Please update your R (and probably your RGtk2: you did not tell us its
version), as the posting guide asked you to do before posting.
On Sun, 23 Oct 2011, Aref Nammari wrote:
Hello,
I hope this is the right place to ask for help with a problem I am
having with RGtk2 installation with R on
Hi All:
If I download yahoo data by getSymbols() in R, the date column gets
accompanied along with the downloaded data. There is no column header for
the date column to access separately.
What is the way to eliminate the date column?
If I want to draw a xy scatter plot with the downloaded price
Dear list members,
what is the reason that one obviously can't do arithmetic operations on
zoo members with different index positions?
require(zoo)
z - zoo(c(1,1,1),order.by=c(1,2,3))
z
1 2 3
1 1 1
z[1] + z[1]
1
2
z[1:2] + z[1:2]
1 2
2 2
z[1] + z[2]
n - 10
P1 - runif(n)
P2 - runif(n)
P3 - P1 + P2 + runif(n)/100
P4 - P1 + P2 + P3 + runif(n)/100
mydata - data.frame(cbind(P1,P2,P3,P4))
mydata[1,1] - 8
mydata[3,1] - -5
mydata[2,3] - -6
mydata[7,3] - 7
f=function(z){quantile(z, c(0.01, 0.99)) }
temp1 - lapply(mydata, f)
temp1
$P1
1%
Dear All,
I am facing a problem in calling an user defined R function from Java
through JRI. The user defined R function does a loess normalization on micro
array data ( find in the limma package of BioConductor) and the last 2 lines
of the R Code is :
MA - normalizeWithinArrays(RG,
Hi
how do we get intercepts standard error. I'm using the package pls.
i got the coefficient but not able to get the stabdard error
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-get-intecerpt-standard-error-in-PLS-tp3932104p3932104.html
Sent from the R help mailing
On Oct 23, 2011, at 9:35 PM, Hugo Mildenberger wrote:
Dear list members,
what is the reason that one obviously can't do arithmetic operations
on
zoo members with different index positions?
zoo-objects are designed to merged by their indices before applying
arithmetic operations.
You
The date column is very unlikely to interfere with your plots if you don't
specify it. I suggest you follow the instructions at the bottom of this and
every other message on this mailing list. (Hint: reproducible code and identify
your software platform.)
On Mon, 24 Oct 2011, Hugo Mildenberger wrote:
Dear list members,
what is the reason that one obviously can't do arithmetic operations on
zoo members with different index positions?
It's a _feature_ that zoo first matches the index positions, uses only
their intersection, and then performs
Insufficient problem specification. Read the posting guide and try again with
reproducible code and platform identification.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live
I would like to add a legend to my Pie chart but I don't seem to get this
working:
Below is my code, this is done on Rpy2.
grdevices = importr('grDevices')
grdevices.png(file=FileNameLocation, width=1400, height=1000)
Robjects.r.par(mar=[1, 0, 2, 12], cex=1.0)
Am 24.10.2011 09:07, schrieb Jeff Newmiller:
Insufficient problem specification. Read the posting guide and try again with
reproducible code and platform identification.
---
Jeff Newmiller The . . Go Live...
Hi
This is my data present in a file
Year Y X2 X3 X4 X5 X6
1960 27.8 397.5 42.2 50.7 78.3 65.8
1960 29.9 413.3 38.1 52 79.2 66.9
1961 29.8 439.2 40.3 54 79.2 67.8
1961 30.8 459.7 39.5 55.3 79.2 69.6
1962 31.2 492.9 37.3 54.7 77.4 68.7
my formula for using PLS
fit1 - mvr(formula=Y~X1+X2+X3+X4,
Hello,
Suppose I have the dataset shown below. The amount of observations is too
massive to get a nice geom_point and smoother on top. What I would like to do
is to bin the data first. The data is indexed by Time (minutes from 1 to 120
i.e. two hours of System benchmarking).
Option 1) group
On Thu, Oct 20, 2011 at 12:22 AM, Norm Matloff matl...@cs.ucdavis.eduwrote:
I've developed a new R debugging tool, debugR, available at
http://heather.cs.ucdavis.edu/debugR.html
This basically replaces my edtdbg, which I will no longer be supporting.
The new tool is now decoupled from one's
Hi,
IS is possible to align a chart to the left of the plot area as to make
space for a legend?
--
Regards/Groete/Mit freundlichen GrüÃen/recuerdos/meilleures salutations/
distinti saluti/siong/duì yú/пÑивеÑ
Jurgens de Bruin
[[alternative HTML version deleted]]
On 10/24/2011 12:35 AM, Philipp Fischer wrote:
Hello,
I am just starting with R and I am having a (most probably) stupid problem by
creating a new variable in a data.frame based on a part of another character
variable.
I have a data frame like this one:
A B
See ?par and its mar argument.
Uwe Ligges
On 24.10.2011 11:15, Jurgens de Bruin wrote:
Hi,
IS is possible to align a chart to the left of the plot area as to make
space for a legend?
__
R-help@r-project.org mailing list
On Oct 24, 2011, at 11:15 , Jurgens de Bruin wrote:
Hi,
IS is possible to align a chart to the left of the plot area as to make
space for a legend?
Various possibilities, depending on what you mean precisely:
You can create a larger right hand margin. See the mar= argument under
Yes.
If you want a more detailed answer, follow the posting guide instructions and
provide a reproducible example. There are three plot systems in R, each with
their own approaches to solving this problem.
---
Jeff
Hi,
I am writing a program to plot the correlation coefficients of 45 elements
using the commend plot. Error message Error in plot.new() : figure
margins too large appeared. I think it may because there are too many
elements. But I really need to find their correlation coefficients. What can
I
On 23.10.2011 22:33, Ali Tofigh wrote:
Hi,
When I plot text and use cex to change the text size, I notice that the cex
multiplier is not exact. It looks as if the real size of text can take only
certain discrete values. Is there a workaround to get text to follow the cex
value more closely,
On 24.10.2011 11:35, Maggie Wong wrote:
Hi,
I am writing a program to plot the correlation coefficients of 45 elements
using the commend plot. Error message Error in plot.new() : figure
margins too large appeared. I think it may because there are too many
elements. But I really need to find
On Mon, 24 Oct 2011, Uwe Ligges wrote:
On 23.10.2011 22:33, Ali Tofigh wrote:
Hi,
When I plot text and use cex to change the text size, I notice that the cex
multiplier is not exact. It looks as if the real size of text can take only
certain discrete values. Is there a workaround to get
Hi,
Is there any way or function in R that can find the Connecting
Dominating Set?
Regards,
Amir
--
___
Amir Darehshoorzadeh |Comp. Architecture Dept.
PhD Student |UPC-Campus Nord, C6-221
Email: a...@ac.upc.edu
Good Afternoon,
I am inexperienced in data visualization R. so I wonder if someone can help
me.
I am using the generic database mtcars.
I would like to change the chart plot, instead of appearing the name of the
medicine, I wanted a symbol in the chart drugX arise for example in a yellow
Hi.
I am trying to analyze with BRugs the Box-Tiao variance components example
in WinBUGS. The output from BRugs,
mean sd MC_error val2.5pc median val97.5pc start sample
sigma2.btw 681.9 116110.890.7016 253.8 4232 25001 10
sigma2.with 4266.0 1246 4.92
Many thanks for your replies. I appreciate that.
I tried what you suggested and it did work for the Poisson model (glm,
poisson familly). Unfortunately, the negative binomial (glm.nb) did not
work as I work the following message:
Warning messages:
1: In ifelse(y mu, d.res, -d.res) :
Reached
I'm not implying they should be discarded; however, at the same time I'm not
certain I fully understand why we should check the ordinality assumption if
in the end we're going to include predictors with which the response
variable behaves in a non-ordinal fashion.
--
View this message in context:
I am facing a problem with a function in survey package. The function svyvar
gives the estimated population variance from a given sampling scheme. I am
working with a data having more than four continuous variables. In order to
have have population total for all those cont. variables I have
Hello
I am a new user of R and I need help to merge two large datasets about
stocks with different number of rows and columns.
Both have 2 variable(column) that are have same values (name and date1),
but they are not in same order and data3 contains much more observations.
In the data have
here is some more info:
http://r.789695.n4.nabble.com/file/n3932898/help.jpg
this is what`s happend when I run the function in R
I appreciate if someone can give me some inputs here
Thanks in advance
--
View this message in context:
A few issues -
Don't let the overall unimportance of a predictor make you worry about
non-ordinality (e.g., when scale of plot.xmean.ordinaly has a low range on
the y-axis).
We frequently have to face the issue of using an imperfect model by fitting
a few variables that don't exactly meet our
Hi:
On Mon, Oct 24, 2011 at 2:01 AM, Giovanni Azua brave...@gmail.com wrote:
Hello,
Suppose I have the dataset shown below. The amount of observations is too
massive to get a nice geom_point and smoother on top. What I would like to do
is to bin the data first. The data is indexed by Time
Thank you. This works pretty well. I am having some trouble with the text on
the left-hand side getting cut off. I have tried haling=center with not
luck. How can I make the text seem wider than it really is to avoid the
truncation. It is only a few characters but still it is annoying. Thank you.
I get the following error trying to install the package:
install.packages(grid)
--- Please select a CRAN mirror for use in this session ---
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
package grid is not available (for R version 2.13.2)
-Original
grid is an R base package that always ships with R and is not
distributed in the repository of contributed packages.
Uwe Ligges
On 24.10.2011 14:50, Kevin Burton wrote:
I get the following error trying to install the package:
install.packages(grid)
--- Please select a CRAN mirror for use
Hi
If you want to get rid of regular expressions at all and your A values
start AWI for Arctic and UFT for boreal you can
DF$D - ifelse(substr(DF$A, 1,1) == A, Arctic, Boreal)
Regards
Petr
Hello,
I am just starting with R and I am having a (most probably) stupid
problem
by creating a
Hi
Hello
I am a new user of R and I need help to merge two large datasets about
stocks with different number of rows and columns.
Both have 2 variable(column) that are have same values (name and
date1),
but they are not in same order and data3 contains much more
observations.
In the
Hi Kevin,
this should be read as halign=center - so is it a typo just in your
mail or in your program as well?
apart from that, the rules at the bottom lines of every post on this
list also apply here: what have you tried and what went wrong?
Cheers
Am 24.10.2011 14:48, schrieb Kevin Burton:
I would like to split a string into words at its blanks but also to preserve
all blanks.
Example:
c( somewords to split )
should become
c( , some,, words, , to , , split, )
I was not able to achieve this via strsplit() .
But I am not familiar with regular
On Mon, Oct 24, 2011 at 9:46 AM, Mark Heckmann mark.heckm...@gmx.de wrote:
I would like to split a string into words at its blanks but also to preserve
all blanks.
Example:
c( some words to split )
should become
c( , some, , words, , to , , split, )
I was not able
Hi Mark,
here is a way using gsub to insert a split marker and strsplit.
strsplit(gsub(([[:alnum:]]+),|\\1|,c( somewords to split ))[[1]]
cheers
Am 24.10.2011 15:46, schrieb Mark Heckmann:
I would like to split a string into words at its blanks but also to preserve
all blanks.
If you don't mind using a loop, e.g. like this:
x - mydata
qs - sapply(x, function(x) quantile(x, c(0.01, 0.99)))
for (i in 1:ncol(x))
x - subset(x, x[ ,i] = qs[1,i] x[ ,i] = qs[2,i])
--
View this message in context:
On Sun, 2011-10-23 at 12:00 +0200, r-help-requ...@r-project.org wrote:
The results by the survfit routine do not agree with the
results of
these formulae as obtained by SAS.
The next question should be is SAS correct. The answer in this case
is no.
For survival data the mean
Hi,
Is there any way to find the Connected Dominating Set (CDS) of a graph in R?
Regards,
Amir
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Thank you Petr!
I have read on the merge help page, but I cant figure out how to write this
function.
When I use your function it includes all data from data3, but all columns
in data has NA(without name and date. I hoped to keep these values
to.
I try to explain it more precise:
In data with
One last thing. At the outset of this discussion I provided the results of a
validation procedure on a model (see below). As discussed previously, the
model overall seems to fair well, with the exception of the slope. With
that in mind, is there a way to correct the coefficients of the model to
Dear experts,
I am trying to create a data frame from the residuals I get after
having applied a linear regression to each column of a data frame, but
I don't know how to create this data frame from the resulting list
since the list has differing numbers of rows.
So for example:
age-
Comments inline.
On Oct 24, 2011, at 1:43 AM, ajc arkos...@gmail.com wrote:
Hi All:
If I download yahoo data by getSymbols() in R,
from the quantmod package
the date column gets
accompanied along with the downloaded data. There is no column header for
the date column to access
Hello Terri,
Thank you very much. This is the answer I needed. Could you also
tell me how I can calculate 25 and 50% quantiles in R? I can only get median
as far as I know.
Cem
Cem Girit
-Original Message-
From: Terry Therneau [mailto:thern...@mayo.edu]
Sent: Monday,
Hello,
I am trying to do a nonlinear model using the nls command in R software.
The data I am using is as follows:
A-c(7.132000,8.668667,9.880667,8.168000,10.86,10.381333,11.059333,7.589333,4.716667,4.268667,7.265333,10.309333,8.456667,13.359333,8.624000,13.571333,12.52,4.084667
,NaN,NaN)
This is very basic but I have not been able to find an answer. Basically I
want to find the length of a string.
length(Text)
returns 1 so I know that is not right.
Thank you.
Kevin
__
R-help@r-project.org mailing list
I was able to solve it by supplying the 'mar' argument. Even when I spell it
right this argument did not solve it.
Thanks again.
Kevin
-Original Message-
From: Eik Vettorazzi [mailto:e.vettora...@uke.de]
Sent: Monday, October 24, 2011 8:46 AM
To: Kevin Burton
Cc: r-help@r-project.org
On Mon, Oct 24, 2011 at 4:21 AM, amitava amtv.statpr...@gmail.com wrote:
I am facing a problem with a function in survey package. The function svyvar
gives the estimated population variance from a given sampling scheme. I am
working with a data having more than four continuous variables. In
Hello (Heinrich),
I did not know I could do this. It doesn't seem to be documented anywhere.
Thought this would be helpful to the fraction of the community using package
R.oo. Note the call of a setMethodS3 method, xOne, in the setConstructorS3.
This is extremely useful if xOne (in this case) is
On Oct 24, 2011, at 11:14 AM, Kevin Burton wrote:
This is very basic but I have not been able to find an answer.
Basically I
want to find the length of a string.
length(Text)
?nchar
returns 1 so I know that is not right.
Thank you.
Kevin
Hello,
I am trying to get a linear model of y ~ log(x).
* lm (y~log(x))*
However, I always get an error report:
/Error in model.frame.default(formula = y ~ log(x), drop.unused.levels =
TRUE) :
variable lengths differ (found for 'log(x)')/
*Here was my y:*
y
[1]0.4500.050
Hi, I am new to R so I would appreciate any help. I have some data that has
passenger flight data between city pairs. The way I got the data, there are
multiple rows of data for each city pair; the number of passengers needs to
be summed to get a TOTAL annual passenger count for each city pair.
Dear Kevin,
# this works
nchar(read the help for length”)
[1] 24
On 24 Oct 2011, at 4:14 PM, Kevin Burton wrote:
This is very basic but I have not been able to find an answer. Basically I
want to find the length of a string.
length(Text)
returns 1 so I know that is not right.
Thank
__
R-help@r-project.org 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.
On Fri, 21 Oct 2011, David Winsemius wrote:
The first thing I would try would be
with(subset(chemdata, param %in% c('TDS', 'Cond', 'Mg', 'SO4', 'Cl', 'Na',
and 'Ca') , 1:4) ,
xtabs(quant ~ site + sampdate + param) )
David,
Need to remove the 'and' from the above.
The results
Mitra
You have a couple of problems. First, if you plot the data (plot(A~Pot))
you'll see there's little chance that an asymptotic, exponential
equation describes your data because they don't rise from zero to Amax
as your model assumes. If Pot is your only independent variable, the
best
On Oct 24, 2011, at 11:34 AM, Rich Shepard wrote:
On Fri, 21 Oct 2011, David Winsemius wrote:
The first thing I would try would be
with(subset(chemdata, param %in% c('TDS', 'Cond', 'Mg', 'SO4',
'Cl', 'Na', and 'Ca') , 1:4) ,
xtabs(quant ~ site + sampdate + param) )
David,
Need to
You are trying to regress ~372 observations of the dependent against
~4 observations of the independent variable. Ask yourself again if
this makes sense.
A further hint might be given by this
y = rnorm(5); x = y[1:4]
lm(y~x)
Michael
On Mon, Oct 24, 2011 at 11:13 AM, Julie
Hi all,
I have been trying to run a bestglm in R for a while now and am struggling
to get it to run. When I thought I had succeeded, the output it gave me
was NULL and that's it. Below is my code:
bestglmtest-read.table(C:\\Documents and
It would be good to follow the posting guide and at least supply a
sample of the data.
Most likely 'tapply' is one way of doing it:
tapply(df$passenger, list(df$orig, df$dest), sum)
On Mon, Oct 24, 2011 at 11:27 AM, asindc siiri...@eastwestcenter.org wrote:
Hi, I am new to R so I would
Hi all
I'm trying to use the 'by' function to extract the co-efficients from a
mixed model which is performed on multiple individuals. I basically have a
group of individuals and for each individual I want the co-efficient for
there change in 'pots_hauled' in response to a change in 'vpue' with
how do I code the following in R. I want to produce a vector where dx=log(
(d(x))/(d(x-1)) ). I can do it for dx=diff(log(x)). I am learning/trying to
model log returns of a stock market index. But instead of using the
difference of the closing values of two consecutive days, i want to use the
log
... Well, this works in this simple case, but is too clumsy for a general
formulation of this problem: given a dictionary consisting of two
character vectors of unique names (or two columns in a data frame), x and
y, how does one convert a factor z with levels in x into the corresponding
See the count() function in the plyr package; it does fast summation.
Something like
library('plyr')
count(passengerData, c('ORIGIN_WAC', 'DEST_WAC'), 'npassengers')
HTH,
Dennis
On Mon, Oct 24, 2011 at 8:27 AM, asindc siiri...@eastwestcenter.org wrote:
Hi, I am new to R so I would appreciate
Thanks so much, this is very very helpful.
I do have one remaining question here. I definitely see the value of
making a list of the datasets, an advise I will definitely follow.
However, for educational purposes, I would still like to know how to
automate the following without using a list:
On Mon, 24 Oct 2011, David Winsemius wrote:
The appearance of levels with all zeroes is probably because I didn't include
drop.unused.levels = FALSE in the xtabs specification.
OK. Adding 'drop.unused.levels' does make a huge difference.
Thanks,
Rich
Assuming that d(x) is equal to x, (I don't know a d() function in R)
these should be the same.
log(a/b) = log(a) - log(b) = diff(log(c(a,b))
If you mean simple returns instead of continuous/log returns, perhaps try this:
x[-1]/x[-length(x)] - 1
Michael
On Mon, Oct 24, 2011 at 11:44 AM,
The merge function seems perfect for your problem. If you can't get it to work,
the problem may be in the data you are working with, which you have not
supplied (read the posting guide, and use the head and dput functions to make
your example small and reproducible).
One common mistake by
Write a function that encapsulates the following three lines:
city1997- dataCleaning(read.csv2(C:\\city\\year1997.txt))
city1997- wasteCalculations(city1997, year = 1997)
if (city1997[1,1] == Time) {city1997- timeCalculations(city1997)}
and then pass in the appropriate parameters.
On Mon, Oct
Hi:
bestglmtest is your input data frame, is it not? From the names() line,
you can see that it has no variable named BestModel that corresponds
to a list containing a component named coefficients.
Were you perhaps looking for
output$BestModel$coefficients ??
Dennis
On Mon, Oct 24, 2011 at
X and y must have the same number of elements, and NA values must be removed
(?na.omit)
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
On Mon, Oct 24, 2011 at 06:10, Prof Brian Ripley rip...@stats.ox.ac.uk wrote:
On Mon, 24 Oct 2011, Uwe Ligges wrote:
On 23.10.2011 22:33, Ali Tofigh wrote:
Hi,
When I plot text and use cex to change the text size, I notice that the
cex
multiplier is not exact. It looks as if the real
You might also need the assign() function which is sort of the opposite of get()
Michael
On Mon, Oct 24, 2011 at 12:15 PM, jim holtman jholt...@gmail.com wrote:
Write a function that encapsulates the following three lines:
city1997- dataCleaning(read.csv2(C:\\city\\year1997.txt))
city1997-
On Oct 24, 2011, at 12:10 PM, Rich Shepard wrote:
On Mon, 24 Oct 2011, David Winsemius wrote:
The appearance of levels with all zeroes is probably because I
didn't include drop.unused.levels = FALSE in the xtabs specification.
OK. Adding 'drop.unused.levels' does make a huge difference.
tom_pienkowski tom.pienkowski at blueyonder.co.uk writes:
I'm trying to use the 'by' function to extract the co-efficients from a
mixed model which is performed on multiple individuals. I basically have a
group of individuals and for each individual I want the co-efficient for
there change in
On Mon, 24 Oct 2011, David Winsemius wrote:
You could also have saved the subsetted data, applied `factor` to the
subsetted column and then used `xtabs`.
temp - subset(chemdata, your subset criteria, your column selection)
temp$param - factor(temp$param)
(Now only levels that exist are in the
Try it yourself:
x = seq(1, 11, by = 2)
diff(log(x))
log(x[-1]/x[-length(x)])
all.equal(diff(log(x)), log(x[-1]/x[-length(x)]))
It seems like you don't really understand logs / log returns and why
they are used by some in quant finance: might I suggest you read this:
Hi All,
Its a bit of a beginners question I'm afraid.
I have a looped stepwise regression (using MASS and StepAIC) to take random
predictors out of the total number. For this example a random sample of 5
out of a total of 20. The loop will continue until all combinations of
variables have been
Sorry, I attempted to paste the sample data but it must have been stripped
out when I posted. It is hopefully now listed below.
tapply looks useful. I will check it out further.
Here's the sample data:
flights[1:10,]
PASSENGERS DISTANCE ORIGIN ORIGIN_CITY_NAME ORIGIN_WAC DEST
The variable y is made of four columns, each paired to 20, 200, 2000 or 20
000.
y - c(rdiktator20, rDiktator200, rDikt2000, rDikt2)
So I guess the problem is in the fact that I did not specify it correctly,
is it so? How can I tell R properly that one part of y matches to one part
of x?
The
I am new to R coding and I am trying to model the returns on the ftse100
since 1990. I have got a vector with all the closing values on each trading
day. however, instead of using the difference in the closing values of two
consecutive days, (ie dx=diff(x) where x is the vector containing the
The count() function in the plyr package works beautifully. Thanks to Jim,
Rainer and Dennis for your help.
Best.
-Original Message-
From: Dennis Murphy [mailto:djmu...@gmail.com]
Sent: Monday, October 24, 2011 12:05 PM
To: asindc
Cc: r-help@r-project.org
Subject: Re: [R] How to
Hello all,
I'm trying to make a histogram of the data contained in my dataframe.
The summary of the data gives me exactly what I want
summary (Data)
V1 V2
first001: 3 last001: 9
first002: 3 last002: 7
first003: 2 last003: 6
(Other) :52(Other): 27
But how do I capture the names and
Hello all,
I have the following. Two sets of p x 3 matrices where p is say relatively
large.
My data looks like :
mat1
2 3 4
2 3 4
1 2 3
mat2
2 3 4
12 12 4
10 12 3
when p = 3.
I form a 2 x 3 contingency table using the i^th row from each matrix.
Its Fisher's exact
What about?
hist(Data$V1)
hist(Data$V2)
Cheers,
Josh
On Mon, Oct 24, 2011 at 9:38 AM, Joe Carl joseph.w.carl...@gmail.com wrote:
Hello all,
I'm trying to make a histogram of the data contained in my dataframe.
The summary of the data gives me exactly what I want
summary (Data)
V1
It surely can be done. One way is to keep track of selected variables
in a set. If a new variable is selected, you expand the selected set
and set the frequency to be one, otherwise just increase the freqency
of the selected variable (if... else).
Also, you might want to have a look at glmulti
Hi,
I'm trying to follow the suggestions given by Deepayan Sarkar in this
message:
http://tolstoy.newcastle.edu.au/R/help/05/11/16135.html
to plot 3-D points on a wireframe plot. The problem is that I keep getting a
partly formed plot- with the colored lattice visible but no axis labels or
OK, what is the trick to extracting the overall p value from an lm object?
It shows up in the summary(lm(model)) output but I can't seem to extract it:
test2 = apply(aa, 1, function(x) summary(lm(x[,1] ~ 0 + x[,3] + x[,6])))
test2[[1]]
Call:
lm(formula = x[, 1] ~ 0 + x[, 3] + x[, 6])
[omitted
Hi Jim,
Its a bit of a trick question. There isn't actually any overall p
value stored in the lm object, or even in the summary.lm object. It
is calculated from the f statistic by the print methods for
summary.lm. Of course, none of that helps, per se. Try this:
summary(lm(mpg ~ hp, data =
It's not directly extractable since it's calculated on the fly in the
printing method. If you type stats:::print.summary.lm, you can see
the code the leads to the calculation: It's basically (I'm leaving out
some formatting stuff):
pf(x$fstatistic[1L], x$fstatistic[2L], x$fstatistic[3L],
On 24/10/2011 1:47 PM, Jim Bouldin wrote:
OK, what is the trick to extracting the overall p value from an lm object?
It shows up in the summary(lm(model)) output but I can't seem to extract it:
It's not part of the object, it is computed when the object is printed.
To see the print method,
Hi:
Aren't V1 and V2 factors in this data frame? If so, you should be
plotting bar charts rather than histograms (they're not the same). Do
you want separate graphs for V1 and V2, do you want them stacked, or
do you want them dodged (side-by side for each level 001, 002, 003)?
In the absence of
I see the problem, I fixed this bug for version 2.8 of TeachingDemos, but have
not submitted the new version to CRAN yet (I thought that I had fixed this
earlier, but apparently it is still only in the development version). An easy
fix is to install version 2.8 from R-forge
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