Hi, Gundala
pretty much u can do is adding them directly , that is , treat them
like two numbers instead of matrix.
z - (x+y)/2 .
Is that what u want ? or u have another meaning.
regards
On 2008-6-25, at 上午10:19, Gundala Viswanath wrote:
Hi,
I have two matrices x and y (same
Hi all,
A while back I sent a message concerning working with tiff-files, and
mentioned that I encountered a bug in 2.7.0.
This bug still occurs in 2.7.1, and is reproducable on a separate
computer (both running WinXP professional):
tiff()
plot(1:1000)
dev.off()
This causes R to show the window
Gustaf Rydevik wrote:
Hi all,
A while back I sent a message concerning working with tiff-files, and
mentioned that I encountered a bug in 2.7.0.
This bug still occurs in 2.7.1, and is reproducable on a separate
computer (both running WinXP professional):
tiff()
plot(1:1000)
dev.off()
This
Gregory Warnes wrote:
Hello Hans,
Do you need *both* a categorical *and* a continuous variable, or just a
continuous variable?
-Greg
Hello Greg,
I want to visualize survival data above the columns. So my continuous
variable is follow up time and my categorical variable indicates the
On Wed, Jun 25, 2008 at 10:16 AM, Uwe Ligges
[EMAIL PROTECTED] wrote:
Gustaf Rydevik wrote:
Hi all,
A while back I sent a message concerning working with tiff-files, and
mentioned that I encountered a bug in 2.7.0.
This bug still occurs in 2.7.1, and is reproducable on a separate
Hi , Gustaf
i don't know why but it works pretty well on a mac.
regards .
On 2008-6-25, at 下午4:16, Uwe Ligges wrote:
Gustaf Rydevik wrote:
Hi all,
A while back I sent a message concerning working with tiff-files, and
mentioned that I encountered a bug in 2.7.0.
This bug still occurs in
Hi , yan
I don't think that is a R specific problem. Treating missing value
is not that trivial,
it depends too much on your raw data and your essential problem.
regards.
On 2008-6-25, at 上午2:14, 程燕 wrote:
Hi,Dear all R experts,
I am trying to do the 2-way contingency table
On 25 июн, 09:28, Jim Porzak [EMAIL PROTECTED] wrote:
The user of your R script sees only the outputs you create. The R source is
hidden.
But you may expose R sources via the web server if you wish.
Andrey
__
R-help@r-project.org mailing list
Hi,
I came across with a tiny problem.
E.g.:
pdf()
plot(1:5)
points(2, 3, cex=10, pch=21, bg=grey, lwd=0.3)
points(2, 4, cex=1, pch=21, bg=grey, lwd=0.3)
dev.off()
If I execute this I'll get a nice PDF. Fine.
But I want to edit this PDF with let's say by using Adobe Illustrator.
If I try
On Tue, 2008-06-24 at 11:39 -0300, Megan J Bellamy wrote:
Hello,
I am constructing a boxplot but have a very wide range of values (zero -
~28000). I have placed an axis break at the 8000 mark but would like to have
a different scale above the break that ranges from 8000-28000. Right now my
Hi
[EMAIL PROTECTED] napsal dne 25.06.2008 10:24:35:
I'm trying to make a loopo with some file
library(dplR)
files.rwl - system(ls *.rwl, intern=TRUE) files.rwl
[1] cimfasy.rwl rocquce.rwl
for (i in files.rwl) assign(gsub(\\.rwl$, _rwl, i), read.rwl(i,
header=0))
There are 70
Hi,
This is more of a general question than a pure R one, but I hope that is OK.
I want to combine one-tailed independent p-values using the weighted version
of fisher's inverse chi-square method. The unweighted version is pretty
straightforward to implement. If x is a vector with p-values, then
Are you sure you want to use Fisher's method?
http://www.burns-stat.com/pages/Working/perfmeasrandport.pdf
page 13 shows an example where Fisher's method
is not attractive, but Stouffer's method gives reasonable
results. Weighting in Stouffer's method is straightforward.
Patrick Burns
[EMAIL
Hi Patrick,
I have actually read your paper before and I am aware of the objection. I
have also implemented the weighted Stouffer's in R. I nevertheless want to
try the weighted fisher's if possible (if for no other reason than
curiosity)
Andreas Cederfeldt
On Wed, Jun 25, 2008 at 12:53 PM,
Hi Paul,
Duncan has shown you how to do it. There is often a simpler route that is
worth knowing about. Whether it works depends on how the function was coded.
In this case it works:
## Example
par(cex.main = 3)
spineplot (table (tbl$DAY, tbl$SEX), main='TIPS')
par(cex.main = 1.2)
spineplot
On Tue, Jun 24, 2008 at 7:43 PM, Deepayan Sarkar
[EMAIL PROTECTED] wrote:
On 6/24/08, Douglas Bates [EMAIL PROTECTED] wrote:
I am indebted to Deepayan Sarkar for lattice graphics, Fritz Leisch
for Sweave and the authors of the beamer package for LaTeX. For me
these are a killer app
I can't figure this one out- I am the administrator, The file that I want
to install is a .tar.gz which is located on my desktop. How do I get it
into my packages directory- through the GUI or through brute force?
thanks
stephen
--
Let's not spend our time and resources thinking about things
Hello,
When I put in the following script line:
gap.boxplot(CLI3, CLI4, CLI5, CLI6, CLI7, gap=list(top=c(8000,28),
bottom=c(0,250)), range=50, outline=TRUE, par(ask=FALSE)
I get a '+' telling me I am missing something. I have tried adding ')',
'width=NULL', etc and then I get this error:
install.packages(file.choose(), repos=NULL)
thought this would work, but it didn't
the package is the sowas package - this doesn't seem to be a CRAN package,
and it can be found at :
http://tocsy.agnld.uni-potsdam.de/wavelets/
On Wed, Jun 25, 2008 at 8:25 AM, stephen sefick [EMAIL PROTECTED]
the par(ask=FALSE) doesn't belong in the code (I think).
gap.boxplot(CLI3, CLI4, CLI5, CLI6, CLI7, gap=list(top=c(8000,28),
bottom=c(0,250)), range=50, outline=TRUE)
but I haven't tried this.
Stephen
On Wed, Jun 25, 2008 at 10:49 AM, Megan J Bellamy [EMAIL PROTECTED]
wrote:
Hello,
When I
Dear Everybody,
install.packages(Cairo)
returns
[...]
* Installing *source* package 'Cairo' ...
checking for gcc... gcc-4.2 -std=gnu99
checking for C compiler default output file name...
configure: error: C compiler cannot create executables
See `config.log' for more details.
ERROR: configuration
Just a small correction to my previous reply.
I meant to say probabilities of success for each type add up to one.
thanks Frank for noticing the mistake.
Bill.Venables wrote:
It looks like A*B*C*D is a complete, totally saturated model, (the
residual deviance is effectively zero, and the
From the (terminal window) command line try:
R CMD INSTALL path.to.file/filename.tar.gz
Does that do it?
Richard
stephen sefick wrote:
install.packages(file.choose(), repos=NULL)
thought this would work, but it didn't
the package is the sowas package - this doesn't seem to be a CRAN
no it didn't- am I missing something
(Mac OS X 10.5.3, R 2.7)
new-host-3:~ sefick$ R CMD INSTALL /Users/Desktop/sowas.tar.gz
WARNING: invalid package '/Users/Desktop/sowas.tar.gz'
* Installing to library '/Library/Frameworks/R.framework/Resources/library'
ERROR: no packages specified
On Wed,
Hi Stephen,
I tried what you suggested and got a different error message instead... any
ideas?
gap.boxplot(CLI3, CLI4, CLI5, CLI6, CLI7, gap=list(top=c(8000,28),
bottom=c(0,250)), range=50, outline=TRUE)
Error in bxgap$out[bxgap$out gap$top[2]] - bxgap$out[bxgap$out gap$top[2]]
- :
is there an na.rm argument in gap.boxplot?
On Wed, Jun 25, 2008 at 11:24 AM, Megan J Bellamy [EMAIL PROTECTED]
wrote:
Hi Stephen,
I tried what you suggested and got a different error message instead... any
ideas?
gap.boxplot(CLI3, CLI4, CLI5, CLI6, CLI7, gap=list(top=c(8000,28),
On Wed, 25 Jun 2008, Peng Jiang wrote:
Hi , Gustaf
i don't know why but it works pretty well on a mac.
with completely different code.
Gustaf Rydevik has mentioned this before -- it never fails for me on
Windows and hence one would not expect there to be a change in 2.7.1.
Only if someone
Please look at the NEWS for R-devel, which was an option to
work around this known bug in Adobe Illustrator.
(Of course, the R posting guide suggested this for before posting.)
On Wed, 25 Jun 2008, Hans-Joerg Bibiko wrote:
Hi,
I came across with a tiny problem.
E.g.:
pdf()
plot(1:5)
On Wed, 25 Jun 2008, stephen sefick wrote:
install.packages(file.choose(), repos=NULL)
thought this would work, but it didn't
the package is the sowas package - this doesn't seem to be a CRAN package,
and it can be found at :
You forgot a lot of things asked for in the posting guide, but if
I don't believe so. Sorry if I am missing something (I am a beginner user of
R). I tried just putting in all the defaults as see on
http://bm2.genes.nig.ac.jp/RGM2/R_current/library/plotrix/man/gap.boxplot.html
Here is a copy of my script and how the graph looks right now (without the
break
A short update that may be of help:
The snippet of code does not crash R if i run under vanilla, nor if I
change R to MDI-mode.
It does crash R infallibly if I set it to SDI-mode in the Rprofile
file. Strange...
/Gustaf
On Wed, Jun 25, 2008 at 3:16 PM, Prof Brian Ripley
[EMAIL PROTECTED]
On 25 Jun 2008, at 15:22, Prof Brian Ripley wrote:
Please look at the NEWS for R-devel, which was an option to work
around this known bug in Adobe Illustrator.
Thanks a lot for the hint.
(Of course, the R posting guide suggested this for before posting.)
I looked at several mailing
Macintosh OS X 10.5.3
R version 2.7 (binary distribution)
#I tried this in R
install.packages(file.choose(), repos=NULL, type=source)
WARNING: omitting pointless dependence on 'R' without a version requirement
Warning message:
In install.packages(file.choose(), repos = NULL, type = source) :
Jim Porzak wrote:
The user of your R script sees only the outputs you create. The R source
is hidden.
Ah .. that sounds great .. I wish I had known about this a month ago!
I'll have to check it out - thanks!
Esmail
HTH,
Jim Porzak
..
Would the R script that is being run be hidden from
Chad Junkermeier junkermeier at byu.edu writes:
I have a set of data in the form
x1, y1, z1
x1, y2, z2
...
x1, yN, zN
x2, y1, z(N+1)
x2, y2, z(N+2)
...
x2, yN, z(2N)
...and so on...
xM, yN, val(M*N)
Do you have any suggestions?
Not at all carefully tested, but:
## make up
If you would have tried
help.search(t-test)
It would have listed the function t.test()
?t.test
Will give you the helpfile with instructions.
ir. Thierry Onkelinx
Instituut voor natuur- en
Hello EveryBody,
I have some problems to connect R with Excel. It there somebody can help me
to understand how to proceed.
See you very soon
--
View this message in context:
http://www.nabble.com/Working-with-R-in-a-Excel-VBA-tp18109068p18109068.html
Sent from the R help mailing list archive
Hi list!
I am trying to use xyplot to plot some graphs.
The data I have looks like this:
alldata[1:10,]
breaks numbers moltypetype
1 0.0006598 5S Between species
2 0.407 0 5S Between species
3 0.8135228 5S Between species
4
heya,
i am fitting linear mixed effect model to a response Y. Y shows an s-shaped
distribution when using QQ-plots (some zero values and some very high values).
hence, which transformation should i apply that Y follows a normal
distribution? any r-function/package available to do this?
thanks
Hey,
I just like to know how to extract details from the naiveBayes model
(package e1071). I mean, for each possible value the model defines how much
it influences the outcome. I want to sort those probabilities and show the
values with the highest impact.
How could I do that?
PS: I tried using
On Mon, Jun 23, 2008 at 9:12 PM, Gundala Viswanath [EMAIL PROTECTED]
wrote:
I've also tried with
data$var
But still fail to access var column
you can't use $ with matrices (see help($) !), but you can use [ with
names, or convert you matrix to a data frame:
myData[, var]
Hi,
I am not sure this is what you want …. But
alist - list.files(path = whatever path you have for your files, full.names
= TRUE)
Now you have a list with the names of your files …. Form each name you can make
a variable name and assign that file ….
fname - substring(alist[i], first,
Following code bugs with Memory allocation failed: Copying Node error after
parsing n thousand files. I have included the main code(below) and
functions(after the main code).
I am not sure which lines are causing the copying Node which results in
memory failure. Please advise.
#Beginning
If you want to hide the fact that you are using R -- especially if
you charge people for your software that uses R clandestinely -- that's
a violation of the license (GPL). I doubt if anyone associated with R
would bother with a lawsuit, but a competitor who offers related
software
I recommend avoid loop in this case
I'm working with many files...
I must utilise a loop necessarily.
And you shall know that R is pretty smart in handling missing values. How
do you know after change NA - 0 that 0 is from NA and not
like in 1994 row and second column.
This isn't a
To the R colleagues,
I do not have any problems when running this code, everything
works well:
tiff()
plot(1:1000)
dev.off()
I am using R 2.7.1 and Windows XP Professional (service pack 2).
Regards,
Paulo Barata
---
Paulo
Hi,
I don't know if this matters but it worked for me with no problems ….
sessionInfo()
R version 2.7.0 (2008-04-22)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
I am trying to create a boxplot that has a gap with different scales so that my
boxes actually show (compare attachments). I have referred to the help pages
for gap.boxplot, gap.plot, list with no luck so far. Here is my script and the
resulting error message:
# Import *.csv files containing
Spencer Graves wrote:
If you want to hide the fact that you are using R -- especially
if you charge people for your software that uses R clandestinely --
that's a violation of the license (GPL). I doubt if anyone associated
with R would bother with a lawsuit, but a competitor who offers
Hi,
help(t.test)
x = rnorm(10, mean = 10, sd = 1) #some data
mytest = t.test(x, mu = 10) #is mu = 10
attributes(mytest) #what's in the test
mytest$p.value #obtain the p-value
mytest$statistic #obtain the t-value aka statistic
HTH
Thomas
Hello
I'm trying to calculate 2d confindence bounds into a scatterplot using the
function kde2d (package MASS) and a contour plot.
I found a similar post providing a solution - unfortunatly I do not realy
understand which data I have to use to calculated the named quantile:
Post URL:
Some clarifications.
R's license (GPL v2) is not about money,
you can charge anyone as much as you wish.
If you create an R program (and don't modify R itself), then
you can distribute that program according to any license you wish.
If you modify R itself _and_ distribute the modified
I am not an expert nor a lawyer but ...
On Wed, Jun 25, 2008 at 5:28 PM, Michael Conklin
[EMAIL PROTECTED] wrote:
Spencer Graves wrote:
If you want to hide the fact that you are using R -- especially
if you charge people for your software that uses R clandestinely --
that's a violation
Hello R users,
I'm student and I'm actually having a lecture introducing repeated mesures
analysis. Unfortunately, all examples use SAS system...
I'm working with lme function (package nlme), and I'm using
extract.lme.cov (package mgcv) to extract covariance structure of models.
One example is
I don't understand this. Why not just get hist() to plot on the
density scale,
thereby making its output commensurate with the output of density()?
The hist() function will plot on the density scale if you ask it
to. Set freq=FALSE
(or prob=TRUE) in the call to hist.
I would like to make clear that the SAS unstructured correlation matrix in
the second model was :
Estimated R Correlation Matrix for id 1
1 1. 0.8575 0.6984 0.4657 0.3165
2 0.8575 1. 0.8557 0.5670 0.4039
3 0.6984 0.8557
Dear r-help
I am trying to run LDA on a training data set, and test it on another data set
with the same variables. I found examples using crossvalidation, and using
training and testing data sets set up with sample, but not when they are
preassigned.
Here is what I tried
# FIRST SET UP A
on 06/25/2008 11:19 AM Daren Tan wrote:
unique(c(1:10,1)) gives 1:10 (i.e. unique values), is there any
method to get only 2:10 (i.e. values that are unique) ?
The easiest might be:
Vec
[1] 1 2 3 4 5 6 7 8 9 10 1
Vec[table(Vec) == 1]
[1] 2 3 4 5 6 7 8 9 10
HTH,
Marc
on 06/25/2008 11:29 AM Marc Schwartz wrote:
on 06/25/2008 11:19 AM Daren Tan wrote:
unique(c(1:10,1)) gives 1:10 (i.e. unique values), is there any
method to get only 2:10 (i.e. values that are unique) ?
The easiest might be:
Vec
[1] 1 2 3 4 5 6 7 8 9 10 1
Vec[table(Vec) ==
I think this line
mafdiscpred - predict(mafdisc, data = test)
needs to be
mafdiscpred - predict(mafdisc, newdata = test)
Michael Conklin
Chief Methodologist - Advanced Analytics
MarketTools, Inc.
6465 Wayzata Blvd. Suite 170
Minneapolis, MN 55426
Tel: 952.417.4719 |
Hmmm, this is not very good:
Vec - c(10:1,1)
Vec[ table(Vec) == 1 ]
[1] 9 8 7 6 5 4 3 2 1
and these are obviously not the unique values.
This one is better:
Vec [ ! duplicated(Vec) ! duplicated(Vec, fromLast=TRUE) ]
Gabor
On Wed, Jun 25, 2008 at 11:29:31AM -0500, Marc Schwartz wrote:
On Wed, 25 Jun 2008, Marc Schwartz wrote:
on 06/25/2008 11:19 AM Daren Tan wrote:
unique(c(1:10,1)) gives 1:10 (i.e. unique values), is there any
method to get only 2:10 (i.e. values that are unique) ?
The easiest might be:
Vec
[1] 1 2 3 4 5 6 7 8 9 10 1
Vec[table(Vec) == 1]
I'm sorry to say, but this one is wrong, too.
Maybe coffee really helps, I just had one. :)
Vec - c(20:30,20)
which(table(Vec) == 1)
21 22 23 24 25 26 27 28 29 30
2 3 4 5 6 7 8 9 10 11
You would actually need the names, but that would involve
some numberic - character - numeric
on 06/25/2008 11:44 AM Gabor Csardi wrote:
I'm sorry to say, but this one is wrong, too.
Maybe coffee really helps, I just had one. :)
Vec - c(20:30,20)
which(table(Vec) == 1)
21 22 23 24 25 26 27 28 29 30
2 3 4 5 6 7 8 9 10 11
You would actually need the names, but that would
On Wed, Jun 25, 2008 at 12:19 PM, Daren Tan [EMAIL PROTECTED] wrote:
unique(c(1:10,1)) gives 1:10 (i.e. unique values), is there any method to get
only 2:10 (i.e. values that are unique) ?
Try this:
setdiff(x, x[duplicated(x)])
__
Wow, that is smart, although is seems to be overkill.
I guess 'duplicated' is better than O(n^2), is it really?
Gabor
On Wed, Jun 25, 2008 at 05:43:30PM +0100, Prof Brian Ripley wrote:
On Wed, 25 Jun 2008, Marc Schwartz wrote:
on 06/25/2008 11:19 AM Daren Tan wrote:
unique(c(1:10,1))
Spencer Graves wrote:
If you want to hide the fact that you are using R -- especially if
you charge people for your software that uses R clandestinely -- that's
a violation of the license (GPL).
No on both accounts .. but thanks for pointing this out none the less.
On Wed, 25 Jun 2008, Gabor Csardi wrote:
Wow, that is smart, although is seems to be overkill.
I guess 'duplicated' is better than O(n^2), is it really?
Yes as it hashes, but the overhead on short vectors is high since it
always hashes.
Gabor
On Wed, Jun 25, 2008 at 05:43:30PM
DeaR list,
I'm a bit lost in the behavior of substitute and co.
I often use fairly long axis labels in my graphs (long to write, that
is). Typically, they would contain some greek letters and units with
exponents, as in:
xlab=expression(paste(text , alpha, / , V,., m^{-3}, .,
kg^{-2},
I tried the following
data(promotergene)
## train svm using custom kernel
stringkernel-stringdot(length = 4, lambda = 0.5, type = sequence, normalized
= TRUE)
gene - ksvm(Class~.,data=promotergene,kernel=stringkernel,C=10,cross=5)
I get the following error
cannot allocate vector of size 1.2
Hello, everyone.
I'm hoping to prevent myself from doing a lot of pointing and clicking
in Excel. I have a dataframe of bird nest check observations, in which
I know the date of the first check, the date of the second check (both
currently in Julian date format), the status of the nest at the
Dear HelpeRs,
i have a data.frame df as follows:
df - data.frame(id=rep(1:3,rep(10,3)),
emoqu=as.factor(rep(c(0,0,1,1,2,2,3,3,4,4),3)),
x=rnorm(30), y=runif(30))
Now, I would like to rearrange the data and it works - regarding the
variables/columns I would
Try this:
plot(1, xlab = ~ alpha / V * m^-3 * kg ^-2 * l^4)
On Wed, Jun 25, 2008 at 1:06 PM, baptiste Auguié [EMAIL PROTECTED] wrote:
DeaR list,
I'm a bit lost in the behavior of substitute and co.
I often use fairly long axis labels in my graphs (long to write, that is).
Typically, they
The description of your problem is too complex for me to analyze
in the time I have available for this. Could you develop a very brief,
simple example that illustrates your question? A very brief Monte
Carlo, estimating 2 parameters in a very simple formula with one of the
parameters at
Our July *** New York City *** R/S-Plus Fundamentals and Programming
Techniques course is scheduled for:
New York City / July 28-29, 2008 ***
Please direct enquiries to Sue Turner: [EMAIL PROTECTED]
Payment due AFTER the class
Ask for Group Discount ---
Looking for R
Hello,
In WinBUGS 1.4 manual
(http://www.mrc-bsu.cam.ac.uk/bugs/winbugs/manual14.pdf), the gamma
density is presented as dgamma(r,mu) where r and mu are the shape and
rate parameters, respectively. In JAGS (rjags) manual version 1.0.2,
May 9, 2008
I realize questions about packages should go to the package maintainer,
but perhaps I have an old email address ([EMAIL PROTECTED])
Also I have both a general, and a specific, question.
1) General question: i've used pvclust before to assess significance of
clusters and got reasonable results.
Hi, I'm trying to use sapply to compute the min of several variables, each
of them stored in data.frames, grouped as a list:
Is it normal that mean() and min() produce different objects dimensions?
str(dats)
List of 5
$ log20:'data.frame': 83 obs. of 5 variables:
..$
Sorry for flooding this forum, but I think I've realised that I need to do
multinomial logistic regression for my problem...
Would be interested in your opinion as to whether this is actually any
better than running three binomial logistic regressions separately..
Thanks
M
On Wed, Jun 25, 2008
Hi there,
I am trying to write a function to perform GOF test of data to a
zero-truncated Poisson distribution. I am facing 2 problems.
1) How can I obtain a frequency table for all values within the range of
observed values?
For instance if the observations are
obs - c(A, A, A, A, B, C, C, D,
Dear helpeRs:
I am writting you because I have some doubts in relation to the equivalence
test. I am interested in compare seed set of treated plants and control plants.
I used the tost command in R, and the value of epsilon that I put was the
standard deviation of the control sample. In all
R has wonderful graphics but I am wondering whether there is anything in R to
provide customization of tables like PROC Report does in SAS.
A Smile costs Nothing
But Rewards Everything
Happiness is not perfected until it is shared
If I analyze a client's data using an R script I created then I can
charge the client a $20,000 consulting fee, but, if I let the client
push the button to execute the R script and charge him 10 cents for the
privilege then I can be sued for violating the GPL? Or are my
I think you cannot be
It seems that just recently my plot outputs are getting messed up.
Currently I am only getting the plotted points and the bottom axis. I am
getting no text and no side axis. As far as I know nothing got changed on
the server. I have tested this with simple plots and with plots that have
heading
Example- I don't know SAS well, or at all at this point. But, if I knew
what you wanted maybe I could help you.
thanks
Stephen
On Wed, Jun 25, 2008 at 3:32 PM, Philip Twumasi-Ankrah
[EMAIL PROTECTED] wrote:
R has wonderful graphics but I am wondering whether there is anything in R
to
Thanks to all of the responders. It looks like dreams may come true
after all :-)
The application is as follows. Respondents fill out a questionnaire that
has been posted online. Output is in excel or text format. My client
(typically working for a government agency that provides subsidies
On Fri, 20-Jun-2008 at 01:08PM +0100, Gavin Simpson wrote:
| But it is not too difficult to do this yourself:
|
| in2cm - function(x) return(x * 2.54)
|
| cm2in - function(x) return(x * 0.3937008)
|
| then use in2cm() or cm2in() depending on what you want to achieve. If
| you want a margin of
On Wed, Jun 25, 2008 at 6:00 PM, Victor Homar [EMAIL PROTECTED] wrote:
Hi, I'm trying to use sapply to compute the min of several variables,
each
of them stored in data.frames, grouped as a list:
Is it normal that mean() and min() produce different objects dimensions?
it is exactly
The answer to your question is ``yeah, sort of''. The reason for the
difference is that mean() is generic and has a method for data frames,
according to which the mean of each column of the data frame is found
in some ``appropriate'' manner. (Essentially the columns of the data
frame must be
As for your first problem, try:
xyplot(numbers~breaks|moltype, groups = type, data = alldata, type = l)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Karin Lagesen
Sent: Wednesday, June 25, 2008 2:13 AM
To: r-help@r-project.org
Subject: [R] xyplot
hi,
I have a question could please help me.
Geneset.ls - list()
for(i in 1: 5452)
{
Geneset.ls[[i]] - read.delim(c.all.v2.5.symbols.gmt, header=T, skip=1,
sep=\t)
}
I have a file named c.all.v2.5.symbols.gmt.it is a huge file with about 30
columns and 5452 rows.
I want to write single rows
apply max to columns f1...f4 and assign it to rs$f:
rs$f - apply(rs[,paste(f,1:4,sep=)],1,max)
or
rs$f - apply(rs[,2:5],1,max)
On Wed, Jun 25, 2008 at 1:41 AM, Anh Tran [EMAIL PROTECTED] wrote:
Hi,
Here's the data we have:
rs[1:5,]
probe_id f1 f2 f3 f4 MA f
Rolf Turner wrote:
The answer to your question is ``yeah, sort of''. The reason for the
difference is that mean() is generic and has a method for data frames,
according to which the mean of each column of the data frame is found
in some ``appropriate'' manner. (Essentially the columns of
Mark Kimpel wrote:
Ben and Duncan,
Thanks for your helpful suggestions. Im having some difficulty
navigating this really good package using my normal learning
techniques. When I do 'help(package = rgl) it seems only a very
small subset of functions available show up.
I think the full list
On Wed, Jun 25, 2008 at 01:06:48PM -0700, [EMAIL PROTECTED] wrote:
Geneset.ls - list()
for(i in 1: 5452)
{
Geneset.ls[[i]] - read.delim(c.all.v2.5.symbols.gmt, header=T, skip=1,
sep=\t)
}
I have a file named c.all.v2.5.symbols.gmt.it is a huge file with
about 30 columns and 5452 rows.
If you like using IRC, then you might like to try the channel devoted to R:
irc://irc.freenode.net/#R
You can go there to ask all your R related questions!
--
MetaBase - http://BioDatabase.Org
[[alternative HTML version deleted]]
__
Hello R-help List
I am writing some R scripts to create graphs of water quality trends
that will be called by a web service running R. The axis titles will
need to change as the input data (ie. water quality variable) changes
according to a user's choice made via a web page. The way I am
Thanks for the pointers. I think the package is great, just want to
use it to its full potential without driving the list crazy with
questions. Below is my output to help and sessionInfo(). I don't see
the scaling functions, although I now see that they can be retrieved
with ?matrices. You guys
Thanks all. I got the result I wanted. I appreciate all the pointer. I never
fully understand the apply function, as well as the lapply. And this
occasion is one of those where these functions comes in to play.
On Wed, Jun 25, 2008 at 2:10 PM, Kenn Konstabel [EMAIL PROTECTED] wrote:
apply max
Philip Twumasi-Ankrah wrote:
R has wonderful graphics but I am wondering whether there is anything in R to
provide customization of tables like PROC Report does in SAS.
You can do far better than PROC REPORT. See for example
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