprised if that's the case for you (if you are trying to do
this on some version of Linux).
Andy
From: Steiner, Julien
Hello,
I'm getting confused with my experience of R installing.
I had R installed on January without any trouble. (I just had
to install
gcc4.1.1)
Now I'd like to install a packages which requires tcl/tk. So
basically I
need to reconfigure and re install R right after having installed
tcl/tk.
So I installed tcl/tk I run the process to install R but I
receive this
error :
checking for dummy main to link with Fortran libraries...
none
checking for Fortran name-mangling scheme... configure:
error: cannot compile a simple Fortran program See
`config.log' for more
details.
I checked in the config.log and the fact is that there's no fortran
compiler installed. But don't gcc already have a fortran compiler in
it?
If somebody could help I would be thankful and especially if somebody
has a clue why it worked without any error before and now yes.
Thanks a lot
Julien Steiner
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Re: [R] how to convert the lower triangle of a matrix to a symmetricmatrix [Broadcast]
Ranjan and Prof. Fox,
Similar approach can be found in stats:::as.matrix.dist().
Andy
From: John Fox
Dear Ranjan,
If the elements are ordered by rows, then the following
should do the trick:
X - diag(p)
X[upper.tri(X, diag=TRUE)] - elements
X - X + t(X) - diag(diag(X))
If they are ordered by columns, substitute lower.tri() for
upper.tri().
I hope this helps,
John
John Fox, Professor
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Ranjan Maitra
Sent: Thursday, April 19, 2007 9:28 PM
To: r-help@stat.math.ethz.ch
Subject: [R] how to convert the lower triangle of a matrix to
a symmetricmatrix
Hi,
I have a vector of p*(p+1)/2 elements, essentially the lower
triangle of a symmetric matrix. I was wondering if there is
an easy way to make it fill a symmetric matrix. I have to do
it several times, hence some efficient approach would be
very useful.
Many thanks and best wishes,
Ranjan
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Re: [R] Suggestions for statistical computing course
I really like John Monahan's Numerical Methods of Statistics (Cambridge
University Press).
As to running/editing R scripts, you may want to look into JGR. The
built-in editor is not as smart as ESS in some respect, but smarter
than ESS in others. The only thing that keep me from using it regularly
is the fact that it won't take arguments to R itself (at least on
Windows): I need the --internet2 argument to be able to access the net
from R.
Andy
From: Giovanni Petris
Dear R-helpers,
I am planning a course on Statistical Computing and Computational
Statistics for the Fall semester, aimed at first year Masters students
in Statistics. Among the topics that I would like to cover are linear
algebra related to least squares calculations, optimization and
root-finding, numerical integration, Monte Carlo methods (possibly
including MCMC), bootstrap, smoothing and nonparametric density
estimation. Needless to say, the software I will be using is R.
1. Does anybody have a suggestion about a book to follow that covers
(most of) the topics above at a reasonable revel for my audience?
Are there any on-line publicly-available manuals, lecture notes,
instructional documents that may be useful?
2. I do most of my work in R using Emacs and ESS. That means that I
keep a file in an emacs window and I submit it to R one line at a
time or one region at a time, making corrections and iterating as
needed. When I am done, I just save the file with the last,
working, correct (hopefully!) version of my code. Is there a way of
doing something like that, or in the same spirit, without using
Emacs/ESS? What approach would you use to polish and save your code
in this case? For my course I will be working in a Windows
environment.
While I am looking for simple and effective solutions that do not
require installing emacs in our computer lab, the answer you
should teach your students emacs/ess on top of R is perfecly
acceptable.
Thank you for your consideration, and thank you in advance for the
useful replies.
Have a good day,
Giovanni
--
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Department of Mathematical Sciences
University of Arkansas - Fayetteville, AR 72701
Ph: (479) 575-6324, 575-8630 (fax)
http://definetti.uark.edu/~gpetris/
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Re: [R] Help on averaging sets of rows defined by row name
You might want to check which of the following scales better for the
size of data you have.
## Make up some data to try.
R dat - data.frame(gene=rep(letters[1:3], each=3), s1=runif(9),
s2=runif(9))
R dat
genes1s2
1a 0.9959172 0.9531052
2a 0.2064497 0.4257022
3a 0.4791100 0.5977923
4b 0.1307096 0.8256453
5b 0.7887983 0.8904983
6b 0.7841745 0.6901540
7c 0.3356583 0.7125086
8c 0.5859311 0.0509323
9c 0.7681325 0.8677725
## Use aggregate():
R aggregate(dat[-1], dat[1], mean)
genes1s2
1a 0.5604923 0.6588666
2b 0.5678941 0.8020992
3c 0.5632407 0.5437378
## Do it by hand: need a bit more work if there are Nas.
R rowsum(dat[-1], dat[[1]]) / table(dat[[1]])
s1s2
a 0.5604923 0.6588666
b 0.5678941 0.8020992
c 0.5632407 0.5437378
Andy
From: Booman, M
Dear all,
This is my problem: I have a table of gene expression data,
where 1st column is gene name, and 2nd -39th columns each are
exression data for 38 samples. There are multiple
measurements per sample for each gene, so there are multiple
rows for each gene name. I want to average these measurements
so i end up with one value per sample for each gene name. The
output data frame (table.averaged) is further used in other R
script. The code I use now (see below) takes 20 secs for each
loop, so it takes 45 minutes to average my files of 13500
unique genes. Can anyone help me do this faster?
Cheers, marije
Code I use:
table.imputed[,1] - as.character(table.imputed[,1])
#table.imputed is data.frame,1st column = gene name (class
factor), rest of columns = expression data (class numeric)
genesunique - unique(table.imputed[,1])
#To make list of unique genes in the set
table.averaged - NULL
for (j in 1:length(genesunique)) {
if (j%%100 == 0){
#To report progress
cat(j, genes finished, sep= , fill=TRUE)
}
table.averaged-rbind(table.averaged,givemean(genesunique[j],
table.imputed)) #collects all rows of average values and
binds them back into one data frame
}
givemean - function (gene, table.imputed) {
thisgene-table.imputed[table.imputed[,1]==gene,]
#make a subset containing only
the rows for one gene name
data.frame(gene,t(sapply(thisgene[,2:ncol(thisgene)],mean,
na.rm=TRUE))) #calculates average for each sample
(column) and outputs one row of average values and the gene name
}
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Re: [R] Help on averaging sets of rows defined by row name
Do note that I used dat[1] instead of dat[,1] or dat[[1]] as the second
argument to aggregate(): If dat is a data frame, then dat[1] is also a
data frame with only the first column. Since data frame is also a list,
dat[1] is a one-component list.
My guess is that Tierry didn't try his suggestion, or else he would have
noticed the error.
Andy
From: Booman, M [mailto:[EMAIL PROTECTED]
Sent: Friday, April 20, 2007 10:26 AM
To: Liaw, Andy; r-help@stat.math.ethz.ch
Subject: RE: [R] Help on averaging sets of rows defined by row
name
Thanks for your help everyone!
I had some trouble with the 'aggregate' function because the
'table.impute[,1]' was not a list (which the 'by' argument should be),
and it took a very very long time to coerce it into one. But the
rowmeans method works almost instantly! And I have no problems with NA's
because I used a knn imputer first.
-Original Message-
From: Liaw, Andy [mailto:[EMAIL PROTECTED]
Sent: Fri 4/20/2007 4:09 PM
To: Booman, M; r-help@stat.math.ethz.ch
Subject: RE: [R] Help on averaging sets of rows defined by row
name
You might want to check which of the following scales better for
the
size of data you have.
## Make up some data to try.
R dat - data.frame(gene=rep(letters[1:3], each=3),
s1=runif(9),
s2=runif(9))
R dat
genes1s2
1a 0.9959172 0.9531052
2a 0.2064497 0.4257022
3a 0.4791100 0.5977923
4b 0.1307096 0.8256453
5b 0.7887983 0.8904983
6b 0.7841745 0.6901540
7c 0.3356583 0.7125086
8c 0.5859311 0.0509323
9c 0.7681325 0.8677725
## Use aggregate():
R aggregate(dat[-1], dat[1], mean)
genes1s2
1a 0.5604923 0.6588666
2b 0.5678941 0.8020992
3c 0.5632407 0.5437378
## Do it by hand: need a bit more work if there are Nas.
R rowsum(dat[-1], dat[[1]]) / table(dat[[1]])
s1s2
a 0.5604923 0.6588666
b 0.5678941 0.8020992
c 0.5632407 0.5437378
Andy
From: Booman, M
Dear all,
This is my problem: I have a table of gene expression data,
where 1st column is gene name, and 2nd -39th columns each are
exression data for 38 samples. There are multiple
measurements per sample for each gene, so there are multiple
rows for each gene name. I want to average these measurements
so i end up with one value per sample for each gene name. The
output data frame (table.averaged) is further used in other R
script. The code I use now (see below) takes 20 secs for each
loop, so it takes 45 minutes to average my files of 13500
unique genes. Can anyone help me do this faster?
Cheers, marije
Code I use:
table.imputed[,1] - as.character(table.imputed[,1])
#table.imputed is data.frame,1st column = gene name (class
factor), rest of columns = expression data (class numeric)
genesunique - unique(table.imputed[,1])
#To make list of unique genes in the set
table.averaged - NULL
for (j in 1:length(genesunique)) {
if (j%%100 == 0){
#To report progress
cat(j, genes finished, sep= , fill=TRUE)
}
table.averaged-rbind(table.averaged,givemean(genesunique[j],
table.imputed)) #collects all rows of average values and
binds them back into one data frame
}
givemean - function (gene, table.imputed) {
thisgene-table.imputed[table.imputed[,1]==gene,]
#make a subset containing only
the rows for one gene name
data.frame(gene,t(sapply(thisgene[,2:ncol(thisgene)],mean,
na.rm=TRUE))) #calculates average for each sample
(column) and outputs one row of average values and the gene
name
}
De inhoud van dit bericht is vertrouwelijk en alleen bestemd
voor de geadresseerde(n). Anderen dan de geadresseerde mogen
geen gebruik maken van dit bericht, het openbaar maken of op
enige wijze verspreiden of vermenigvuldigen. Het UMCG kan
niet aansprakelijk gesteld worden voor een incomplete
aankomst of vertraging van dit verzonden bericht.
The contents of this message are confidential and only
intended for the eyes of the addressee(s
Re: [R] How to return more than one variable from function
From: Vincent Goulet
Le Vendredi 20 Avril 2007 07:46, Julien Barnier a écrit :
Hi,
I have written a function which computes variance, sd,
r^2, R^2adj etc. But i am not able to return all of
them in return statement.
You can return a vector, or a list.
For example :
func - function() {
...
result - list(variance=3, sd=sqrt(3))
return(result) # you can omit this
}
Nitpicking and for the record: if you omit the
return(result) line, the
function will return nothing since it ends with an
assignment.
Have you actually checked? Counterexample:
R f - function(x) y - 2 * x
R f(3)
R (z - f(3))
[1] 6
R f2 - function(x) { y - 2 * x; y }
R f2(3)
[1] 6
Furthermore,
explicit use of return() is never needed at the end of a
function. The above
snippet is correct, but this is enough:
func - function() {
...
result -list(variance=3, sd=sqrt(3))
result
}
But then, why assign to a variable just to return its value?
Better still:
func - function() {
...
list(variance=3, sd=sqrt(3))
}
Or, as has been suggested, if all values to be returned are of the same type,
just use a (named) vector:
func - function(...) {
...
c(Variance=3, R-squared=0.999)
}
Andy
a - func()
a$variance
a$sd
HTH,
Julien
--
Vincent Goulet, Professeur agrégé
École d'actuariat
Université Laval, Québec
[EMAIL PROTECTED] http://vgoulet.act.ulaval.ca
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Re: [R] Udpate R under a proxy
This is what I just tried (thanks, Dirk!):
Start R and then Sys.putenv(http_proxy, whatever),
options(download.file.method=wget) doesn't work.
Open up a command prompt, define http_proxy there, then run Rgui. Set
options(download.file.method=wget). This works.
Perhaps you can define http_proxy in Renviron. I have not tried that.
Andy
From: justin bem
dear all,
I get internet via a proxy server when I try to
downlaod package it always fail. Even when i add and
environnment variable for the http proxy server. I use
windows XP SP2
Sincerly
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
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Re: [R] general question about plotting multiple regression results [Broadcast]
I suspect you'll greatly benefit a read of Prof. Fox's book(s) on
regression models, as well as making use of his car package. You may
want to read up on partial residual plots and partial regression plots.
Andy
From: Simon Pickett
Hi all,
I have been bumbling around with r for years now and still
havent come up with a solution for plotting reliable graphs
of relationships from a linear regression.
Here is an example illustrating my problem
1.I do a linear regression as follows
summary(lm(n.day13~n.day1+ffemale.yell+fmale.yell+fmale.chroma
,data=surv))
which gives some nice sig. results
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) -0.739170.43742 -1.690 0.093069 .
n.day11.004600.05369 18.711 2e-16 ***
ffemale.yell 0.224190.06251 3.586 0.000449 ***
fmale.yell0.258740.06925 3.736 0.000262 ***
fmale.chroma 0.235250.11633 2.022 0.044868 *
2. I want to plot the effect of ffemale.yell, fmale.yell
and fmale.chroma on my response variable.
So, I either plot the raw values (which is fine when there is
a very strong relationship) but what if I want to plot the
effects from the model?
In this case I would usually plot the fitted values values
against the raw values of x... Is this the right approach?
fit-fitted(lm(n.day13~n.day1+ffemale.yell+fmale.yell+fmale.ch
roma,data=fsurv1))
plot(fit~ffemale.yell)
#make a dummy variable across the range of x
x-seq(from=min(fsurv1$ffemale.yell),to=max(fsurv1$ffemale.yel
l), length=100)
#get the coefficients and draw the line
co-coef(lm(fit~ffemale.yell,data=fsurv1))
y-(co[2]*x)+co[1]
lines(x,y, lwd=2)
This often does the trick but for some reason, especially
when my model has many terms in it or when one of the
independent variables is only significant when the other
independent variables are in the equation, it gives me strange lines.
Please can someone show me the light?
Thanks in advance,
Simon.
Simon Pickett
PhD student
Centre For Ecology and Conservation
Tremough Campus
University of Exeter in Cornwall
TR109EZ
Tel 01326371852
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Re: [R] R in cron job: X problems
This is in the FAQ, if I remember correctly... However, alternatively:
As Jeff Horner recently pointed out on the list, the Cairo package is a
good way of generating png without needing an X display. You may want
to look into that. I've just installed cairo on our CentOS boxes and
the Cairo package from CRAN.
Andy
From: Mark Liberman
I'd like to use an R CMD BATCH script as part of a chron job
that is set up to run every hour.
The trouble is that the script creates a graphical output in
a file via png(), and apparently this in turn works through X.
When cron invokes the job, no X server is available -- I
suppose that the DISPLAY variable is not set -- and so R
exits with an error message in the output file. (If I run the
same script in an environment where an X server is properly
available, it works as I want it to.)
I tried setting DISPLAY to thecomputername:0.0 (where
thecomputername
is the X.Y.Z form of the computer's name as names it for ssh
etc.), but that didn't work.
Any advice about how to persuade the graphics subsystem to
bypass X, or how to set DISPLAY in a safe way to run in a cron job?
This is a linux system (a recent RedHat server system) with R 2.2.1.
Thanks,
Mark Liberman
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Re: [R] installing new packages
See if 2.19 The Internet download functions fail in the R for Windows
FAQ helps.
Andy
From: Bill Shipley
Hello,
I have just installed the newest version of R (2.4.1) for
Windows XP. I can no longer install new packages. When
trying to connect to a server (I have tried several) I get
the following message:
chooseCRANmirror()
Error in open.connection(file, r) : unable to open connection
In addition: Warning message:
unable to connect to 'cran.r-project.org' on port 80.
Have other people had the same problem with this version, or
is it unique to my computer? Can someone suggest a solution?
Thanks.
Bill Shipley
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Re: [R] Subsetting list of vectors with list of (boolean) vectors?
From: Marc Schwartz
On Thu, 2007-04-12 at 18:12 +0200, Johannes Graumann wrote:
Dear Rologists,
I'm stuck with this. How would you do this efficiently:
aPGI
[[1]]
[1] 864 5576
aPGItest
[[1]]
[1] TRUE FALSE
result - [magic box involving subset)
result
[[1]]
[1] 864
Thanks for any hints,
Joh
lapply(seq(along = length(aPGI)), function(x)
aPGI[[x]][aPGItest[[x]]])
[[1]]
[1] 864
Alternatively:
R mapply([, aPGI, aPGItest, SIMPLIFY=FALSE)
[[1]]
[1] 864
Cheers,
Andy
I think that this should be a generic solution for multiple
(but common) levels in each list.
HTH,
Marc Schwartz
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Re: [R] Random Forest Imputations [Broadcast]
Please provide the information the posting guide asks (version of R, packages
used, version of package used, etc). There are no yaImpute() or yai()
functions in the randomForest package.
Andy
From: [EMAIL PROTECTED] on behalf of Ricky Jacob
Sent: Wed 4/11/2007 5:55 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Random Forest Imputations [Broadcast]
Dear All,
I am not able to run the random forest with my dataset..
X- 280 records with satellite data(28 columns) - B1min, b1max, b1std etc..
y- 280 records with 3 columns - TotBasal Area, Stem density and Volume
yref - y[1:230,] #Keeping 1st 230 records as reference records
want to set 0 to y values for records 231 to 280..
yimp - y[231:280,] #records for which we want to impute the basal area,
stem density and volume
mal1 - yai(x=x, y=yref, method=mahalanobis, k=1, noRefs = TRUE) # This
works fine for mahalanobis, msn, gnn, raw and Euclidean
Want to do a similar thing with random forest where the 1st 230 records
alone should be used for calculating Nearest Neighbours for the records with
number 231 to 280..
What needs to be done.. Went through the yaImpute document.. but all i
could do without any error message was to have NN generated using the yai()
where all 280 records have been used for finding nearest neighbour.
Regards
Ricky
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Re: [R] Random Forest Imputations [Broadcast] [Broadcast]
The package has a doc/ subdirectory (in the pre-compiled package, or inst/doc in the source package), which contains yaImputePaper.pdf. Page 9 of that document may be of some help to you. This is the first time I've seen this package, so can't help you much there. It looks like the package authors would like me to add some feature to the randomForest package (which I maintain). I'll look into that. Andy From: Ricky Jacob [mailto:[EMAIL PROTECTED] Sent: Wednesday, April 11, 2007 7:11 AM To: Liaw, Andy Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Random Forest Imputations [Broadcast] [Broadcast] I am currently using R 2.4.1 version. Am using the yaImpute package for k-NN imputation.. http://forest.moscowfsl.wsu.edu/gems/yaImpute.pdf In yaImpute, i am using the yai function which uses randomForest as a method for finding out the k-Nearest Neighbours.. http://cran.r-project.org/doc/packages/yaImpute.pdf With the help iof the example given I was able to use the other methods available. from the document, and the MoscowMtStJoe exampe, is similar to the work i am trying to do. But the y variable needs to be entered in the form of a factor for random forest. what can be done here?! On 4/11/07, Liaw, Andy [EMAIL PROTECTED] wrote: Please provide the information the posting guide asks (version of R, packages used, version of package used, etc). There are no yaImpute() or yai() functions in the randomForest package. Andy From: [EMAIL PROTECTED] on behalf of Ricky Jacob Sent: Wed 4/11/2007 5:55 AM To: r-help@stat.math.ethz.ch Subject: [R] Random Forest Imputations [Broadcast] Dear All, I am not able to run the random forest with my dataset.. X- 280 records with satellite data(28 columns) - B1min, b1max, b1std etc.. y- 280 records with 3 columns - TotBasal Area, Stem density and Volume yref - y[1:230,] #Keeping 1st 230 records as reference records want to set 0 to y values for records 231 to 280.. yimp - y[231:280,] #records for which we want to impute the basal area, stem density and volume mal1 - yai(x=x, y=yref, method=mahalanobis, k=1, noRefs = TRUE) # This works fine for mahalanobis, msn, gnn, raw and Euclidean Want to do a similar thing with random forest where the 1st 230 records alone should be used for calculating Nearest Neighbours for the records with number 231 to 280.. What needs to be done.. Went through the yaImpute document.. but all i could do without any error message was to have NN generated using the yai() where all 280 records have been used for finding nearest neighbour. Regards Ricky [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu - direct contact information for affiliates is available at http://www.merck.com/contact/contacts.html) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete
Re: [R] Reasons to Use R [Broadcast]
From: Douglas Bates On 4/10/07, Wensui Liu [EMAIL PROTECTED] wrote: Greg, As far as I understand, SAS is more efficient handling large data probably than S+/R. Do you have any idea why? SAS originated at a time when large data sets were stored on magnetic tape and the only reasonable way to process them was sequentially. Thus most statistics procedures in SAS act as filters, processing one record at a time and accumulating summary information. In the past SAS performed a least squares fit by accumulating the crossproduct of [X:y] and then using the using the sweep operator to reduce that matrix. For such an approach the number of observations does not affect the amount of storage required. Adding observations just requires more time. This works fine (although there are numerical disadvantages to this approach - try mentioning the sweep operator to an expert in numerical linear algebra - you get a blank stare) For those who stared blankly at the above: The sweep operator is just a facier version of the good old Gaussian elimination... Andy as long as the operations that you wish to perform fit into this model. Making the desired operations fit into the model is the primary reason for the awkwardness in many SAS analyses. The emphasis in R is on flexibility and the use of good numerical techniques - not on processing large data sets sequentially. The algorithms used in R for most least squares fits generate and analyze the complete model matrix instead of summary quantities. (The algorithms in the biglm package are a compromise that work on horizontal sections of the model matrix.) If your only criterion for comparison is the ability to work with very large data sets performing operations that can fit into the filter model used by SAS then SAS will be a better choice. However you do lock yourself into a certain set of operations and you are doing it to save memory, which is a commodity that decreases in price very rapidly. As mentioned in other replies, for many years the majority of SAS uses are for data manipulation rather than for statistical analysis so the filter model has been modified in later versions. On 4/10/07, Greg Snow [EMAIL PROTECTED] wrote: -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Bi-Info (http://members.home.nl/bi-info) Sent: Monday, April 09, 2007 4:23 PM To: Gabor Grothendieck Cc: Lorenzo Isella; r-help@stat.math.ethz.ch Subject: Re: [R] Reasons to Use R [snip] So what's the big deal about S using files instead of memory like R. I don't get the point. Isn't there enough swap space for S? (Who cares anyway: it works, isn't it?) Or are there any problems with S and large datasets? I don't get it. You use them, Greg. So you might discuss that issue. Wilfred This is my understanding of the issue (not anything official). If you use up all the memory while in R, then the OS will start swapping memory to disk, but the OS does not know what parts of memory correspond to which objects, so it is entirely possible that the chunk swapped to disk contains parts of different data objects, so when you need one of those objects again, everything needs to be swapped back in. This is very inefficient. S-PLUS occasionally runs into the same problem, but since it does some of its own swapping to disk it can be more efficient by swapping single data objects (data frames, etc.). Also, since S-PLUS is already saving everything to disk, it does not actually need to do a full swap, it can just look and see that a particular data frame has not been used for a while, know that it is already saved on the disk, and unload it from memory without having to write it to disk first. The g.data package for R has some of this functionality of keeping data on the disk until needed. The better approach for large data sets is to only have some of the data in memory at a time and to automatically read just the parts that you need. So for big datasets it is recommended to have the actual data stored in a database and use one of the database connection packages to only read in the subset that you need. The SQLiteDF package for R is working on automating this process for R. There are also the bigdata module for S-PLUS and the biglm package for R have ways of doing some of the common analyses using chunks of data at a time. This idea is not new. There was a program in the late 1970s and 80s called Rummage by Del Scott (I guess technically it still exists, I have a copy on a 5.25 floppy somewhere) that used the approach of specify the model you wanted to fit first, then specify the data file. Rummage would then figure out which
Re: [R] Reasons to Use R [Broadcast]
I've probably been away from SAS for too long... we've recently tried to
get SAS on our 64-bit Linux boxes (because SAS on PC is not sufficient
for some of my colleagues who need it). I was shocked by the quote for
our 28-core Scyld cluster--- the annual fee was a few times the total
cost of our hardware. We ended up buying a new quad 3GHz Opterons box
with 32GB ram just so that the fee for SAS on such a box would be more
tolerable. It just boggles my mind that the right to use SAS for a year
is about the price of a nice four-bedroom house (near SAS Institute!).
I don't understand people who rather pay that kind of price for the
software, instead of spending the money on state-of-the-art hardware and
save more than a bundle.
Just my $0.02...
Andy
From: Jorge Cornejo-Donoso
I have a Dell with 2 Intel XEON 3.0 procesors and 2GB of ram
The problem is the DB size.
-Mensaje original-
De: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Enviado el: Lunes, 09 de Abril de 2007 11:28
Para: Jorge Cornejo-Donoso
CC: r-help@stat.math.ethz.ch
Asunto: Re: [R] Reasons to Use R
Have you tried 64 bit machines with larger memory or do you
mean that you can't use R on your current machines?
Also have you tried S-Plus? Will that work for you? The
transition from that to R would be less than from SAS to R.
On 4/9/07, Jorge Cornejo-Donoso [EMAIL PROTECTED] wrote:
tha s9ze of db is an issue with R. We are still using SAS because R
can't handle own db, and of couse we don't want to sacrify
resolution,
because the data collection is expensive (at least in fisheries and
oceagraphy), so.. I think that R need to improve the use of big DBs.
Now I only can use R for graph preparation and some data
analisis, but
we can't do the main work on R, abd that is really sad.
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Re: [R] subset [Broadcast]
From: Thomas Lumley
On Mon, 26 Mar 2007, Marc Schwartz wrote:
Sergio,
Please be sure to cc: the list (ie. Reply to All) with follow up
questions.
In this case, you would use %in% with a negation:
NewDF - subset(DF, (var1 == 0) (var2 == 0) (!var3 %in% 2:3))
Probably a typo: should be !(var3 %in% 2:3) rather than (!var
%in% 2:3)
I used to think so, but found I didn't need the parens:
R a - 1:3; b - c(1, 3, 5)
R ! a %in% b
[1] FALSE TRUE FALSE
R ! (a %in% b)
[1] FALSE TRUE FALSE
Andy
-thomas
See ?%in% for more information.
HTH,
Marc
On Mon, 2007-03-26 at 17:30 +0200, Sergio Della Franca wrote:
Ok, this run correctly.
Another question for you:
I want to put more than one condition for var3, i.e.:
I like to create a subset when:
- var1=0
- var2=0
- var3 is different from 2 and from 3.
Like you suggested, i perform this code:
NewDF - subset(DF, (var1 == 0) (var2 == 0) (var 3 !=
2)) (var
3 != 3))
There is a method to combine (var 3 != 2)) (var 3 != 3)) in one
condition?
Thank you.
Sergio
2007/3/26, Marc Schwartz [EMAIL PROTECTED]:
On Mon, 2007-03-26 at 17:02 +0200, Sergio Della
Franca wrote:
Dear R-Helpers,
I want to make a subset from my data set.
I'd like to perform different condition for subset.
I.e.:
I like to create a subset when:
- var1=0
- var2=0
- var3 is different from 2.
How can i develop a subset under this condition?
Thank you in advance.
Sergio Della Franca.
See ?subset
Something along the lines of the following should work:
NewDF - subset(DF, (var1 == 0) (var2 == 0)
(var 3 != 0))
HTH,
Marc Schwartz
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Thomas Lumley Assoc. Professor, Biostatistics
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Re: [R] frequency tables and sorting by rowSum
1. This is probably overkill, but works:
as.data.frame(table(as.data.frame(m)))
V1 V2 V3 Freq
1 0 0 00
2 1 0 02
3 0 1 03
4 1 1 00
5 0 0 11
6 1 0 10
7 0 1 10
8 1 1 10
You can easily get rid of 0-frequency rows afterward.
2. Not sure what you want, but guessing something like:
m.sorted - m[order(rowSums(m)), order(colSums(m))]
Andy
From: [EMAIL PROTECTED] on behalf of Stefan Nachtnebel
Sent: Sat 3/24/2007 8:41 AM
To: r-help@stat.math.ethz.ch
Subject: [R] frequency tables and sorting by rowSum [Broadcast]
Dear list,
I have some trouble generating a frequency table over a number of vectors.
Creating these tables over simple numbers is no problem with table()
table(c(1,1,1,3,4,5))
1 3 4 5
3 1 1 1
, but how can i for example turn:
0 1 0
0 0 1
0 1 0
1 0 0
0 1 0
1 0 0
into
0 0 1 1
1 0 0 2
0 1 0 3
My second problem is, sorting rows and columns of a matrix by the
rowSums/colSums.
I did it this way, but i think there should be a more efficient way:
sortRowCol-function(taus) {
swaprow - function(rsum) {
taus[(rowSums(taus)==rsum),]
}
for( i in 1:2 )
taus-sapply(sort(rowSums(taus)),swaprow)
}
thanks in advantage, Stefan Nachtnebel
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Re: [R] Get home directory and simple I/O
From: Gabor Grothendieck
See:
?R.home
That's not what Alberto wanted: It gives the location of the R
installation, not where user's home directory is. AFAIK Windows does
not set the HOME environment variable by default.
?dput
On 3/23/07, Alberto Monteiro [EMAIL PROTECTED] wrote:
Is there any generic function that gets the home directory? This
should return /home/user in Linux and x:/Documents and
Settings/user (or whatever) in Windows XP.
Another (unrelated) question: what is the _simplest_ way to
read and
write R variables to/from files such that they are stored in a
human-readable but R-like form? For example, if (say), x is
a vector
defined as x - c(1, 2, 3), can I write (and read) x as a file with
just one line, namely: c(1, 2, 3) ?
Alberto Monteiro
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Re: [R] objects of class matrix and mode list? [Broadcast]
It may help to (re-)read ?sapply a bit more in detail. Simplification
is done only if it's possible, and what possible means is defined
there.
A list is a vector whose elements can be different objects, but a vector
nonetheless. Thus a list can have dimensions. E.g.,
R a - list(1, 1:2, 3, c(abc, def))
R dim(a) - c(2, 2)
R a
[,1] [,2]
[1,] 1 3
[2,] Integer,2 Character,2
That sometimes can be extremely useful (not like the example above!).
Andy
From: Stephen Tucker
Hello everyone,
I cannot seem to find information about objects of class
matrix and mode
list, and how to handle them (apart from flattening the
list). I get this
type of object from using sapply(). Sorry for the long
example, but the code
below illustrates how I get this type of object. Is anyone aware of
documentation regarding this object?
Thanks very much,
Stephen
= begin example
# I am just making up a fake data set
df - data.frame(Day=rep(1:3,each=24),Hour=rep(1:24,times=3),
Name1=rnorm(24*3),Name2=rnorm(24*3))
# define a function to get a set of descriptive statistics
tmp - function(x) {
# this function will accept a data frame
# and return a 1-row data frame of
# max value, colname of max, min value, and colname of min
return(data.frame(maxval=max(apply(x,2,max)),
maxloc=names(x)[which.max(apply(x,2,max))],
minval=min(apply(x,2,min)),
minloc=names(x)[which.min(apply(x,2,min))]))
}
# Now applying function to data:
# (1) split the data table by Day with split()
# (2) apply the tmp function defined above to each data frame from (1)
# using lapply()
# (3) transpose the final matrix and convert it to a data frame
# with mixed characters and numbers
# using as.data.frame(), lapply(), and type.convert()
final -
as.data.frame(lapply(as.data.frame(t(sapply(split(df[,-c(1:2)],
+
f=df$Day),tmp))),
+ type.convert,as.is=TRUE))
Error in type.convert(x, na.strings, as.is, dec) :
the first argument must be of mode character
I thought sapply() would give me a data frame or matrix, which I would
transpose into a character matrix, to which I can apply type.convert()
and get the same matrix as what I would get from these two lines (Fold
function taken from Gabor's post on R-help a few years ago):
Fold - function(f, x, L) for(e in L) x - f(x, e)
final2 - Fold(rbind,vector(),lapply(split(df[,-c(1:2)],f=day),tmp))
print(c(class(final2),mode(final2)))
[1] data.frame list
However, by my original method, sapply() gives me a matrix
with mode, list
intermediate1 - sapply(split(df[,-c(1:2)],f=df$Day),tmp)
print(c(class(intermediate1),mode(intermediate1)))
[1] matrix list
Transposing, still a matrix with mode list, not character:
intermediate2 - t(sapply(split(df[,-c(1:2)],f=day),tmp))
print(c(class(intermediate2),mode(intermediate2)))
[1] matrix list
Unclassing gives me the same thing...
print(c(class(unclass(intermediate2)),mode(unclass(intermediate2
[1] matrix list
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Re: [R] how to get lsmeans?
numbers of observations in the different levels of the factors that are held constant). The obstacle to computing either least-squares means or effect displays in R via predict() is that predict() wants factors in the new data to be set to particular levels. The effect() function in the effects package bypasses predict() and works directly with the model matrix, averaging over the columns that pertain to a factor (and reconstructing interactions as necessary). As mentioned, this has the effect of setting the factor to its proportional distribution in the data. This approach also has the advantage of being invariant with respect to the choice of contrasts for a factor. The only convenient way that I can think of to implement least-squares means in R would be to use deviation-coded regressors for a factor (that is, contr.sum) and then to set the columns of the model matrix for the factor(s) to be averaged over to 0. It may just be that I'm having a failure of imagination and that there's a better way to proceed. I've not implemented this solution because it is dependent upon the choice of contrasts and because I don't see a general advantage to it, but since the issue has come up several times now, maybe I should take a crack at it. Remember that I want this to work more generally, not just for levels of factors, and not just for linear models. Brian is quite right in mentioning that he suggested some time ago that I use critical values of t rather than of the standard normal distribution for producing confidence intervals, and I agree that it makes sense to do so in models in which the dispersion is estimated. My only excuse for not yet doing this is that I want to undertake a more general revision of the effects package, and haven't had time to do it. There are several changes that I'd like to make to the package. For example, I have results for multinomial and proportional odds logit models (described in a paper by me and Bob Andersen in the 2006 issue of Sociological Methodology) that I want to incorporate, and I'd like to improve the appearance of the default graphs. But Brian's suggestion is very straightforward, and I guess that I shouldn't wait to implement it; I'll do so very soon. Regards, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Prof Brian Ripley Sent: Wednesday, March 21, 2007 12:03 PM To: Chuck Cleland Cc: r-help Subject: Re: [R] how to get lsmeans? On Wed, 21 Mar 2007, Chuck Cleland wrote: Liaw, Andy wrote: I verified the result from the following with output from JMP 6 on the same data (don't have SAS: don't need it): set.seed(631) n - 100 dat - data.frame(y=rnorm(n), A=factor(sample(1:2, n, replace=TRUE)), B=factor(sample(1:2, n, replace=TRUE)), C=factor(sample(1:2, n, replace=TRUE)), d=rnorm(n)) fm - lm(y ~ A + B + C + d, dat) ## Form a data frame of points to predict: all combinations of the ## three factors and the mean of the covariate. p - data.frame(expand.grid(A=1:2, B=1:2, C=1:2)) p[] - lapply(p, factor) p - cbind(p, d=mean(dat$d)) p - cbind(yhat=predict(fm, p), p) ## lsmeans for the three factors: with(p, tapply(yhat, A, mean)) with(p, tapply(yhat, B, mean)) with(p, tapply(yhat, C, mean)) Using Andy's example data, these are the LSMEANS and intervals I get from SAS: Ay LSMEAN 95% Confidence Limits 1 -0.071847 -0.387507 0.243813 2 -0.029621 -0.342358 0.283117 By LSMEAN 95% Confidence Limits 1 -0.104859 -0.397935 0.188216 20.003391 -0.333476 0.340258 Cy LSMEAN 95% Confidence Limits 1 -0.084679 -0.392343 0.222986 2 -0.016789 -0.336374 0.302795 One way of reproducing the LSMEANS and intervals from SAS using predict() seems to be the following: dat.lm - lm(y ~ A + as.numeric(B) + as.numeric(C) + d, data = dat) newdat - expand.grid(A=factor(c(1,2)),B=1.5,C=1.5,d=mean(dat$d)) cbind(newdat, predict(dat.lm, newdat, interval=confidence)) A B C d fitlwr upr 1 1 1.5 1.5 0.09838595 -0.07184709 -0.3875070 0.2438128 2 2 1.5 1.5 0.09838595 -0.02962086 -0.3423582 0.2831165
Re: [R] how to get lsmeans?
I verified the result from the following with output from JMP 6 on the
same data (don't have SAS: don't need it):
set.seed(631)
n - 100
dat - data.frame(y=rnorm(n), A=factor(sample(1:2, n, replace=TRUE)),
B=factor(sample(1:2, n, replace=TRUE)),
C=factor(sample(1:2, n, replace=TRUE)),
d=rnorm(n))
fm - lm(y ~ A + B + C + d, dat)
## Form a data frame of points to predict: all combinations of the
## three factors and the mean of the covariate.
p - data.frame(expand.grid(A=1:2, B=1:2, C=1:2))
p[] - lapply(p, factor)
p - cbind(p, d=mean(dat$d))
p - cbind(yhat=predict(fm, p), p)
## lsmeans for the three factors:
with(p, tapply(yhat, A, mean))
with(p, tapply(yhat, B, mean))
with(p, tapply(yhat, C, mean))
Andy
From: Xingwang Ye
Dear all,
I search the mail list about this topic and learn that no
simple way is available to get lsmeans in R as in SAS.
Dr.John Fox and Dr.Frank E Harrell have given very useful
information about lsmeans topic.
Dr. Frank E Harrell suggests not to think about lsmeans,
just to think about what predicted values wanted
and to use the predict function. However, after reading
the R help file for a whole day, I am still unclear how to do it.
Could some one give me a hand?
for example:
A,B and C are binomial variables(factors); d is a continuous
variable ; The response variable Y is a continuous variable too.
To get lsmeans of Y according to A,B and C, respectively, in
SAS, I tried proc glm data=a; class A B C; model Y=A B C d;
lsmeans A B C/cl; run;
In R, I tried this:
library(Design)
ddist-datadist(a)
options(datadist=ddist)
f-ols(Y~A+B+C+D,data=a,x=TRUE,y=TRUE,se.fit=TRUE)
then how to get the lsmeans for A, B, and C, respectively
with predict function?
Best wishes
yours, sincerely
Xingwang Ye
PhD candidate
Research Group of Nutrition Related Cancers and Other Chronic
Diseases
Institute for Nutritional Sciences,
Shanghai Institutes of Biological Sciences,
Chinese Academy of Sciences
P.O.Box 32
294 Taiyuan Road
Shanghai 200031
P.R.CHINA
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Re: [R] package:AlgDesign and .Random.seed [Broadcast]
From: Michael Kubovy
On Mar 21, 2007, at 4:16 AM, Uwe Ligges wrote:
Michael Kubovy wrote:
Dear r-helpers,
Could you please help me solve the following problem: When I run
require(AlgDesign)
trt - LETTERS[1:5]
blk - 10
trtblk - 3
BIB - optBlock(~., withinData = trt, blocksizes =
rep(trtblk, blk))
In response to the last command, R complains:
Error in optBlock(~., withinData = trt, blocksizes = rep(trtblk,
blk)) :
object .Random.seed not found
The documentation of optBlock() in AlgDesign doesn't say that I
needed to set .Random.seed. I thought it was initiated
automatically
at the beginning of a session. What am I missing?
The first line in that function is
seed - .Random.seed
but .Random.seed is generated at the first use of R's RNG, hence
maybe later. This means the function contains a bug which you
should report to the package maintainer, please.
Best,
Uwe Ligges
Bob Wheeler's response:
From: Bob Wheeler [EMAIL PROTECTED]
Date: March 21, 2007 9:19:29 AM EDT
To: Michael Kubovy [EMAIL PROTECTED]
Subject: Re:
Each workspace in R requires you to set a random seed to
start. You
have not done this. It is an R artifact, and has nothing to
do with
AlgDesign.
I do not agree with that assessment (well, it's just my $0.02 anyway).
I don't need a random seed unless I'm doing computations that requires
pseudo-random numbers. There are plenty of times I use R without
needing random seed.
None of the builtin RNGs in R requires explcit seed setting, nor does
any of the ones in the contributed packages that I know of. Thus I
would claim that's a flaw in AlgDesign.
Andy
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ECHIP, Inc. --- Randomness comes in bunches.
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Re: [R] Select the last two rows by id group
Something like the following should work:
last.n - function(x, n) {
last - nrow(x)
x[(last - n + 1):last, , drop=FALSE]
}
## Example: get the last two rows.
do.call(rbind, lapply(split(score, score$id), last.n, 2))
You might want to add a check in last.n() to make sure that there are at
least n rows to extract.
Andy
From: Lauri Nikkinen
Hi R-users,
Following this post
http://tolstoy.newcastle.edu.au/R/help/06/06/28965.html , how
do I get last two rows (or six or ten) by id group out of the
data frame? Here the example gives just the last row.
Sincere thanks,
Lauri
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Re: [R] Bad points in regression [Broadcast]
(My turn on the soapbox ...) I'd like to add a bit of caveat to Bert's view. I'd argue (perhaps even plead) that robust/resistant procedures be used with care. They should not be used as a shortcut to avoid careful analysis of data. I recalled that in my first course on regression, the professor made it clear that we're fitting models to data, not the other way around. When the model fits badly to (some of the) the data, do examine and think carefully about what happened. Verify that bad data are indeed bad, instead of using statistical criteria to make that judgment. A scientific colleague reminded me of this point when I tried to sell him the idea of robust/resistant methods: Don't use these methods as a crutch to stand on badly run experiments (or poorly fitted models). Cheers, Andy From: Bert Gunter (mount soapbox...) While I know the prior discussion represents common practice, I would argue -- perhaps even plead -- that the modern(?? 30 years old now) alternative of robust/resistant estimation be used, especially in the readily available situation of least-squares regression. RSiteSearch(Robust) will bring up numerous possibilities.rrcov and robustbase are at least two packages devoted to this, but the functionality is available in many others (e.g. rlm() in MASS). Bert Gunter Genentech Nonclinical Statistics South San Francisco, CA 94404 650-467-7374 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ted Harding Sent: Friday, March 16, 2007 6:44 AM To: r-help@stat.math.ethz.ch Subject: Re: [R] Bad points in regression On 16-Mar-07 12:41:50, Alberto Monteiro wrote: Ted Harding wrote: alpha - 0.3 beta - 0.4 sigma - 0.5 err - rnorm(100) err[15] - 5; err[25] - -4; err[50] - 10 x - 1:100 y - alpha + beta * x + sigma * err ll - lm(y ~ x) plot(ll) ll is the output of a linear model fiited by lm(), and so has several components (see ?lm in the section Value), one of which is residuals (which can be abbreviated to res). So, in the case of your example, which(abs(ll$res)2) 15 25 50 extracts the information you want (and the 2 was inspired by looking at the residuals plot from your plot(ll)). Ok, but how can I grab those points _in general_? What is the criterium that plot used to mark those points as bad points? Ahh ... ! I see what you're after. OK, look at the plot method for lm(): ?plot.lm ## S3 method for class 'lm': plot(x, which = 1:4, caption = c(Residuals vs Fitted, Normal Q-Q plot, Scale-Location plot, Cook's distance plot), panel = points, sub.caption = deparse(x$call), main = , ask = prod(par(mfcol)) length(which) dev.interactive(), ..., id.n = 3, labels.id = names(residuals(x)), cex.id = 0.75) where (see further down): id.n: number of points to be labelled in each plot, starting with the most extreme. and note, in the default parameter-values listing above: id.n = 3 Hence, the 3 most extreme points (according to the criterion being plotted in each plot) are marked in each plot. So, for instance3, try plot(ll,id.n=5) and you will get points 10,15,25,28,50. And so on. But that pre-supposes that you know how many points are exceptional. What is meant by extremeis not stated in the help page ?plot.lm, but can be identified by inspecting the code for plot.lm(), which you can see by entering plot.lm In your example, if you omit the line which assigns anomalous values to err[15[, err[25] and err[50], then you are likely to observe that different points get identified on different plots. For instance, I just got the following results for the default id.n=3: [1] Residuals vs Fitted: 41,53,59 [2] Standardised Residuals:41,53,59 [3] sqrt(Stand Res) vs Fitted: 41,53,59 [4] Cook's Distance: 59,96,97 There are several approaches (with somewhat different outcomes) to identifying outliers. If you apply one of these, you will probably get the identities of the points anyway. Again in the context of your example (where in fact you deliberately set 3 points to have exceptional errors, thus coincidentally the same as the default value 3 of id.n), you could try different values for id.n and inspect the graphs to see whether a given value of id.n marks some points that do not look exceptional relative to the mass of the other points. So, the above plot(ll,id.n=5) gave me one point, 10 on the residuals plot, which apparently belonged to the general distribution of residuals. Hoping this helps, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 16-Mar-07 Time: 13:43:54 -- XFMail
Re: [R] how to assign fixed factor in lm
Either you did not read docs sufficiently carefully, or the source where
you learn to do this from is questionable. The lm() function has no
argument called fixed, and the warning should have made that clear to
you. It was sheer luck on your part that you happen to put Value as
the first variable in Food, in which case lm() will treat it as the
response and the rest as predictors in the absence of a model formula.
You should try:
lm(Value ~ Gender, Food)
lm() itself has no concept of fixed or random effects. lme() in the
nlme package does, and it has the fixed argument.
Andy
From: [EMAIL PROTECTED]
Hi there,
Value=c(709,679,699,657,594,677,592,538,476,508,505,539)
Lard=rep(c(Fresh,Rancid),each=6)
Gender=rep(c(Male,Male,Male,Female,Female,Female),2)
Food=data.frame(Value,Lard,Gender)
Food
Value Lard Gender
1709 Fresh Male
2679 Fresh Male
3699 Fresh Male
4657 Fresh Female
5594 Fresh Female
6677 Fresh Female
7592 Rancid Male
8538 Rancid Male
9476 Rancid Male
10 508 Rancid Female
11 505 Rancid Female
12 539 Rancid Female
lm(fixed=Value~Gender,data=Food)
Call:
lm(data = Food, fixed = Value ~ Gender)
Coefficients:
(Intercept) LardRancid GenderMale
651.4 -142.8 35.5
Warning message:
extra arguments fixed are just disregarded. in: lm.fit(x, y,
offset = offset, singular.ok = singular.ok, ...)
lm(fixed=Value~Lard+Gender,data=Food)
Call:
lm(data = Food, fixed = Value ~ Lard + Gender)
Coefficients:
(Intercept) LardRancid GenderMale
651.4 -142.8 35.5
Warning message:
extra arguments fixed are just disregarded. in: lm.fit(x, y,
offset = offset, singular.ok = singular.ok, ...)
I wanted to consider only one factor. But why
lm(fixed=Value~Gender,data=Food) return me two estimates of
Gender and Lard. And I found the returning results are the
same as lm(fixed=Value~Lard+Gender,data=Food). Why lm cannot
do analysis of variance according to assigned formula?
Thank you very much.
Fan
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Re: [R] 0 * NA = NA
From: Alberto Monteiro
Is there any way to force 0 * NA to be 0 instead of NA?
For example, suppose I have a vector with some valid values,
while other values are NA. If I matrix-pre-multiply this by a
weight row vector, whose weights that correspond to the NAs
are zero, the outcome will still be NA:
x - c(1, NA, 1)
wt - c(2, 0, 1)
wt %*% x # NA
I don't think it's prudent to bend arthmetic rules of a system,
especially when there are good reasons for them. Here's one:
R 0 * Inf
[1] NaN
If you are absolutely sure that the Nas in x cannot be Inf (or -Inf),
you might try to force the result to 0, but the only way I can think of
is to do something like:
R wt %*% ifelse(wt, x, 0)
[,1]
[1,]3
Andy
Alberto Monteiro
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Re: [R] how to apply the function cut( ) to many columns in a data.frame?
From: Chuck Cleland
ahimsa campos-arceiz wrote:
Dear useRs,
In a data.frame (df) I have several columns (x1, x2, x3xn)
containing data as a continuous numerical response:
df
var x1x2 x3
1143 147 137
2 9393 117
316439 101
4123 11897
5 63 125 97
612983 124
712393 136
812380 79
9 89 107 150
10 7895121
I want to classify the values in the columns x1, x2, etc,
into bins of
fix margins (0-5, 5-10, ). For one vector I can do it
easily with
the function cut:
df$x1 - cut(df$x1, br=5*(0:40), labels=5*(1:40))
df$x1
[1] 145 95 165 125 65 130 125 125 90 80 40 Levels: 5 10
15 20 25
30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 ...
200
However if I try to use a subset of my data.frame:
df[,3:4] - cut(df[,3:4], br=5*(0:40), labels=5*(1:40))
Error in cut.default(df[, 3:4], br = 5 * (0:40), labels = 5
* (1:40)) :
'x' must be numeric
How can I make this work with data frames in which I want
to apply the
function cut( ) to many columns in a data.frame?
You have an answer within your question - use one of the
various apply functions. For example:
lapply(df[,3:4], function(x){cut(x, br=5*(0:40), labels=5*(1:40))})
Or perhaps a bit more simply:
lapply(df[, 3:4], cut, br=5*(0:40), labels=5*(1:40)))
and if a data frame is desired as output, wrap the above in
as.data.frame().
(Just keep in mind that a data frame is like a list.)
Andy
?lapply
?sapply
?apply
I guess that I might have to use something like for ( )
(which I'm not
familiar with), but maybe you know a straight forward method to use
with data.frames.
Thanks a lot!
Ahimsa
*
# data
var - 1:10
x1 - rnorm(10, mean=100, sd=25)
x2 - rnorm(10, mean=100, sd=25)
x3 - rnorm(10, mean=100, sd=25)
df - data.frame(var,x1,x2,x3)
df
# classifying the values of the vector df$x1 into bins of width 5
df$x1 - cut(df$x1, br=5*(0:40), labels=5*(1:40))
df$x1
# trying it a subset of the data.frame df[,3:4] - cut(df[,3:4],
br=5*(0:40), labels=5*(1:40)) df[,3:4]
--
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NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894
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Re: [R] How to read in this data format?
You can't expect general-purpose tools like read.table in R to be able
to deal with highly specialized file format. Here's what I'd start. It
doesn't put data in the format you specified exactly, but I doubt you'll
need that. This might be sufficient for your purpose:
dat - readLines(file(yourdata.dat))
## Get rid of blank lines.
dat - dat[dat != ]
scan.lines - grep(Scan, dat)
## Chop off the header rows.
dat - dat[scan.lines[1]:length(dat)]
scan.lines - scan.lines - scan.lines[1] + 1
lines.per.scan - c(scan.lines[-1], length(dat) + 1) - scan.lines
## Split the data into a list, with each scan taking up one component.
dat - split(dat, rep(seq(along=lines.per.scan), each=lines.per.scan))
## Process the data one scan at a time.
result - lapply(dat, function(x) {
x - strsplit(x, \t)
rtime - x[[2]][2] # second field of second line
t(matrix(as.numeric(do.call(rbind, c(rtime, x[-(1:2)]))), ncol=2))
})
This is what I get from the data you've shown:
R result
$`1`
[,1] [,2] [,3] [,4]
[1,] 0.017 399.8112 399.8742 399.9372
[2,] 0.017 184. 0. 152.
$`2`
[,1] [,2] [,3] [,4]
[1,] 0.021 399.8112 399.8742 399.9372
[2,] 0.021 181. 1. 153.
Note that you probably should avoid using numbers as column names in a
data frame, even if it's possible.
Andy
From: Bart Joosen
Hi,
I recieved an ascii file, containing following information:
$$ Experiment Number:
$$ Associated Data:
FUNCTION 1
Scan 1
Retention Time0.017
399.8112 184
399.8742 0
399.9372 152
Scan 2
Retention Time0.021
399.8112 181
399.8742 1
399.9372 153
.
I would like to import this data in R into a dataframe, where
there is a column time, the first numbers as column names,
and the second numbers as data in the dataframe:
Time 399.8112399.8742399.9372
0.017 184 0 152
0.021 181 1 153
I did take a look at the read.table, read.delim, scan, ...
But I 've no idea about how to solve this problem.
Anyone?
Thanks
Bart
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Re: [R] Package RGtk2, rattle, libatk-1.0.0.dll Errors
The way I have that problem resolved is by installing the rggobi package
using the command shown on http://www.ggobi.org/rggobi/, which is
source(http://www.ggobi.org/download/install.r;)
That will install all the things that Ggobi needs. Since rattle depends
on rggobi, it's probably a good idea to do this anyway. After that,
rattle will start just fine. I have not actually use rattle, though.
Andy
From: j.joshua thomas
Dear Group,
I have followed the instructions from the link
http://datamining.togaware.com/survivor/Installing_GTK.html
However i couldn't fix the libatk01.0.0.dll application error
Here, i did uninstall R-Gui-2.4.0 then did the fresh
installation and still facing the same problem I am using Windows- XP
*Please find the following*
R version 2.4.0 (2006-10-03)
Copyright (C) 2006 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.
Natural language support but running in an English locale
R is a collaborative project with many contributors.
Type 'contributors()' for more information and 'citation()'
on how to cite R or R packages in publications.
Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.
[Previously saved workspace restored]
install.packages(RGtk2)
--- Please select a CRAN mirror for use in this session ---
trying URL '
http://cran.au.r-project.org/bin/windows/contrib/2.4/RGtk2_2.8.7.zip'
Content type 'application/zip' length 13050736 bytes opened
URL downloaded 12744Kb
package 'RGtk2' successfully unpacked and MD5 sums checked
The downloaded packages are in
C:\Documents and Settings\jjoshua\Local
Settings\Temp\RtmpLetwrb\downloaded_packages
updating HTML package descriptions
install.packages(rattle)
trying URL '
http://cran.au.r-project.org/bin/windows/contrib/2.4/rattle_2.2.0.zip'
Content type 'application/zip' length 340875 bytes opened URL
downloaded 332Kb
package 'rattle' successfully unpacked and MD5 sums checked
The downloaded packages are in
C:\Documents and Settings\jjoshua\Local
Settings\Temp\RtmpLetwrb\downloaded_packages
updating HTML package descriptions
library(RGtk2)
Error in dyn.load(x, as.logical(local), as.logical(now)) :
unable to load shared library
'C:/PROGRA~1/R/R-24~1.0/library/RGtk2/libs/RGtk2.dll':
LoadLibrary failure: The specified module could not be found.
In addition: Warning message:
package 'RGtk2' was built under R version 2.4.1
Error: package/namespace load failed for 'RGtk2'
--
Lecturer J. Joshua Thomas
KDU College Penang Campus
Research Student,
University Sains Malaysia
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Re: [R] Multiple conditional without if [Broadcast]
From: bunny, lautloscrew.com
Dear all,
i am stuck with a syntax problem.
i have a matrix which has about 500 rows and 6 columns.
now i want to kick some data out.
i want create a new matrix which is basically the old one
except for all entries which have a 4 in the 5 column AND a 1
in the 6th column.
i tried the following but couldn´t get a new matrix, just some wierd
errors:
newmatrix=oldmatrix[,2][oldmatrix[,5]==4]oldmatrix[,2][oldmatrix[,6]
==1]
all i get is:
numeric(0)
That's not a `weird error', but a numeric vector of length 0.
does anybody have an idea how to fix this one ?
Try:
newmatrix = oldmatrix[oldmatrix[, 5]==4 oldmatrix[, 6] == 1, 2, drop=FALSE]
If you just want a subset of column 2 as a vector, you can leave off the
drop=FALSE part.
Reading An Introduction to R should have save you some trouble in the first
place.
Andy
thx in advance
matthias
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Re: [R] how to use apply with two variables
Yes. Just try it and see.
BTW, your usage of return() is not recommended anymore. This is
probably easier:
myfun-function(x) c(mean=mean(x), sd=sd(x))
out - apply(mat, 1, myfun)
## or...
out2 - cbind(mean=rowMeans(mat), sd=sd(t(mat)))
Andy
From: Serguei Kaniovski
Hi,
this is a made-up example. Function myfun returns two
arguments. Can apply be used so that myfun is called only once?
Thanks
Serguei
mat-matrix(runif(50),nrow=10,ncol=5)
myfun-function(x) {
mymean-mean(x)
mysd-sd(x)
return(mymean,mysd)
}
out1-t(apply(mat,1,function(x) myfun(x)$mymean))
out2-t(apply(mat,1,function(x) myfun(x)$mysd))
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Re: [R] JGR launcher for linux
Isn't it right in front of you? I get:
JGR()
Starting JGR ...
(You can use /usr/local/lib64/R/library/JGR/cont/run to start JGR directly)
^^^
Andy
From: Ronaldo Reis Junior
Hi,
anybody have a JGR launcher for linux? Maybe a script that
launch JGR directly without open R then library(JGR) and JGR().
Thanks
Ronaldo
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--
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| .''`. UNIMONTES/Depto. Biologia Geral/Lab. Ecologia Evolutiva
| : :' : Campus Universitário Prof. Darcy Ribeiro, Vila Mauricéia `.
| `'` CP: 126, CEP: 39401-089, Montes Claros - MG - Brasil
| `- Fone: (38) 3229-8190 | [EMAIL PROTECTED] |
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| ICQ#: 5692561 | LinuxUser#: 205366
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Re: [R] memory management uestion [Broadcast]
I don't see why making copies of the columns you need inside the loop is
better memory management. If the data are in a matrix, accessing
elements is quite fast. If you're worrying about speed of that, do what
Charles suggest: work with the transpose so that you are accessing
elements in the same column in each iteration of the loop.
Andy
From: Federico Calboli
Charles C. Berry wrote:
Whoa! You are accessing one ROW at a time.
Either way this will tangle up your cache if you have many rows and
columns in your orignal data.
You might do better to do
Y - t( X ) ### use '-' !
for (i in whatever ){
do something using Y[ , i ]
}
My question is NOT how to write the fastest code, it is
whether dummy variables (for lack of better words) make the
memory management better, i.e. faster, or not.
Best,
Fede
--
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Department of Epidemiology and Public Health Imperial
College, St Mary's Campus Norfolk Place, London W2 1PG
Tel +44 (0)20 7594 1602 Fax (+44) 020 7594 3193
f.calboli [.a.t] imperial.ac.uk
f.calboli [.a.t] gmail.com
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Re: [R] rpart tree node label [Broadcast]
Try the following to see:
library(rpart)
iris.rp(Sepal.Length ~ Species, iris)
plot(iris.rp)
text(iris.rp)
Two possible solutions:
1. Use text(..., pretty=0). See ?text.rpart.
2. Use post(..., filename=).
Andy
From: Wensui Liu
not sure how you want to label it.
could you be more specific?
thanks.
On 2/14/07, Aimin Yan [EMAIL PROTECTED] wrote:
I generate a tree use rpart.
In the node of tree, split is based on the some factor.
I want to label these node based on the levels of this factor.
Does anyone know how to do this?
Thanks,
Aimin
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Re: [R] [BioC] Outlook does threading [Broadcast]
This is really off-topic for both BioC and R-help, so I'll keep it short. From: Kimpel, Mark William See below for Bert Gunter's off list reply to me (which I do appreciate). I'm putting it back on the list because it seems there is still confusion regarding the difference between threading and sorting by subject. I thought the example I will give below will serve as instructional for other Outlook users who may be similarly confused as I was (am?). Per Bert's instructions, I just set up my inbox to sort by subject. I sent one email to myself with the subject test1 and then replied to it without changing the subject. The reply correctly went to test1 in the inbox sorter. I then changed the subject heading in the test1 reply to test2 and sent it to myself. This time Outlook re-categorized it and put it in a separate compartment in the view called test2. If Outlook can do threading the way the R mail server does, I don't think this is the way to do it. AFAIK there's no proper way to get the correct threading in Outlook. What I do is group by conversation topic, but that doesn't solve the problem. This is only problem on your (and all Outlook users'?) end, though. The bigger problem that affects the lists is that some versions of MS Exchange Server do not include the In-reply-to header field that many mailing lists rely on for proper threading. As a result, when I reply to other people's post, it may show up in Outlook as having been threaded properly (because the subject is fine), but it throws everything else that does proper threading off. Unless someone has an idea of how to correctly set up Outlook to do threading in the manner that the R mail server does, Maybe some VBA coding can be done to get it right, but short of that, I very much doubt it. I think the message for us Outlook users is to just create, from scratch, a new message when initiating a new subject. That message ought to be clear for everyone. You should never reply to a message when you really mean to start a new topic, regardless what you are using. Andy Thanks for all your help. Mark -Original Message- From: Bert Gunter [mailto:[EMAIL PROTECTED] Sent: Wednesday, January 31, 2007 7:03 PM To: Kimpel, Mark William Subject: Outlook does threading Mark: No need to bother the R list with this. Outlook does threading. Just sort on Subject in the viewer. Bert Gunter Genentech Nonclinical Statistics South San Francisco, CA 94404 650-467-7374 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Kimpel, Mark William Sent: Wednesday, January 31, 2007 3:36 PM To: Peter Dalgaard Cc: r-help@stat.math.ethz.ch; [EMAIL PROTECTED] Subject: Re: [R] possible spam alert Peter, Thanks you for your explanation, I had taken Mr. Connolly's message to me to imply that I was not changing the subject line. I use MS Outlook 2007 and, unless I am just not seeing it, Outlook does not normally display the in reply to header, I was under the mistaken impression that that was what the Subject line was for. See, for example, the header to your message to me below. Outlook will, however, sort messages by Subject, and that is what I thought was meant by threading. Well, I learned something today and apologize for any inconvenience my posts may have caused. BTW, I use Outlook because it is supported by my university server and will synch my appointments and contacts with my PDA, which runs Windows CE. If anyone has a suggestion for me of a better email program that will provide proper threading AND work with a MS email server and synch with Windows CE, I'd love to hear it. Thanks again, Mark Mark W. Kimpel MD (317) 490-5129 Work, Mobile (317) 663-0513 Home (no voice mail please) 1-(317)-536-2730 FAX -Original Message- From: Peter Dalgaard [mailto:[EMAIL PROTECTED] Sent: Wednesday, January 31, 2007 6:25 PM To: Kimpel, Mark William Cc: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch Subject: Re: [R] possible spam alert Kimpel, Mark William wrote: The last two times I have originated message threads on R or Bioconductor I have received the message included below from someone named Patrick Connolly. Both times I was the originator of the message thread and used what I thought was a unique subject line that explained as best I could what my question was. Patrick seems to be implying that I am abusing the R and BioC help newsgroups in this fashion. When I emailed him to give me a specific example, he did not reply. The most recent thread that he seems concerned about was to the R list and was entitled regexpr and parsing question . I believe the previous post of mine that he had problems with was to the BioC list but I can't remember its subject. Is this spam? No. Breach of netiquette, yes. The
Re: [R] Memory problem on a linux cluster using a large data set [Broadcast]
In addition to my off-list reply to Iris (pointing her to an old post of mine that detailed the memory requirement of RF in R), she might consider the following: - Use larger nodesize - Use sampsize to control the size of bootstrap samples Both of these have the effect of reducing sizes of trees grown. For a data set that large, it may not matter to grow smaller trees. Still, with data of that size, I'd say 64-bit is the better solution. Cheers, Andy From: Martin Morgan Iris -- I hope the following helps; I think you have too much data for a 32-bit machine. Martin Iris Kolder [EMAIL PROTECTED] writes: Hello, I have a large data set 320.000 rows and 1000 columns. All the data has the values 0,1,2. It seems like a single copy of this data set will be at least a couple of gigabytes; I think you'll have access to only 4 GB on a 32-bit machine (see section 8 of the R Installation and Administration guide), and R will probably end up, even in the best of situations, making at least a couple of copies of your data. Probably you'll need a 64-bit machine, or figure out algorithms that work on chunks of data. on a linux cluster with R version R 2.1.0. which operates on a 32 This is quite old, and in general it seems like R has become more sensitive to big-data issues and tracking down unnecessary memory copying. cannot allocate vector size 1240 kb. I've searched through use traceback() or options(error=recover) to figure out where this is actually occurring. SNP - read.table(file.txt, header=FALSE, sep=)# read in data file This makes a data.frame, and data frames have several aspects (e.g., automatic creation of row names on sub-setting) that can be problematic in terms of memory use. Probably better to use a matrix, for which: 'read.table' is not the right tool for reading large matrices, especially those with many columns: it is designed to read _data frames_ which may have columns of very different classes. Use 'scan' instead. (from the help page for read.table). I'm not sure of the details of the algorithms you'll invoke, but it might be a false economy to try to get scan to read in 'small' versions (e.g., integer, rather than numeric) of the data -- the algorithms might insist on numeric data, and then make a copy during coercion from your small version to numeric. SNP$total.NAs = rowSums(is.na(SN # calculate the number of NA per row and adds a colum with total Na's This adds a column to the data.frame or matrix, probably causing at least one copy of the entire data. Create a separate vector instead, even though this unties the coordination between columns that a data frame provides. SNP = t(as.matrix(SNP)) # transpose rows and columns This will also probably trigger a copy; snp.na-SNP R might be clever enough to figure out that this simple assignment does not trigger a copy. But it probably means that any subsequent modification of snp.na or SNP *will* trigger a copy, so avoid the assignment if possible. snp.roughfix-na.roughfix(snp.na) fSNP-factor(snp.roughfix[, 1])# Asigns factor to case control status snp.narf- randomForest(snp.roughfix[,-1], fSNP, na.action=na.roughfix, ntree=500, mtry=10, importance=TRUE, keep.forest=FALSE, do.trace=100) Now you're entirely in the hands of the randomForest. If memory problems occur here, perhaps you'll have gained enough experience to point the package maintainer to the problem and suggest a possible solution. set it should be able to cope with that amount. Perhaps someone has tried this before in R or is Fortram a better choice? I added my R If you mean a pure Fortran solution, including coding the random forest algorithm, then of course you have complete control over memory management. You'd still likely be limited to addressing 4 GB of memory. I wrote a script to remove all the rows with more than 46 missing values. This works perfect on a smaller dataset. But the problem arises when I try to run it on the larger data set I get an error cannot allocate vector size 1240 kb. I've searched through previous posts and found out that it might be because i'm running it on a linux cluster with R version R 2.1.0. which operates on a 32 bit processor. But I could not find a solution for this problem. The cluster is a really fast one and should be able to cope with these large amounts of data the systems configuration are Speed: 3.4 GHz, memory 4GByte. Is there a way to change the settings or processor under R? I want to run the function Random Forest on my large data set it should be able to cope with that amount. Perhaps someone has tried this before in R or is Fortram a better choice? I added my R script down
Re: [R] plot.svm
Try
debug(e1071:::plot.svm)
and then re-run your plot command, stepping through one line at a time
and see where it fails.
Andy
From: Aimin Yan
where is plot.svm method?
I just find plot(svm, data, formula) method
Aimin
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Re: [R] any way to make the code more efficient ?
I don't know about efficiency, but at least for readability, you may
want to do the following:
1. Indent your code.
2. Create a list of appropriate length, and populate the list with
objects you're creating in the loop.
3. After the loop, use do.call(rbind, list).
HTH,
Andy
From: Leeds, Mark (IED)
ravi : I appreciate your help but could you be a little more
specific about what you mean ? I can just stack aggfxdata
below the current full one ( the rbind works out the
ordrering by date because it's a zoo object ) so it's not a
question of where to put the new one. It's a question of how
to avoid rbind ? I apologize because I don't think I
understand what you are saying. Or maybe it's not possible to
avoid rbind ? Thanks.
-Original Message-
From: Ravi Varadhan [mailto:[EMAIL PROTECTED]
Sent: Friday, December 08, 2006 5:21 PM
To: Leeds, Mark (IED); r-help@stat.math.ethz.ch
Subject: RE: [R] any way to make the code more efficient ?
Using rbind almost always slows things down significantly.
You should
define the objects aggfxdata and fullaggfxdata before the loop and
then assign appropriate values to the corresponding rows
and/or columns.
Ravi.
--
--
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: [EMAIL PROTECTED]
Webpage:
http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
--
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-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Leeds,
Mark (IED)
Sent: Friday, December 08, 2006 4:17 PM
To: r-help@stat.math.ethz.ch
Subject: [R] any way to make the code more efficient ?
The code bekow works so this is why I didn't include the data to
reproduce it. The loops about 500 times and each time, a zoo object
with 1400 rows and 4 columns gets created. ( the rows
represent minutes
so each file is one day worth of data). Inside the loop, I
keep rbinding
the newly created zoo object to the current zoo object so that it gets
bigger and bigger over time.
Eventually, the new zoo object, fullaggfxdata, containing
all the days
of data is created.
I was just wondering if there is a more efficient way of doing this. I
do know the number of times the loop will be done at the beginning so
maybe creating the a matrix or data frame at the beginning and putting
the daily ones in something like that would Make it be
faster. But, the
proboem with this is I eventually do need a zoo object. I ask this
question because at around the 250 mark of the loop, things start to
slow down significiantly and I think I remember reading somewhere that
doing an rbind of something to itself is not a good idea. Thanks.
#=
==
===
start-1
for (filecounter in (1:length(datafilenames))) {
print(paste(File Counter = , filecounter)) datafile=
paste(datadir,/,datafilenames[filecounter],sep=)
aggfxdata-clnaggcompcurrencyfile(fxfile=datafile,aggminutes=a
ggminutes,
fillholes=1)
logbidask-log(aggfxdata[,bidask])
aggfxdata-cbind(aggfxdata,logbidask)
if ( start == 1 ) {
fullaggfxdata-aggfxdata
start-0
} else {
fullaggfxdata-rbind(fullaggfxdata,aggfxdata)
}
}
#=
==
==
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Re: [R] from table to dataframe
Like this?
R data.frame(unclass(df2))
a b c d
s1 8 4 4 4
s2 8 4 4 4
s3 8 4 4 4
Andy
From: Milton Cezar Ribeiro
Hi there,
I have a two-entrance dataframe, and I would like generate
a new dataframe with its frequency. I tryed this
site-rep(c(s1,s2,s3),20)
species-rep(c(a,b,a,c,d),12)
df-data.frame(cbind(site,species))
df2-table(df)
But when I convert df2 to data.frame I miss the square
format. I would like have my data.frame like this:
site a b c d
s1 8 4 4 4
s2 8 4 4 4
s3 8 4 4 4
Any help?
Miltinho
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Re: [R] Count cases by indicator
I might be missing something, but the data you showed don't seem to
match your expectation. Firstly, 1 in binary is 511 in decimal,
so your coordinates are off by 1. Secondly, for the data you've
shown, the matrix equivalent look like:
m - matrix(df$x, ncol=9, byrow=TRUE)
rownames(m) - levels(df$cases)
print(m)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
093/0188001011111
093/0206000000000
093/0216011111011
093/0305011111111
093/0325000000000
093/0449000000000
093/0473001111111
093/0499001111111
The counts of unique occurances are:
table(do.call(paste, c(as.data.frame(m), sep=)
0 00101 00111 01011 0
3 1 2 1 1
which do not agree with yours.
If I understood what you wanted, I would do:
R table(rowSums(matrix(2^(0:8) * df$x, ncol=9, byrow=TRUE)))
0 446 500 508 510
3 1 1 2 1
Andy
From: Serguei Kaniovski
Hi,
In the data below, case represents cases, x binary
states. Each case has exactly 9 x, ie is a binary vector
of length 9.
There are 2^9=512 possible combinations of binary states in a
given case, ie 512 possible vectors. I generate these in
the order of the decimals the vectors represent, as:
cmat-as.matrix(expand.grid(rep(list(0:1),9)))
cmat-cmat[nrow(cmat):1,ncol(cmat):1]
cmat contains the binary vectors as rows.
QUESTION: I would like to know how often each of the 512
vectors occurs in case.
With these data, the output should be a vector with 2^9=512
coordinates, having 2,2,1,3, as, respectively, the coordinate
number 129, 193, 449, 512, and zeros in all other coordinates.
Thank you for your help,
Serguei
df-read.delim(clipboard,sep=;)
DATA:
case;x
093/0188;0
093/0188;0
093/0188;1
093/0188;0
093/0188;1
093/0188;1
093/0188;1
093/0188;1
093/0188;1
093/0206;0
093/0206;0
093/0206;0
093/0206;0
093/0206;0
093/0206;0
093/0206;0
093/0206;0
093/0206;0
093/0216;0
093/0216;1
093/0216;1
093/0216;1
093/0216;1
093/0216;1
093/0216;0
093/0216;1
093/0216;1
093/0305;0
093/0305;1
093/0305;1
093/0305;1
093/0305;1
093/0305;1
093/0305;1
093/0305;1
093/0305;1
093/0325;0
093/0325;0
093/0325;0
093/0325;0
093/0325;0
093/0325;0
093/0325;0
093/0325;0
093/0325;0
093/0449;0
093/0449;0
093/0449;0
093/0449;0
093/0449;0
093/0449;0
093/0449;0
093/0449;0
093/0449;0
093/0473;0
093/0473;0
093/0473;1
093/0473;1
093/0473;1
093/0473;1
093/0473;1
093/0473;1
093/0473;1
093/0499;0
093/0499;0
093/0499;1
093/0499;1
093/0499;1
093/0499;1
093/0499;1
093/0499;1
093/0499;1
--
___
Austrian Institute of Economic Research (WIFO)
Name: Serguei Kaniovski P.O.Box 91
Tel.: +43-1-7982601-231 Arsenal Objekt 20
Fax: +43-1-7989386 1103 Vienna, Austria
Mail: [EMAIL PROTECTED]
http://www.wifo.ac.at/Serguei.Kaniovski
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Re: [R] automatic cleaning of workspace
You can avoid loading .Rdata at start-up without deleting the .Rdata
file by adding the --no-restore option to the R command. I have that,
and additionally, --no-save, in my shortcut for the Rgui.exe command. I
use explicit save() and load() in my scripts to save objects that are
expensive to compute.
Andy
From: Leeds, Mark (IED)
I'm having that problem where I am sometimes using an object
that's from a previous workspace when I don't want to be
doing that. I was thinking of putting rm(objects=ls()) in my
first.R function But, the problem with doing this, is that it
doesn't prompt you with are you sure and there could be
very rare cases where I don't want to delete the workspace ?
Is there a way to make the cleaning of the workspace
automatic but still prompt you ? I guess I can always just
try to remember to manually do rm(objects=ls())when I start
up in whatever workspace I am in but I don't think I can
trust my memory. Thanks.
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Re: [R] random forest regression
One way is to graft the stratified sampling code from the classification
part onto the regression part. I will get to it eventually, but just
not now.
Andy
From: Naiara Pinto
Dear all,
I am doing a regression in ramdomForest, using the option
sampsize reduce the number of records used to produce the
randomForest object.
The manual says For classification, if sampsize is a vector
of the length the number of strata, then sampling is
stratified by strata, and the elements of sampsize indicate
the numbers to be drawn from the strata. I need my sampling
to be done with factors, but I am doing a regression. Does
anyone know a way to do that?
Thanks,
Naiara.
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Re: [R] CPU or memory
My understanding is that it doesn't have much to do with 32- vs. 64-bit,
but what the instruction sets of the CPUs. If I'm not mistaken, at the
same clock speed, a P4 would run slower than PIII simply because P4 does
less per clock-cycle. Also, I believe for the same architecture, single
core chips are available at higher clock speeds than their multi-core
counterparts. That's why we recently went for a box with four
single-core Opterons instead of two dual-core ones.
64-bit PCs should be really affordable: I've seen HP laptops based on
the Turion chip selling below $500US.
Andy
From: John C Frain
I would like to thank all who replied to my question about
the efficiency of various cpu's in R.
Following the advice of Bogdan Romocea I have put a sample
simulation and the latest version of R on a USB drive and
will go to a few suppliers to try it out. I will report back
if I find anything of interest.
With regard to 64-bit and 32-bit I thought that the 64-bit
chip might require less clock cycles for a specific machine
instruction than a 32-bit.
This was one of the advantages of moving from 8 to 16 or from
16 to 32 bit chips. Thus a slower, in terms of clock speed,
64-bit chip might run faster than a somewhat similar 32-bit
chip. I fully realize that the full advantage of a 64-bit
chip is available only with a 64-bit operating system and I
am preparing to switch some work to Linux in case I acquire a
64-bit PC. If I do I will time the simulations on that also.
I already do some coarse-grained parallelism as described
by *Brian Ripley
* but on two separate PC's. This is not ideal but allows the
processing time to be halved without the overheads.
FORTRAN 2 was my first programming language and I agree that
I should try to use C or FORTRAN to speed up things. Finally
Rprof could be a great help.
There are lots of utilities in the utils package with which I
was not familiar.
Again Many Thanks to all who made various suggestions.
bogdan romocea[EMAIL PROTECTED] to *r-help*, me
More options 07-Nov (1 day ago) Does any one know of
comparisons of
the Pentium 9x0, Pentium(r)
Extreme/Core 2 Duo, AMD(r) Athlon(r) 64 , AMD(r) Athlon(r)
64 FX/Dual
Core AM2 and similar chips when used for this kind of work.
On 08/11/06, Prof Brian Ripley [EMAIL PROTECTED] wrote:
On Wed, 8 Nov 2006, Christos Hatzis wrote:
Prof. Ripley,
Do you mind providing some pointers on how
coarse-grained parallelism
could be implemented on a Windows environment? Would it be as
simple as running two R-console sessions and then (manually)
combining the results
of
these simulations. Or it would be better to run them as batch
processes.
That is what I would do in any environment (I don't do such things
under Windows since all my fast machines run Linux/Unix).
Suppose you want to do 1 simulations. Set up two batch scripts
that each run 5000, and save() the results as a list or
matrix under
different names, and set a different seed at the top. Then
run each
via R CMD BATCH simultaneously. When both have finished, use an
interactive session to load() both sets of results and merge them.
RSiteSearch('coarse grained') did not produce any hits so
this topic
might
have not been discussed on this list.
I am not really familiar with running R in any mode other than the
default
(R-console in Windows) so I might be missing something really
obvious. I
am
interested in running Monte-Carlo cross-validation in
some sort of a
parallel mode on a dual core (Pentium D) Windows XP machine.
Thank you.
-Christos
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Prof Brian
Ripley
Sent: Wednesday, November 08, 2006 5:29 AM
To: Stefan Grosse
Cc: r-help@stat.math.ethz.ch; Taka Matzmoto
Subject: Re: [R] CPU or memory
On Wed, 8 Nov 2006, Stefan Grosse wrote:
64bit does not make anything faster. It is only of use
if you want
to use more then 4 GB of RAM of if you need a higher
precision of
your variables
The dual core question: dual core is faster if programs
are able to
use that. What is sure that R cannot make (until now) use of the
two cores if you are stuck on Windows. It works excellent if you
use Linux. So if you want dual core you should work with
linux (and
then its faster of course).
Not necessarily. We have seen several examples in which using a
multithreaded BLAS (the only easy way to make use of
multiple CPUs
under Linux for a single R process) makes things many
times slower.
For tasks that are do not make heavy use of linear algebra, the
advantage
[R] graphics ignore tabs in text
Dear R-help,
I seem to recall that I can use \t to get tab in a string on a
graphics device, but it doesn't seem to work. Try:
lab - a\tb\tc
cat(lab, \n) # works in the console output
plot(1:5, main=lab) # no tabs in the title
text(3, 3, lab) # no tabs in the text
I get the same result both in the windows() and pdf() devices. Any
ideas?
This is R-patched Windows binary just downloaded from CRAN.
R version 2.4.0 Patched (2006-10-29 r39744)
Best,
Andy
Andy Liaw, PhD
Biometrics ResearchPO Box 2000 RY33-300
Merck Research LabsRahway, NJ 07065
andy_liaw(a)merck.com 732-594-0820
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Re: [R] one problem about how to hold graphic with R
I'm not familiar with Matlab, but from what I know, hold on is used to
overlay more stuff on the existing plot. In R such things are
accomplished a bit differently: One put up a plot, then use things like
lines(), points(), abline(), etc. to add to the existing plot. The
closest thing to hold on in Matlab, I think, is par(new=TRUE).
Andy
From: Gavin Simpson
On Tue, 2006-10-31 at 21:36 +0800, yang baohua wrote:
Sorry to disturb you, but can you help me to solve one
little problem?
I want to draw a graphic after another with R but I cannot find the
first one after that.
Do you know the command to hold the graphic with R?
I remember with Matlab you may use hold on.
Thanks.
You don't say which OS. On MS Windows one can turn on a
history of plots to the graphics device and replay your plots
- it is in the menu bar for example.
In all OSes, you can start up a new device to take the plot -
which is what Matlab does IIRC, so you have two or more plot
windows on screen at any one time. This is done like this:
plot(1:10)
x11()
plot(1:20)
x11()
plot(rnorm(100))
see ?Devices
You can set a device to be active, i.e. switch around between
plotting windows using dev.set(), e.g.:
dev.cur() # example from above leaves device 4 active
X11
4
dev.set(3) # switch to device
dev.cur() # check
X11
3
plot(sort(rnorm(100))) # plot something new on this device
Is this what you were looking for?
HTH
G
--
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Gavin Simpson [t] +44 (0)20 7679 0522
ECRC ENSIS, UCL Geography, [f] +44 (0)20 7679 0565
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Re: [R] regarding large csv file import [Broadcast]
TFM (R Data Import/Export manual in this case) can be a better place
to look than the archive. Try specifying colClasses in read.csv()
might help.
Andy
From: [EMAIL PROTECTED]
hi All,
i have a .csv of size 272 MB and a RAM of 512MB and working
on windows XP.
I am not able to import the csv file.
R hangs means it stops responding even SciViews hangs.
i am using read.csv(FILENAME,sep=,,header=TRUE). Is there
any way to import it.
i have tried archives already but i was not able to sense much.
thanks in advance
Sayonara With Smile With Warm Regards :-)
G a u r a v Y a d a v
Assistant Manager,
Economic Research Surveillance Department,
Clearing Corporation Of India Limited.
Address: 5th, 6th, 7th Floor, Trade Wing 'C', Kamala City,
S.B. Marg, Mumbai - 400 013
Telephone(Office): - +91 022 6663 9398 , Mobile(Personal)
(0)9821286118
Email(Office) :- [EMAIL PROTECTED] , Email(Personal)
:- [EMAIL PROTECTED]
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Re: [R] how to improve the efficiency of the following lapply codes [Broadcast]
Make good use of Rprof(): It has helped me a great deal in pinpointing
bottlenecks where I would not have suspected.
Cheers,
Andy
From: Weiwei Shi
object.size(intersect.matrix)
41314204
but my machine has 4 G memory, so it should be ok since after
12 hours, it finishes 16k out of 60k but still slow non-linearly.
I am thinking to chop 60k into multiple 5k data.frames to run
the program. but just wondering is there a way around it?
version
_
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major 2
minor 3.1
year 2006
month 06
day01
svn rev38247
language R
version.string Version 2.3.1 (2006-06-01)
[EMAIL PROTECTED] ox]$ more /proc/meminfo
total:used:free: shared: buffers: cached:
Mem: 4189724672 3035549696 11541749760 282836992 2057129984
Swap: 4293586944 645042176 3648544768
[EMAIL PROTECTED] ox]$ more /proc/cpuinfo
processor : 0
vendor_id : GenuineIntel
cpu family : 15
model : 4
model name : Intel(R) Xeon(TM) CPU 3.60GHz
stepping: 3
cpu MHz : 3591.419
cache size : 2048 KB
thanks.
On 10/25/06, Weiwei Shi [EMAIL PROTECTED] wrote:
Hi,
I have a series of lda analysis using the following lapply function:
n - dim(intersect.matrix)[1]
net1.lda - lapply(1:(n), function(k) i.lda(data.list,
intersect.matrix, i=k, w))
i.lda is function to do the real lda analysis.
intersect.matrix is a nx1026 matrix, n can be a really huge number
like 60k. The target is perform a random search. Building a n=120k
matrix is impossible for my machine. When n=5k, the task
can be done
in 30 min while n=60k, it is estimated to take 5 days. So I am
wondering where my coding problem is, which causes this to be a
nonlinearity.
If more info is needed, I will provide.
thanks
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
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Re: [R] binom.test [Broadcast]
To quote one of the previous answers you've got: The formula you're
using is the TV. The one binom.test() uses is the ballpark. Take your
pick.
Andy
From: Ethan Johnsons
Thank you for the info. It helps.
After all, it would be:
0.1304348-1.96*(sqrt((0.1304348*(1-0.1304348))/46))
[1] 0.03310968
0.1304348+1.96*(sqrt((0.1304348*(1-0.1304348))/46))
[1] 0.2277599
Does R have a function for the calculation above?
ej
On 10/20/06, Francisco J. Zagmutt [EMAIL PROTECTED] wrote:
Ethan,
You need to explain why you think this is not the right
function to
use. R is doing exactly what you are asking it to do. Now
is up to
you to choose the methodology you feel is correct.
For a good discussion on your particular issue I recommend you the
following
reference:
A. Agresti and B. A. Coull, Approximate is better than exact for
interval estimation of binomial proportions, The American
Statistician, vol. 52, no.
2, pp. 119-126, 1998.
Once you figure out the right function to use see if the
function is
available in R. If not readily available, and if after
searching through
R's documentation and the forum archives you still can't
find a way to
perform the calculation, then is time to get back to this forum.
Regards,
Francisco
Dr. Francisco J. Zagmutt
College of Veterinary Medicine and Biomedical Sciences
Colorado State
University
From: Ethan Johnsons [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject: [R] binom.test
Date: Fri, 20 Oct 2006 17:18:02 -0400
A quick question, please.
46 e coli lab samples are tested, 6 of them returned positive.
So, the best point estimate for p is 6/46 = 0.1304348.
For a 95% CI for p, I thought binom.test would give me the correct
result, but it seems it is not the right function to use. What is
the R function for this?
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Re: [R] Box M test [Broadcast]
See
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/0.html
Andy
From: GRAHAM LEASK
Dear List
I am looking for a script that will calculate the Box M
test to test the homogeneity of the variance/covariance
matrix between two matrices.
If anyone could send me the script I would appreciate it.
I am aware of the scepticism about this test, where due to
extreme sensitivity a p value of 0.01 is recommended. Despite
this however Box's M test is the established method for
identification of stable strategic time periods within the
strategic management literature and I would like the
opportunity to use this method within either R or S plus.
Any help would be gratefully received.
Kind regards
Graham
Kind regards
Dr Graham Leask
Economics and Strategy Group
Aston Business School
Aston University
Aston Triangle
Birmingham
B4 7ET
Tel: Direct line 0121 204 3150
email [EMAIL PROTECTED]
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Re: [R] Getting group size in a data frame [Broadcast]
Is this sort of what you want?
R aggregate(df[2:3], df[1], function(x) sum(!is.na(x)))
factor val1 val2
1 2421
Andy
From: Ulrik Stervbo
Hi all,
I have a data frame with some measured values of some
animals. Sometimes the
measurement failed, resulting in a NA for a measurement and
sometimes the
animal died, resulting in NA for all measurements.
I have several groups of animals. How do I find the size of
each group with
only alive animals? And how do I find the size of the groups for each
measurement?
An example:
l1 - list(factor=c(24,24,24), val1=c(2, 3, NA), val2=c(4, NA, NA))
df - as.data.frame(l1)
df$factor - factor(df$factor)
The size of factors should be 2 and not 3. The number of
measurement in
val1 should be 2 and the number of measurements in val2 should be 1
Thanks in advance for any help and suggestions
Ulrik
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Re: [R] Question about random sampling in R
When sampling with replacement (like ordinary bootstrap), each draw is
done independently, and in each draw every point has equal probability
of being drawn. When sampling without replacement (random permutation),
all possible sequences (permutations) have equal probability of
occurring. E.g., if the data is 1:2, then (1, 2) has the same
probability of occurring as (2, 1).
Andy
From: tom soyer
Hi,
I looked up the help file on sample(), but didn't find the
info I was looking for.
When sample() is used to resample from a distribution, e.g.,
bootstrap, how does it do it? Does it use an uniform
distribution, e.g., runif(), or something else? And, when the
help file says:sample(x) generates a random permutation of
the elements of x (or 1:x), would I be correct if I
translate the statement as follows: it means that the order
of sequence, which was generated from a uniform distribution,
would look like a random normal distribution.
Thanks,
Tom
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Re: [R] [Q] How to fit data to HORIZONTAL line [Broadcast]
The horizontal line can be fitted by lm(y ~ 1).
Andy
From: Young-Jin Lee
Dear R users
I posted a question about how to fit data to a straight
line this afternoon. But I realized that my question was not
correct because I needed to fit data to a HORIZONTAL line,
not a ordinary straight line.
I looked at lm method, but could not figure out how to fix
the regression coefficient to 0. I also tried nls, but it
did not work.
The reason I wanted to fit the data to a horizontal line is
that I want to compare AIC/BIC values of two models (a simple
straight line mode vs a nonlinear curve model). I thought
that I can call
aic(horizontal_fit_model) and aic (nonlinear_fit_model) to
achieve this goal.
If I can compute AIC/BIC value of a horizontal fit model
without doing acutal fitting, that would be fine, too.
Thank in advance.
Young-Jin
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Re: [R] MARS help?
Spencer,
MARS fits splines, not disconnected lines. Perhaps the strucchange package has
facility to fit your data better.
Cheers,
Andy
From: [EMAIL PROTECTED] on behalf of Spencer Graves
Sent: Tue 10/17/2006 11:43 PM
To: R-help; Kurt Hornik
Subject: [R] MARS help? [Broadcast]
I'm trying to use mars{mda} to model functions that look fairly
close to a sequence of straight line segments. Unfortunately, 'mars'
seems to totally miss the obvious places for the knots in the apparent
first order spline model, and I wonder if someone can suggest a better
way to do this. The following example consists of a slight downward
trend followed by a jump up after t1=4, following by a more marked
downward trend after t1=5:
Dat0 - cbind(t1=1:10,
x=c(1, 0, 0, 90, 99, 95, 90, 87, 80, 77))
library(mda)
fit0 - mars(Dat0[, 1, drop=FALSE], Dat0[, 2],
penalty=.001)
plot(Dat0, type=l)
lines(Dat0[, 1], fit0$fitted.values,
lty=2, col=red)
Are there 'mars' options I'm missing or other software I should be
using?
I've got thousands of traces crudely like this of different
lengths, and I want an automated way of summarizing similar traces in
terms of a fixed number of knots and associated slopes for each linear
spline segment max(0, t1-t.knot).
Thanks,
Spencer Graves
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Re: [R] Automatic File Reading [Broadcast]
Works on all platforms:
flist - list.files(path=file.path(somedir, somewhere),
pattern=[.]csv$)
csvlist - lapply(flist, read.csv, header=TRUE)
whateverList - lapply(csvlist, whatever)
Andy
From: Richard M. Heiberger
Wensui Lui asks:
is there a similar way to read all txt or csv files with same
structure from a folder?
On Windows I use this construct to find all files with the
specified wild card name.
I used the \\ in the file paths with the translate=FALSE,
because the / in
the DOS switches /w/B must not be translated. On Windows
this picks up
both lower and upper case filenames
A similar construct can be written for Unix.
tmp - shell('dir c:\\HOME\\rmh\\tmp\\*.R /w/B', intern=TRUE,
translate=FALSE) ##msdos
for (i in tmp) source(paste(c:\\HOME\\rmh\\tmp\\, i, sep=))
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Re: [R] Nested source() errors [Broadcast]
I've seen people doing that without problem. Not something I'd like to
do myself, precisely because when problems occur, it's difficult to
figure out what went wrong. Such practice usually indicate that you
ought to organize your functions better. (You _are_ writing functions,
instead of just scripts?)
Andy
From: Pierce, Ken
Does anyone know of any issues with nesting source() calls
within multiple scripts? I have at least one script which
always finds errors when I source it but runs fine when run
on its own. It containd source() calls to other scripts and
it seems to fail during the first nested
source() command.
Ken
Kenneth B. Pierce Jr.
Research Ecologist
Landscape Ecology, Modeling, Mapping and Analysis Team
PNW Research Station - USDA-FS
3200 SW Jefferson Way, Corvallis, OR 97331
[EMAIL PROTECTED]
541 750-7393
http://www.fsl.orst.edu/lemma/gnnfire
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Re: [R] CI
Here's one way:
R x - c(6,11,5,14,30,11,17,3,9,3,8,8)
R confint(lm(x~1), level=.9)
5 %95 %
(Intercept) 6.546834 14.2865
Andy
From: Ethan Johnsons
I have a quick question, please.
Does R have function to compute i.e. a 90% confidence
interval for the mean for these numbers?
mean (6,11,5,14,30,11,17,3,9,3,8,8)
[1] 6
I thought pt or qt would give me the interval, but it seems not.
thx much.
ej
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Re: [R] CI
You did ask for CI of mean, so that's what you got. If you want CI for
proportion, here are two (non-bootstrap) ways:
R confint(lm(I(x == 1) ~ 1), level=.9)
5 % 95 %
(Intercept) 0.2666456 0.6133544
R binom.test(sum(x == 1), length(x), conf.level=.9)
Exact binomial test
data: sum(x == 1) and length(x)
number of successes = 11, number of trials = 25, p-value = 0.69
alternative hypothesis: true probability of success is not equal to 0.5
90 percent confidence interval:
0.2698531 0.6213784
sample estimates:
probability of success
0.44
I hope these are not HW problems?
Andy
From: Ethan Johnsons
Thank you so much for the feedback.
The random numbers are working great. I have tried
non-random numbers, and the outcome is not correct with confint.
Is there a way to compute i.e. a 90% confidence interval for
percent of 1?
i.e. where 1 = apple; 2 = orange
x
[1] 2 2 2 2 2 1 1 2 2 1 2 1 2 2 2 1 1 1 1 1 1 1 2 2 2
table (x)
x
1 2
11 14
x =11
confint(lm(x~1), level=0.90)
5 % 95 %
(Intercept) NaN NaN
ej
On 10/18/06, Liaw, Andy [EMAIL PROTECTED] wrote:
Here's one way:
R x - c(6,11,5,14,30,11,17,3,9,3,8,8) confint(lm(x~1), level=.9)
5 %95 %
(Intercept) 6.546834 14.2865
Andy
From: Ethan Johnsons
I have a quick question, please.
Does R have function to compute i.e. a 90% confidence
interval for
the mean for these numbers?
mean (6,11,5,14,30,11,17,3,9,3,8,8)
[1] 6
I thought pt or qt would give me the interval, but it seems not.
thx much.
ej
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Re: [R] Help on Random forest [Broadcast]
Do provide a reproducible example, as the Posting Guide suggests.
Try:
library(randomForest)
example(predict.randomForest)
iris.pred - predict(iris.rf, iris[ind == 2,], nodes=TRUE)
str(iris.pred)
attr(iris.pred, nodes)
Andy
From: Rupendra
Hello all,
I am trying to explore random forest in R. What I want to do
is get the node number in which the case falls in the tree of
random forest. For that I am calling the predict method as:
learn.pred - predict (learn.rf,
newdata=learn.data.x,norm.votes= TRUE,predict.all = TRUE,
nodes= TRUE,type=response)
Studying the manual of random forest, I suppose that
learn,pred$nodes should contain the node numbers, but there
is no attributes called nodes in learn.pred object.
I am not much experienced with R. Please help me to resolve
this issue.
Thanks in advance,
Rupendra
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Re: [R] how to convert all columns of a data frame into factors
Alternatively:
x[] - lapply(x, factor)
Recall that a data frame is a list, so lapply() is a natural choice.
Andy
From: Gabor Grothendieck
Try this:
replace(BOD, TRUE, lapply(BOD, factor))
On 10/4/06, Weiwei Shi [EMAIL PROTECTED] wrote:
Hi,
I use apply
apply(x, 2, factor)
but it does not work. please help. thanks.
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
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[R] how do I tell configure where to find Java?
Dear R-help,
I'm trying to build R-2.4.0 on our Opteron-based Scyld cluster. The system
has gcj (the GNU Java compiler, part of GCC) stuff in /usr/bin. When I
installed jdk 1.5.08, the install script placed it in /usr/java (I didn't
have a choice, as the script didn't offer that option). Now when I run
configure in R-2.4.0, it finds gcj, which is not what I want to use. Is
there a way to tell configure where to look for Java? I tried configure
--help but didn't see anything related to Java.
Best,
Andy
Andy Liaw, PhD
Biometrics ResearchPO Box 2000 RY33-300
Merck Research LabsRahway, NJ 07065
andy_liaw(a)merck.com 732-594-0820
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Re: [R] how do I tell configure where to find Java? [Broadcast]
Before I do that, I would need to remove the gcj stuff that are in /usr/bin.
If I know how to remove gcj, I'd gladly do that. However, for the
particular version of the OS, the entire GCC seems to be bundled into one
rpm, and I could not remove just the gcj component. Neither do I wish to
mess with files that are part of some RPMs--- in my experience that's
invitation for trouble later.
Best,
Andy
From: mike waters
I'm not familiar with gcj, but my initial reaction would be a
ln -s for the relevant compiler executable from /usr/java
into /usr/bin.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Liaw, Andy
Sent: 03 October 2006 19:40
To: r-help
Subject: [R] how do I tell configure where to find Java?
Dear R-help,
I'm trying to build R-2.4.0 on our Opteron-based Scyld
cluster. The system has gcj (the GNU Java compiler, part of
GCC) stuff in /usr/bin. When I installed jdk 1.5.08, the
install script placed it in /usr/java (I didn't have a
choice, as the script didn't offer that option). Now when I
run configure in R-2.4.0, it finds gcj, which is not what I
want to use. Is there a way to tell configure where to look
for Java? I tried configure --help but didn't see anything
related to Java.
Best,
Andy
Andy Liaw, PhD
Biometrics ResearchPO Box 2000 RY33-300
Merck Research LabsRahway, NJ 07065
andy_liaw(a)merck.com 732-594-0820
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Re: [R] how do I tell configure where to find Java?
Thanks to everyone who provided the info. I tried Martin Morgan's
suggestion (adding JAVA_HOME=/where/jdk/install/itself) to the
list of variables defined after `configure', and config.log shows
that the desired Java is found.
The Scyld system is based on RH, but I believe it lags far
behind FC. The JDK is from Sun, and didn't come as a RPM.
Best,
Andy
From: Peter Dalgaard
Logan Lewis [EMAIL PROTECTED] writes:
Andy,
On Tuesday 03 October 2006 3:30 pm, Liaw, Andy wrote:
Before I do that, I would need to remove the gcj stuff
that are in
/usr/bin. If I know how to remove gcj, I'd gladly do
that. However,
for the particular version of the OS, the entire GCC seems to be
bundled into one rpm, and I could not remove just the gcj
component.
Neither do I wish to mess with files that are part of
some RPMs---
in my experience that's invitation for trouble later.
The Red Hat way of dealing with different packages
providing the same
binaries is alternatives. You will see a bunch of links in
/etc/alternatives, and the command /usr/sbin/alternatives
allows you
to switch between options that provide the same binaries.
The trouble
is that the Sun JDK package does not interface into this
system, and
doesn't show up as an option when you execute
/usr/sbin/alternatives --config java.
Hmm... I actually have it, but how did it get there?
[EMAIL PROTECTED] R]$ /usr/sbin/alternatives --config java
There are 2 programs which provide 'java'.
SelectionCommand
---
1 /usr/lib/jvm/jre-1.4.2-gcj/bin/java
*+ 2 /usr/lib/jvm/jre-1.5.0-sun/bin/java
Enter to keep the current selection[+], or type selection number:
failed to create /var/lib/alternatives/java.new: Permission denied
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c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph:
(+45) 35327918
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(+45) 35327907
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Re: [R] levels of factor when subsetting the factor
You have at least two choices:
R factor(fact[1:6])
[1] A A A B B B
Levels: A B
R fact[1:6, drop=TRUE]
[1] A A A B B B
Levels: A B
HTH,
Andy
From: Afshartous, David
All,
When I take a subset of a factor the reduced factor still
maintains all
the original levels of the factor when say forming the key in a plot.
The data is correct, but the variable still remembers the original
levels. See below for reproducible code. Does anyone know how to fix
this?
cheers,
dave
fact = as.factor(c(rep(A, 3),rep(B, 3), rep(C, 3)))
new.fact = fact[1:6]
new.fact
[1] A A A B B B
Levels: A B C## should only show A B
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Re: [R] 4^2 factorial help
If you really want the quadratic terms, you need to keep those variables as numeric, instead of factors. (You might also want to look into something like the central composite designs.) summary() and coef() on the resulting fitted object should give you want you need. Things like these are covered in the An Introduction to R manual... Andy From: [EMAIL PROTECTED] To whom it may concern: I am trying a factorial design a system of mine that has two factors. Each factor was set at four different levels, with one replication for each of the combinations. My data is as follows: A B Response 16002.5 0.0257 26002.5 0.0254 36005 0.0217 46005 0.0204 5600100.0191 6600100.0210 7600200.0133 8600200.0139 98002.5 0.0312 10 800 2.5 0.0317 11 800 5 0.0307 12 800 5 0.0309 13 800 100.0330 14 800 100.0318 15 800 200.0225 16 800 200.0234 17 1000 2.5 0.0350 18 1000 2.5 0.0352 19 1000 5 0.0373 20 1000 5 0.0361 21 1000 100.0432 22 1000 100.0402 23 1000 200.0297 24 1000 200.0306 25 1200 2.5 0.0324 26 1200 2.5 0.0326 27 1200 5 0.0353 28 1200 5 0.0353 29 1200 100.0453 30 1200 100.0436 31 1200 200.0348 32 1200 200.0357 I am able to enter my data into R and obtain an ANOVA table (which I have been able to verify as correct using an excel spreadsheet), using the following syntax: Factorial-data.frame(A=c(rep(c(600, 600, 600, 600, 800, 800, 800, 800, 1000, 1000, 1000, 1000, 1200, 1200, 1200, 1200), each=2)), B=c(rep(c(2.5, 5, 10, 20, 2.5, 5, 10, 20, 2.5, 5, 10, 20, 2.5, 5, 10, 20), each=2)), Response = c(0.0257, 0.0254, 0.0217, 0.0204, 0.0191, 0.021, 0.0133, 0.0139, 0.0312, 0.0317, 0.0307, 0.0309, 0.033, 0.0318, 0.0225, 0.0234, 0.035, 0.0352, 0.0373, 0.0361, 0.0432, 0.0402, 0.0297, 0.0306, 0.0324, 0.0326, 0.0353, 0.0353, 0.0453, 0.0436, 0.0348, 0.0357)) anova(aov(Response~A*B, data=Factorial)) However, this is as far as I am able to go. I would like to obtain the coefficients of my model, but am unable. I would also like to use other non-linear models as these factors are not linear. Also would like to add A^2 and B^2 into the ANOVA and modeling. Please can you help with regard and offer some advice. Your help is much appreciated. Yours sincerely, Leslie Correia Department of Process Engineering University of Stellenbosch Private Bag X1 Matieland, 7602 Stellenbosch Tel: 0837012017 E-mail: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
