cex works for me.
On Tue, 5 Sep 2006, Nair, Murlidharan T wrote:
Which is the parameter that is used to decrease the size of ylabs
plotted in biplot? I tried playing with cex and cex.lab I am not getting
it right
pc - princomp(USArrests)
biplot(pc, xlabs = rep(,
Dear list,
I'm trying to find out the following in an AffyBatch.
To get the indices from a loction on a chip I use the function xy2i() for
the hgu133plus2 by Affymetrix.
But now I want to know the name of the probe located at this spot. How can
this be done?
And what about the locations used as
Thanks a lot for your help.
tong
- Original Message -
From: Christos Hatzis [EMAIL PROTECTED]
Date: Monday, September 4, 2006 10:54 pm
Subject: RE: [R] Quick question about lm()
To: 'Tong Wang' [EMAIL PROTECTED], r-help@stat.math.ethz.ch
Say,
my.lm - lm(y ~ x, data=my.data)
Then
Yes, I am using R in windows, sorry for not being specific.
You are right, I have stored too much stuff, and I need to trash some data in
the process.
Problem solved for me, Thank you very much.
tong
- Original Message -
From: Prof Brian Ripley [EMAIL PROTECTED]
Date: Tuesday,
Hi,
Is there a function which operates on a matrix and return a vector of
min/max of each rol/col ?
say, X= 2, 1
3, 4
min.col(X)=c(2,1)
thanks a lot.
tong
__
R-help@stat.math.ethz.ch mailing list
On Tuesday 05 September 2006 20:32, David Reiss wrote:
I am trying to fit a GAM for a simple model, a simple model, y ~ s(x0) +
s(x1) ; with a constraint that the fitted smooth functions s(x0) and
s(x1) have to each always be 0.
From the library documentation and a search of the R-site
Tong
you need to use apply(). The second argument specifies whether
you want to work with rows or columns. The point of this is that
min() and max() operate on vectors and give a single value,
and you want to apply this function to all rows or all columns:
a - matrix(rnorm(30),5,6)
you could use something like:
# for row min and max
apply(X, 1, min)
apply(X, 1, max)
# for column min and max
apply(X, 2, min)
apply(X, 2, max)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Tong Wang a écrit :
Hi,
Is there a function which operates on a matrix and return a vector of
min/max of each rol/col ?
say, X= 2, 1
3, 4
min.col(X)=c(2,1)
thanks a lot.
tong
see ?pmin, which.min, which.max, max.col
hih
Dear R-users,
I'd like to announce the release of the new version of package 'ltm'
for analyzing multivariate dichotomous and polytomous data under the
Item Response Theory approach.
New features:
* function tpm() (along with supporting methods, i.e., anova, plot,
margins, factor.scores,
see ?apply
min.row - apply(X, 1, min)
min.col - apply(X, 2, min)
JeeBee
On Wed, 06 Sep 2006 01:37:22 -0700, Tong Wang wrote:
Hi,
Is there a function which operates on a matrix and return a vector of
min/max of each rol/col ?
say, X= 2, 1
3, 4
On Wed, 6 Sep 2006, Robin Hankin wrote:
Tong
you need to use apply(). The second argument specifies whether
you want to work with rows or columns. The point of this is that
min() and max() operate on vectors and give a single value,
and you want to apply this function to all rows or all
Hi,
THANK YOU ALL for the prompt reply.
cheers.
- Original Message -
From: Robin Hankin [EMAIL PROTECTED]
Date: Wednesday, September 6, 2006 1:42 am
Subject: Re: [R] What is the matrix version of min()
To: Tong Wang [EMAIL PROTECTED]
Cc: R help r-help@stat.math.ethz.ch
Tong
you
Usually the y-axis is shown on the left-hand-side of a graph, is it possible
to artifically creat one more y-axis on the right-hand-side in R? What is
the main reference? Thank you in advance.
[[alternative HTML version deleted]]
__
I intend to draw a plot of y against x. In the background of this graph I
wish to creat a histogram of the horizontal variable x. Does any expert know
how to produce such a plot?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch
Yes, there are many.
To give one example, you could consider using tcltk.
library(tcltk)
tkmessageBox(title=This is terrible,
message=What did you do?\nPromise not to do this again!,
icon=error, type=ok)
On Tue, 05 Sep 2006 23:10:41 +0200, Richard Müller wrote:
Hi all,
How about this?
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=78
JeeBee
On Wed, 06 Sep 2006 18:19:28 +0800, gallon li wrote:
I intend to draw a plot of y against x. In the background of this graph I
wish to creat a histogram of the horizontal variable x. Does any expert
Look at:
?axis (try the examples)
Further, a nice example I found on this mailing lists archive,
from somebody who says this has been asked many times already :)
x - 1:10
y1 - 1:10
y2 - rev(seq(1,1000, length=10))
plot(x,y1,ann=FALSE)
axis(2, at=c(2,4,6,8), labels=as.character(c(2,4,6,8)))
gallon li wrote:
I have found this one before. However, my intension is slightly differing
from this plot: I wish to plot the histogram in the backgroun instead of
in the margin. Thanks anyway!
On Wed, 06 Sep 2006 12:29:51 +0200, JeeBee wrote:
How about this?
-- Forwarded message --
From: gallon li [EMAIL PROTECTED]
Date: Sep 6, 2006 7:48 PM
Subject: Re: [R] plot axises on both sides of a graph
To: Jim Lemon [EMAIL PROTECTED]
Both of your suggestions are so helpful. By combining what you told me, now
I am able to produce a second
I'm using winDialog and winDialogString in scripts running on a XP-machine.
Since we're using some Linux-machines (Suse 10.0 and 10.1 on x86) I'm
Thanks to the respondents. Use of TCl/Tk is a fine idea, because you can use
the same scripts on Win and Linux OS.
Richard
--
Richard Müller - Am
Hi
I use the following code and it stores the results of density() in the
list dr:
dens - function(run) { density( positions$X[positions$run==run], bw=3,
cut=-2 ) }
dr - lapply(1:5, dens)
but the results are stored in dr[[i]] and not dr[i], i.e. plot(dr[[1]])
works, but plot([1]) doesn't.
Is
See these two examples.
plot(1:2)
axis(4)
mtext(right y axis, side=4, line=-1.5)
par(mar=c(5,4,4,5)+.1)
plot(1:2)
axis(4)
mtext(right y axis, side=4, line=3)
Good luck finding the right combination again ;)
On Wed, 06 Sep 2006 20:08:43 +0800, gallon li wrote:
-- Forwarded message
--- JeeBee [EMAIL PROTECTED] wrote:
gallon li wrote:
I have found this one before. However, my intension
is slightly differing
from this plot: I wish to plot the histogram in the
backgroun instead of
in the margin. Thanks anyway!
On Wed, 06 Sep 2006 12:29:51 +0200, JeeBee wrote:
Hi Gunther,
Gunther Höning wrote:
Dear list,
I'm trying to find out the following in an AffyBatch.
This question is related specifically to a Bioconductor package, so
should be asked on the bioconductor listserv rather than R-help.
Best,
Jim
To get the indices from a loction on a
I intend to draw a plot of y against x. In the background of this graph I
wish to creat a histogram of the horizontal variable x. Does any expert know
how to produce such a plot?
When constructing such a plot, you need to be careful that you don't
end up constructing a pretty picture instead
Use 'sapply' instead of 'lapply'. Type
?lapply
for details
Antonio, Fabio Di Narzo.
University of Bologna, Italy
2006/9/6, Rainer M Krug [EMAIL PROTECTED]:
Hi
I use the following code and it stores the results of density() in the
list dr:
dens - function(run) { density(
Antonio, Fabio Di Narzo wrote:
Use 'sapply' instead of 'lapply'. Type
If I use sapply it seems to simplify / collapse to much.
?lapply
for details
Antonio, Fabio Di Narzo.
University of Bologna, Italy
2006/9/6, Rainer M Krug [EMAIL PROTECTED]:
Hi
I use the following code and it
Dear all,
I am a newbie in R and need some help please. (I do apologise if my
email is not as informative as it should be, I've tried to include
the relevant details without overcrowding it with the rest of the code)
I would like to sample (without replacement) Y objects based on the
Very sorry if this mail is sent out twice to the list, I wasn't sure
if the email address I used in the first go was correct.
From: [EMAIL PROTECTED]
Subject:how to loop through 2 lists with different indexes
Date: 6 September 2006 15:16:21 BDT
To find what names are common to both, use 'intersect'
x.bin.size - list('3'=1, '4'=4, '5'=10)
y.bin.size - list('2'=4, '3'=42, '4'=253, '5'=954)
sameNames - intersect(names(x.bin.size), names(y.bin.size))
sameNames
[1] 3 4 5
lapply(sameNames, function(x) sample(seq(y.bin.size[[x]]),
Hello everybody,
I need your help about the package SN and the skew student distribution. Il
will be very grateful if I have the solution.
I construct a stochastic model with a white noise not gaussian but following a
skew student distribution. I fit the noise on monthly data to obtain the
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have
Dear all,
I arrive to do density plots using the function kde2d , and from this do
a countour plot. My problem is that I do not really understand what the
labels for the different levels mean??? What I would like to obtain is a
surface encompassing the 95 percentile of my values. In other
See ?sweep
sweep(a, 2, a[1,],/)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Wednesday, September 06, 2006 11:49 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Matrix multiplication using apply() or lappy() ?
I am
All,
I have a repeated measures data frame and was wondering if the
covariance matrix can be
calculated via some created indexing or built-in R function.
Specifically, say there are 3 variables, where potassium concentration
is measured 6 times on each patient.
Patient number (discrete)
Time
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in the
try the following:
dat - data.frame(id = rep(1:100, each = 6), time = rep(1:6, 100), pot
= rnorm(600))
# for a balanced data-set
mat - matrix(dat$pot, ncol = 6, byrow = TRUE)
cor(mat)
# for a unbalanced data-set
dat - dat[-sample(600, 100), ]
mat - t(sapply(split(dat, dat$id), function(x){
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On
On Wed, 6 Sep 2006, Christos Hatzis wrote:
See ?sweep
sweep(a, 2, a[1,],/)
That is less efficient than
a/rep(a[1,], each=nrow(a))
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Wednesday, September
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a -
Yet another one using the idempotent apply in reshape package
that eliminates the transpose:
library(reshape)
iapply(a, 1, /, a[1,])
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor
Dear all,
I get an error message when I run my model and I am not sure what to do
about it.
I try to determine what factors influence the survival of voles. I use a
mixed-model because I have several voles per site (varying from 2 to 19
voles).
Here is the model:
###
fm5 -lmer(data=cdrgsaou2,
Prof. Brian Ripley wrote:
On Wed, 6 Sep 2006, Christos Hatzis wrote:
See ?sweep
sweep(a, 2, a[1,],/)
That is less efficient than
a/rep(a[1,], each=nrow(a))
*My* first instinct was to use
t(t(a)/a[1,])
(which has not heretofore been suggested).
This seems to be more
The apply was exactly what I was after. And, I will check out the others
as well. great tips!
Gabor Grothendieck [EMAIL PROTECTED]
09/06/2006 11:11 AM
To
[EMAIL PROTECTED] [EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] Matrix multiplication using apply() or lappy() ?
In terms of speed Toby's original idea was actually the fastest.
Here they are decreasing order of the largest timing in each
row of system.time. I also tried it with a 100x10 matrix and
got almost the same order:
library(reshape)
system.time(for(i in 1:1000) iapply(a, 1, /, a[1,]))
[1] 11.51
What version of R was this?
In 2.4.0 alpha
a - matrix(1:24,4)
system.time(for(i in 1:1000) junk - a / rep(a[1,], each = 4))
[1] 0.014 0.000 0.014 0.000 0.000
system.time(for(i in 1:1000) junk - t(t(a)/a[1,]))
[1] 0.057 0.000 0.058 0.000 0.000
shows a large margin the other way, which
Are there any functions available to do a factor analysis with
fewer observations than variables? As long as you have more than 3
observations, my computations suggest you have enough data to estimate a
factor analysis covariance matrix, even though the sample covariance
matrix is
I suspect you are not thinking about the list and the
subsetting/extraction operators in the right way.
A list contains a number of components.
To get a subset of the list, use the '[' operator. The subset can
contain zero or more components of the list, and it is a list itself.
So, if x is
I know this has got to be simple, but I have a added an arrow to a graph
with:
arrows(5,8,8, predict(lmfit,data.frame(x=8)), length=0.1)
but its in the wrong position, correcting it and running again adds an new
arrow (which is what you would expect) so how do I
a) edit the existing arrow, and
On 9/6/2006 3:04 PM, Graham Smith wrote:
I know this has got to be simple, but I have a added an arrow to a graph
with:
arrows(5,8,8, predict(lmfit,data.frame(x=8)), length=0.1)
but its in the wrong position, correcting it and running again adds an new
arrow (which is what you would
This does not actually remove it but you could overwrite it with
an arrow the same color as the background and then plot a
new arrow:
x - 1:10
plot(x ~ x)
arrows(1, 1, 2, 2)
# revise it
arrows(1, 1, 2, 2, col = white)
arrows(2, 2, 3, 3)
On 9/6/06, Graham Smith [EMAIL PROTECTED] wrote:
I know
I literally copy/pasted your second version (without the + signs) in a file
and the sourced that file. It loaded without error and the a matrix was
as I expected.
On 9/6/06, Evan Cooch [EMAIL PROTECTED] wrote:
When I have to enter a very large matrix into the R console, I can make
use of the
Are your sure your second solution does not work? Try again...
Evan Cooch wrote:
When I have to enter a very large matrix into the R console, I can make
use of the continuation feature in the console to enter the matrix in
pieces (e.g., on a row by row basis). So, for example, the console
pierre clauss pierreclauss at yahoo.fr writes:
Hello everybody,
I need your help about the package SN and the skew student distribution. Il
will be very grateful if I have the solution.
I construct a stochastic model with a white noise not gaussian but following a
skew student
Hi,
I'm very new to R, and am not at all a software
programmer of any sort.I appreciate any help you
may have. I have figured out how to get my data into
a dataframe and order it alphabetically according to a
particular column. Now, I would like to seperate out
certain rows based on
Try using 'grep' and regular expressions:
x - 72 5S_F_1501 567
+ 7700 5S_F_2338 611
+ 7517 5S_F_3412 467
+ 10687 5S_F_4380 428
+ 4870 5S_F_5315 368
+ 6035 5S_F_6300
Could you try this model fit again adding control = list(usePQL =
FALSE, msVerbose=TRUE) to the argument list of the call to lmer? By
default PQL iterations are used at the beginning of a generalized
linear mixed model fit followed by optimization of the Laplace
approximation to the
On 9/4/06, Toby Gardner [EMAIL PROTECTED] wrote:
Dear list,
I am having some problems with extracting Variance Components from a
random-effects model:
I am running a simple random-effects model using lme:
model-lme(y~1,random=~1|groupA/groupB)
which returns the output for the StdDev of
I would like to join repeated measures for patients across two visits using
a line. The program below uses symbols to represent each patient. Basically,
I would like to join each pair of symbols.
library(lattice)
patient - c(1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9)
var -
Make each pair of points a separate group using group= and specify
that both points and lines be used via type = b. Also set the
symbols in par.settings= so that they are accessed by both
the main plot and the legend:
xyplot(var ~ visit, group = symbols[patient], type = b,
auto.key =
Just one correction (although in this case it does not change the
output) -- use group = patient rather than group = symbol[patient]:
xyplot(var ~ visit, group = patient, type = b, auto.key = list(space
= right),
par.settings = list(superpose.symbol = list(pch = symbols)))
On 9/6/06, Gabor
Hello,
I'm fitting poisson regression to mortality data and wish to stratify by age.
Is there any way to perform this stratification and use the glm function in R?
Thanks,
Hannah Murdoch
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