I have a following matrix and wish to define a variable based the variable
A=matrix(0,5,5)
A[1,]=c(30,20,100,120,90)
A[2,]=c(40,30,20,50,100)
A[3,]=c(50,50,40,30,30)
A[4,]=c(30,20,40,50,50)
A[5,]=c(30,50,NA,NA,100)
A
[,1] [,2] [,3] [,4] [,5]
[1,] 30 20 100 120 90
[2,] 40 30
Hi
Something to get you started
? as.list
a data.frame can be regarded as a 2 dimensional array of list vectors
df = data.frame(a=1:2,b=2:1,c=4:5,d=9:10)
as.list(df[,1:3])
$a
[1] 1 2
$b
[1] 2 1
$c
[1] 4 5
see also
http://cran.ms.unimelb.edu.au/doc/contrib/Burns-unwilling_S.pdf
Regards
On Tue, 9 Aug 2011, Peter Langfelder wrote:
Assuming you need the full distance matrix at one time (which you do not for
hierarchical clustering, itself a highly dubious method for more than a few
hundred points).
Apologies if this hijacks the thread, but why is hierarchical
clustering highly
Hi R users.
sorry for missing example and if question is to general but I am wondering
if it is possible to execute subqueries in function sqlQuery (package RODBC)
with opened connection with Excel or SQL server 2000. I couldn't find any
example of this.
And if it is possible what should be a
Hi Andra.
I wonder how you come about trying to use LASSO without knowing what lambda
is. I'd advise you to read up on it. In the help (?glmnet) you can find
several paper references, but for a more gentle introduction, you can read
http://www-stat.stanford.edu/~tibs/ElemStatLearn/
In a
Hi,
thanks a lot for pointing me at conditional plotting!
I have to confess that I'm still not really convinced whether this type of
philosophy holds true in each and every situation, especially when there
appears to be a common sense in literature (even if it is not optimal) to
depict such data
Maybe this kind of usage of lavaan is not very common, but in order to
help others in my situation, is this documented somewhere? My
understanding of latent variable analysis is indeed limited, but I did
not understand that lavaan worked liked this when I read the
documentation.
This is not
In what sense is this a 'subquery'? It is just an SQL command (write
it on one line, no terminating ;, which is not part of the query).
On Wed, 10 Aug 2011, andrija djurovic wrote:
Hi R users.
sorry for missing example and if question is to general but I am wondering
if it is possible to
Hi,
I was irritated about your printed last row of A, which apart from
definition contains a 20. Anyway, how about this:
y-x-rep(NA,nrow(A))
#its not clear, whether multiple values of 100 can occur in a single
#row, and what to do, when 100 is found before and after 20, so you may
#alter the
The kappa2() function in the irr library takes an n x 2 matrix as input,
where the two columns are the ratings by two raters. Let x and y below be
the ratings of the two raters:
x-sample(c(0,1,2),100,replace=T)
x
o-sample(c(0,0,0,1),100,replace=T)
y-x+o
y
#Then kappa is computed as:
Dear all
It is difficult to use round(..., digits=2) on a data frame since one
has to first take care to remove non-numeric variables such as
'character' or 'factor':
head(round(iris, 2))
Error in Math.data.frame(list(Sepal.Length = c(5.1, 4.9, 4.7, 4.6, 5, :
non-numeric variable in data
One approach is the following:
numVars - sapply(iris, is.numeric)
iris[numVars] - lapply(iris[numVars], round, digits = 2)
head(iris)
You can also put it in one lapply() call if you like.
I hope it helps.
Best,
Dimitris
On 8/10/2011 11:34 AM, Liviu Andronic wrote:
Dear all
It is difficult
Hello
On Wed, Aug 10, 2011 at 11:41 AM, Dimitris Rizopoulos
d.rizopou...@erasmusmc.nl wrote:
One approach is the following:
numVars - sapply(iris, is.numeric)
iris[numVars] - lapply(iris[numVars], round, digits = 2)
head(iris)
That's interesting, but still doesn't do what I need. Since it's
Hi Listers,
Is it possible to produce an ordination plot in 2d, where bubbles represent
the location of sites (this part is easy enough) and the size of the bubbles
is proportional to the sites location in 3d space (I am stuck on this
option). So sites that are very near the 2d plane of the xy
On 08/10/2011 10:02 AM, Andrew Halford wrote:
Hi Listers,
Is it possible to produce an ordination plot in 2d, where bubbles represent
the location of sites (this part is easy enough) and the size of the bubbles
is proportional to the sites location in 3d space (I am stuck on this
option).
Very easy if you note that cex in plot can be a vector.
example:
x - runif(100)
y-runif(100)
z-runif(100)
#shift and scale z for convenience 9the scaling is based on range 'cos we know
this is in [0,1]
#your mileage may vary but the principle is )
z.scaled - 0.05 + (z-min(z))/diff(range(z))
I thought subqueries in sense of commad inside the command (in my example
two select commands). It works as you proposed an I thought in this case
(subqueries) that I need different syntax for sqlQury function combining SQL
query and paste.
But now I have another problem and again sorry if it is
Xts is an extension of zoo that has some other nice features: character
subsetting, periodic apply functions, good built in time conversions, etc.
More importantly for my work, the authors put a lot of work into making sure it
plays well with all of R's many ts classes so I almost always start
I would use the tapply function (which is designed for the case in which
data exists for most pairs of the levels of A and B) or the
reshape::sparseby function, or something else in the reshape package.
These won't give you exactly the structure you were asking for, but they
will separate the
On 08/10/2011 10:02 AM, Andrew Halford wrote:
Hi Listers,
Is it possible to produce an ordination plot in 2d, where bubbles represent
the location of sites (this part is easy enough) and the size of the bubbles
is proportional to the sites location in 3d space (I am stuck on this
option).
Dear List,
I'm fairly new in R. I'd like to see how glm() uses the argument family in
fitting a model. Specifically, I'd like to see how a glm with a gamma family
is fitted.
Thanks for any help,
Axel.
[[alternative HTML version deleted]]
__
Just type glm at the prompt.
glm
function (formula, family = gaussian, data, weights, subset,
na.action, start = NULL, etastart, mustart, offset, control = list(...),
model = TRUE, method = glm.fit, x = FALSE, y = TRUE, contrasts = NULL,
...)
{
call - match.call()
if
On 08/10/2011 03:00 AM, Nick Sabbe wrote:
Finally, to avoid downward bias, you could run a normal glm with only the
variables selected in the previous step.
At the cost, of course, of introducing upward bias
--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of
On Tue, 2011-08-09 at 22:57 +1000, Andrew Halford wrote:
Hi Listers,
I am trying to reflect a PCA biplot in the x-axis (i.e. PC1) but am not
having much success. In theory I believe all I need to do is multiply the
site and species scores for the PC1 by -1, which would effectively flip the
Hi,
I'm working on some social networks and I managed to create the graphs with
labels and edges weight, but I would also like to change the size of the
vertices according to the age of the persons in the network and the shape
according to the gender. Now for the age, I have people with ages
Hi all,
I need to convert a floating point value from Microsoft Basic format to IEEE
format.
There's a simple way to achieve this in R or I have to write my own
function?
(e.g. convert the C code below)
thanks
t
#include string.h/* for strncpy */
int _fmsbintoieee(float *src4, float
Hello,
I have a problem with using the following design with ANCOVA in R.
There are two groups (control + treatment), each with ten subjects.
The subjects show a response that is monitored over time (four time
points). For a single given subject, the response can be analysed with
linear
Hi, I just started learning R, and one of the most frequent thing that I need
to calculate is cohen kappa in my psychology lab and I figure being able to
do inter rater reliability is a great way for me to explore R. There are
two different scenario in which I need help with. (By the way, I have
Wir sind bis am 20. August in den Ferien und werden keine e-mails beantworten.
Bei dringenden Fällen melden Sie sich bei Stefanie von Felten
steffi.vonfel...@oikostat.ch
We are on vacation until 20. August. In urgent cases, please contact Stefanie
von Felten steffi.vonfel...@oikostat.ch
Dear all,
I would like to do variance partitioning of community dissimilarity matrix
(Y) using 4 explanatory tables:
X1=environmental characteristics
X2=species traits related to dispersal
X3=species characteristics (abundances and richnness)
X4= xy spatial coordinates
The problem is that I have
Hello,
I'd like to perform a regression using MCMCregress (MCMCpack).
One variable therefore should be a function rather than a variable:
I want to use X as an input and X should be defined as a random number between
to values. Therefore I want to use the function runif like:
X -(1, Xa, Xb) but
The function format() might serve your needs.
format(head(iris), digits=1)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
15 41 0.2 setosa
25 31 0.2 setosa
35 31
Dear list,
I wonder if there a better way to have rbind/cbind/append to create
the first element (if it is empty) instead of doing the following in a
loop?
for (i in 1:10) {
if (i == 1) {
aRow = SomeExpression(i)
} else {
aRow = rbind(aRow,SomeExpression(i))
}
}
Thanks
Anthony
Gabe,
Since you didn't provide a small example of your data, I can't test out
your code. However, I used an example from the ?ordiellipse help page to
draw different colored ellipses (using the show.groups= argument) with
labels (using the label= argument).
Hope this helps.
library(vegan)
Hi:
Try this:
## Function that takes a data frame as input and outputs a data frame:
chrSumm - function(d) { # d is a data frame
colnames(d) - c(chr,start,end,base1,base2,
totalreads,methylation,strand)
TR - nrow(d)
RG1 - sum(d['totalreads'] = 1)
percent -
Dear List,
I wonder why when using read.csv(), if the column name contains a
numeric i.e. a stock symbols-0001.HK, it will automatically insert
an X character to the column names - X0001.HK.
Now I have to manually do a loop and use substring() to remove the X
character. Any advice? Thanks
Johannes,
You have the loop set up right, you just need to add indexing to refer to
the looping variable, i.
lT - sample(1:10)
uT - sample(21:30)
X - numeric(length(lT))
for (i in 1:length(lT)) X[i] - runif(1, lT[i], uT[i])
X
Note that I changed the name of the result from T to X, because T
Anthony,
See ?make.names for a description of valid names. Here's an excerpt:
A syntactically valid name consists of letters, numbers and the dot or
underline characters and starts with a letter or the dot not followed by a
number. ... The character X is prepended if necessary.
There is an
To borrow shamelessly from one of the prominent helpers on this list:
What is the problem you're trying to solve?(attribution: Jim Holtman)
I have the sense you want to do something over many subsets of your
data frame. If so, breaking things up into lists of lists of lists is
not
On 10/08/2011 5:58 AM, taraxacum wrote:
Hi all,
I need to convert a floating point value from Microsoft Basic format to IEEE
format.
There's a simple way to achieve this in R or I have to write my own
function?
(e.g. convert the C code below)
You'll need to write your own function. It can be
On 10/08/2011 7:28 AM, Johannes Radinger wrote:
Hello,
I'd like to perform a regression using MCMCregress (MCMCpack).
One variable therefore should be a function rather than a variable:
I want to use X as an input and X should be defined as a random number between
to values. Therefore I want
On Aug 10, 2011, at 9:08 AM, Anthony Ching Ho Ng wrote:
Dear list,
I wonder if there a better way to have rbind/cbind/append to create
the first element (if it is empty) instead of doing the following in a
loop?
for (i in 1:10) {
if (i == 1) {
aRow = SomeExpression(i)
} else {
aRow
Can't you use sapply?
sapply(seq_len(10), function(i){SomeExpression(i)})
Best regards,
Thierry
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens David Winsemius
Verzonden: woensdag 10 augustus 2011 15:50
Aan: Anthony Ching Ho
Hi,
I must be doing something silly here, because I can't get the studentised
and standardised residuals from r output of a linear model to agree with
what I think they should be from equation form.
Thanks in advance,
Jennifer
x = seq(1,10)
y = x + rnorm(10)
mod = lm(y~x)
rstandard(mod)
On Wed, Aug 10, 2011 at 2:34 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 10/08/2011 5:58 AM, taraxacum wrote:
Hi all,
I need to convert a floating point value from Microsoft Basic format to
IEEE
format.
There's a simple way to achieve this in R or I have to write my own
function?
So far we have received over 70 responses to our survey. If you have NOT
responded already and are likely to attend useR! 2012, please take the
extremely short survey today. The link is below.
More information about Nashville may be seen at
http://biostat.mc.vanderbilt.edu/UseR-2012
In a day
On 10/08/2011 10:16 AM, Barry Rowlingson wrote:
On Wed, Aug 10, 2011 at 2:34 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 10/08/2011 5:58 AM, taraxacum wrote:
Hi all,
I need to convert a floating point value from Microsoft Basic format to
IEEE
format.
There's a simple way
R version 2.13.1
OS X (or Windows)
Colleagues,
I received a number of files with a .xls extension. These files open in XL
and, by all appearances, are XL files. However, it appears to me that the
files are actually XML:
readLines(dir()[16])[1:10]
[1] ?xml version=\1.0\?
On 08/10/2011 10:03 AM, Jen wrote:
Hi,
I must be doing something silly here, because I can't get the studentised
and standardised residuals from r output of a linear model to agree with
what I think they should be from equation form.
x = seq(1,10)
y = x + rnorm(10)
mod = lm(y~x)
rstandard(mod)
Following the suggestion by Duncan Murdoch, this should work for you.
X - runif(length(lT), lT, uT)
Jean
From:
Johannes Radinger jradin...@gmx.at
To:
Jean V Adams jvad...@usgs.gov
Cc:
r-help@r-project.org
Date:
08/10/2011 08:40 AM
Subject:
Re: [R] function runif in for loop
Jean,
thank
Thanks Patrick - at least I know I wasn't being too silly :-)
Jen
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
__
Dear R-help,
I wonder if you could give me some suggestions in how to do a union
join of two data frames as follow:
- union join the common column, and insert a 0 if one is missing.
I made a function to perform the following, and I know it may not that
quite welly written, but it works.
Any
On Aug 10, 2011, at 11:04 AM, Anthony Ching Ho Ng wrote:
Dear R-help,
I wonder if you could give me some suggestions in how to do a union
join of two data frames as follow:
- union join the common column, and insert a 0 if one is missing.
I made a function to perform the following, and I
January,
It looks like you will need an interaction effect, perhaps
g - lm( response ~ subject + group*time)
Please see the ancova function in the HH package.
install.packages(HH) ## if necessary
library(HH)
?ancova
Rich
On Wed, Aug 10, 2011 at 5:15 AM, January Weiner
Hello
On Wed, Aug 10, 2011 at 2:31 PM, Jean V Adams jvad...@usgs.gov wrote:
The function format() might serve your needs.
This looks very promising, but yields some strange results. See below:
x - data.frame(a=rnorm(10), b=rnorm(10), c=rnorm(10), d=letters[1:10])
x
a b
We have released to CRAN a new version of the (recommended)
package Matrix, and of package lme4 yesterday.
Anyone who gets the new version of Matrix *MUST* re-install lme4
-- if (s)he is using lme4 at all. Technical details about that further below.
The fact that yesterday's version number of
Dear Jen,
Actually you can check out what R does by looking at the source.
# first type the name of the function
rstandard
function (model, ...)
UseMethod(rstandard)
environment: namespace:stats
# ?methods will list you the corresponding functions
methods(rstandard)
[1] rstandard.glm
Try this:
merge(q1, q2, all = TRUE)
On Wed, Aug 10, 2011 at 12:04 PM, Anthony Ching Ho Ng
anthony.ch...@gmail.com wrote:
Dear R-help,
I wonder if you could give me some suggestions in how to do a union
join of two data frames as follow:
- union join the common column, and insert a 0 if one
As it says in
?format
the digits argument specifies ... how many significant digits are to be
used ... enough decimal places will be used so that the smallest (in
magnitude) number has this many significant digits ...
In your example, the last value in column a is 0.06348058 which is
written
Hi list,
I used to work with RExcel in excel 2003. Now in 2007, I tried the same
RExcel, but it did not work. I got R version 12. I downloaded/installed the
latest version of RExcel 3.2.0 from http://sunsite.univie.ac.at/rcom/. It
has added the RExcel add-ins, but when I click on starting R in
Hello
I am trying to learn the spatial panel data analysis (newbie). I have the R
version 2.13.1 and I did download the spml package required for the spatial
panel data analysis. However, when I tried the analysis, I get the following
error message. could not find function spfeml. Can
Hello Duncan,
Here is a small example to illustrate what I am trying to do.
# Example data.frame
df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4))
# A B C
# 1 a X 1
# 2 a X 2
# 3 b Y 3
# 4 b Z 4
### First way of getting the list structure (ls1) using imbricated lapply
loops:
# Get
Hi Duncan,
I have tried to install a tar.gz package following your instructions
(https://stat.ethz.ch/pipermail/r-help/2008-August/169599.html) but without
success. Here are the steps I followed:
I installed the last version of Rtools and ran Rcmd INSTALL rJava_0.8-8.tar.gz
and got the error
Jean,
thank you for your answer.
especially the line X - numeric(length(lT)) helped me a lot.
Anyway, in my case I'd like to get a dynamic variable or better a
function for X. I mean if i try to call X I'd like that this drawing of random
number is performed. In the case now if I call X several
Original-Nachricht
Datum: Wed, 10 Aug 2011 09:38:38 -0400
Von: Duncan Murdoch murdoch.dun...@gmail.com
An: Johannes Radinger jradin...@gmx.at
CC: r-help@r-project.org
Betreff: Re: [R] function runif in for loop
On 10/08/2011 7:28 AM, Johannes Radinger wrote:
Hello,
Hello Denis,
To borrow shamelessly from one of the prominent helpers on this list:
What is the problem you're trying to solve? Â Â (attribution: Jim Holtman)
I'm trying to connect two sets of legacy R tools: the output of the
first one can be transformed in a data.frame without loss of
Hi,
this command gives you all possible encoding options on your platform:
iconvlist()
hope it answers your question,
T
--
View this message in context:
http://r.789695.n4.nabble.com/scan-file-encoding-tp840838p3733327.html
Sent from the R help mailing list archive at Nabble.com.
Don't they have their own support mailing list? You should review
their documentation for specifics.
--
David.
On Aug 10, 2011, at 12:24 PM, Dr. Alireza Zolfaghari wrote:
Hi list,
I used to work with RExcel in excel 2003. Now in 2007, I tried the
same
RExcel, but it did not work. I got R
On 10/08/2011 10:30 AM, Frederic F wrote:
Hello Duncan,
Here is a small example to illustrate what I am trying to do.
# Example data.frame
df=data.frame(A=c(a,a,b,b), B=c(X,X,Y,Z), C=c(1,2,3,4))
# A B C
# 1 a X 1
# 2 a X 2
# 3 b Y 3
# 4 b Z 4
### First way of getting the list structure
On 10/08/2011 11:59 AM, Lippel, Anna wrote:
Hi Duncan,
I have tried to install a tar.gz package following your instructions
(https://stat.ethz.ch/pipermail/r-help/2008-August/169599.html) but without
success. Here are the steps I followed:
I installed the last version of Rtools and ran Rcmd
On 10/08/2011 9:40 AM, Johannes Radinger wrote:
Jean,
thank you for your answer.
especially the line X- numeric(length(lT)) helped me a lot.
Anyway, in my case I'd like to get a dynamic variable or better a
function for X. I mean if i try to call X I'd like that this drawing of random
number
You might want to look into the packages bigmemory and biganalytics.
Corey
On Tue, Aug 9, 2011 at 8:38 PM, Chris Howden
ch...@trickysolutions.com.auwrote:
Hi,
Im trying to do a hierarchical cluster analysis in R with a Big Data set.
Im running into problems using the dist() function.
Hi Frederic,
shouldn't there be an result for the 3rd row as well, eg ls1$b$Y?
Maybe this will do what you want?
dtf-within(dtf,index-factor(A:B))
tapply(dtf$C,dtf$index,list)
Hth.
Am 10.08.2011 16:30, schrieb Frederic F:
Hello Duncan,
Here is a small example to illustrate what I am
I was going to suggest
AB - df[c(A,B)]
ls2 - array(split(df$C, AB), dim=sapply(AB, nlevels), dimnames=sapply(AB,
levels))
which produces a matrix very similar to what Duncan's by() call produces
ls1 - by(df$C, df[,1:2], identity)
E.g.,
ls2[[a,X]]
[1] 1 2
ls1[[a,X]]
[1] 1 2
Duncan et. al:
Inline below.
On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 10/08/2011 9:40 AM, Johannes Radinger wrote:
Jean,
thank you for your answer.
especially the line X- numeric(length(lT)) helped me a lot.
Anyway, in my case I'd like to get a
HI everyone,
I'm plotting a histogram in R and within that histogram i need to
demonstrate the percentage of another variable (Percentage of MutStatus)
within the bins plotted inthe histogramI don't know how to do that!
Data:Validation_Status Mutation_Status TvarRatio
WildtypeNone
Perhaps you could shade the bars as appropriate?
I'm not going to use your data because it's not an easily paste-able but how
about this:
x = rnorm(100)
y = sample(c(A,B),100,replace=T,prob=c(0.7,0.3))
d = data.frame(level = x, status = y)
n = 10 # Number of bins
breaks = quantile(d$level,
On 10/08/2011 1:16 PM, Bert Gunter wrote:
Duncan et. al:
Inline below.
On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 10/08/2011 9:40 AM, Johannes Radinger wrote:
Jean,
thank you for your answer.
especially the line X- numeric(length(lT)) helped
Assuming your data is in a data.frame called df, try this:
attach(df)
TR.groups - cut(TvarRatio, seq(0.07, 0.11, 0.01))
m - table(Mutation_Status, TR.groups)
mut.no - dim(m)[1]
barplot(m, col=seq(mut.no), xlab=TvarRatio, ylab=Frequency)
legend(topleft, dimnames(m)[[1]], fill=seq(mut.no),
Dear All,
I have vn variable
vn
[1] V300 V376
What I want to get is
300 376
without V and from vn variable.
Could you help me about this issue?
Thank you,
Soyeon
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
On Wed, Aug 10, 2011 at 11:22 AM, Soyeon Kim yunni0...@gmail.com wrote:
Dear All,
I have vn variable
vn
[1] V300 V376
What I want to get is
300 376
as.numeric(substring(vn, 2))
HTH
Peter
__
R-help@r-project.org mailing list
or,
gsub('V','',vn)
On 8/10/2011 2:23 PM, Peter Langfelder wrote:
On Wed, Aug 10, 2011 at 11:22 AM, Soyeon Kimyunni0...@gmail.com wrote:
Dear All,
I have vn variable
vn
[1] V300 V376
What I want to get is
300 376
as.numeric(substring(vn, 2))
HTH
Peter
The see also potion of paste gives you the functions you can use for this
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Soyeon Kim
Sent: Wednesday, August 10, 2011 2:22 PM
To: r-help@r-project.org
Subject: [R] Opposite of
Duncan:
Yup, you're right. Can't assign, just print.
-- Bert
On Wed, Aug 10, 2011 at 11:02 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 10/08/2011 1:16 PM, Bert Gunter wrote:
Duncan et. al:
Inline below.
On Wed, Aug 10, 2011 at 9:48 AM, Duncan Murdoch
murdoch.dun...@gmail.com
Dear all
How does one convert a non-symmetric list to a vector? See below:
x - list()
x[[1]] - letters[1:5]
x[[2]] - letters[6:10]
x[[3]] - letters[11:12]
x
[[1]]
[1] a b c d e
[[2]]
[1] f g h i j
[[3]]
[1] k l
paste(x)
[1] c(\a\, \b\, \c\, \d\, \e\) c(\f\, \g\, \h\,
\i\, \j\)
[3] c(\k\,
Hello,
I am a R beginner and hoping to obtain some hints or suggestions about
using permutations to sort a data set I have.
Here is an example dataset:
Ind1 11 00 12 15 28
Ind2 21 33 22 67 52
Ind3 22 45 21 22 56
Ind4 11 25 74 77 42
Ind5 41 32 67 45 22
This will
unlist()
Michael Weylandt
On Wed, Aug 10, 2011 at 2:58 PM, Liviu Andronic landronim...@gmail.comwrote:
Dear all
How does one convert a non-symmetric list to a vector? See below:
x - list()
x[[1]] - letters[1:5]
x[[2]] - letters[6:10]
x[[3]] - letters[11:12]
x
[[1]]
[1] a b c d e
Hi,
For some reason I'm finding that my table caption is disappearing if I
print xtable output with the floating argument set to FALSE. Below is a
very simple Sweave file that produces two tables the first has no
caption and the second has a caption (if you want to see it
Check function unlist().
Best,
Dimitris
On 8/10/2011 8:58 PM, Liviu Andronic wrote:
Dear all
How does one convert a non-symmetric list to a vector? See below:
x- list()
x[[1]]- letters[1:5]
x[[2]]- letters[6:10]
x[[3]]- letters[11:12]
x
[[1]]
[1] a b c d e
[[2]]
[1] f g h i j
[[3]]
[1]
unlist(x)
r-help-boun...@r-project.org wrote on 08/10/2011 01:58:57 PM:
[image removed]
[R] convert 'list' to 'vector'?
Liviu Andronic
to:
r-help@r-project.org Help
08/10/2011 02:02 PM
Sent by:
r-help-boun...@r-project.org
Dear all
How does one convert a non-symmetric
To pick random elements to sample, you can just use the sample function
sample(1:5,3,replace=T/F) # pick true or false as needed for your data.
If you replicate this, you should have no problem.
replicate(100,function() return(sample(1:5,3,replace=T/F)))
This will be plenty fast, but if you
Hello all,
I am using the clustering functions in R in order to work with large
masses of binary time series data, however the clustering functions do not
seem able to fit this size of practical problem. Library 'hclust' is good
(though it may be sub par for this size of problem, thus doubly
On Wed, Aug 10, 2011 at 9:02 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
unlist()
Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.
Cheers
Liviu
__
R-help@r-project.org mailing list
Sorry, that second line of code won't work: do it in 2.
f - function() {return(sample(1:5,3,replace=T/F))}
replicate(100,f())
Michael
On Wed, Aug 10, 2011 at 3:06 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
To pick random elements to sample, you can just use the sample
On Aug 10, 2011, at 2:02 PM, Zev Ross wrote:
Hi,
For some reason I'm finding that my table caption is disappearing if I print
xtable output with the floating argument set to FALSE. Below is a very simple
Sweave file that produces two tables the first has no caption and the second
has a
On Aug 10, 2011, at 3:10 PM, Liviu Andronic wrote:
On Wed, Aug 10, 2011 at 9:02 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
unlist()
Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.
Well, since `unlist` is linked in the See Also on the
Here is code to transform the matrix that by() or array(split())
produces, along with an example of the speed of the various
approaches. Using split(), either directly or via by() or tapply(),
saves a lot of time.
f0 - function(df) {
# original code with typos fixed.
list_structure -
On Wed, Aug 10, 2011 at 9:32 PM, David Winsemius dwinsem...@comcast.net wrote:
Thanks all! This is perfect, and very R-ish: never where a novice
would expect it to be.
Well, since `unlist` is linked in the See Also on the help page for `list`,
I can only hope you meant that in complete jest.
dear R-experts---can someone please refer me to the latest
installation instructions for graphics fonts in R (the pdf device)?
(I would like to install the Charter font from the texlive 2011
distribution under OSX.)
sincerely,
/iaw
Ivo Welch (ivo.we...@gmail.com)
Try the flow cytometry clustering functions in Bioconductor.
-thomas
On Thu, Aug 11, 2011 at 7:07 AM, Ken Hutchison vicvoncas...@gmail.com wrote:
Hello all,
I am using the clustering functions in R in order to work with large
masses of binary time series data, however the clustering
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