I just ran a test with a couple of my scripts and they took the same
amount of CPU time in both Rterm and RGUI. (37.2 CPU sec on Rterm and
37.1 CPU sec on RGUI). Elapsed times were the same and memory usage
was similar.
It would be helpful to provide some more information. For example, do
the fo
Hi
I send commands to R using an IDE. Configured a few hotkeys to send the
commands to R. I am using Tinn-R and it has options to open a RGUI or RTerm.
My datasets are in the range of 600 MB - 2GB and I generally use scan() or
filehash to work on such data sets. However there are times when I
Dear Folks,
Jim Holtman and I are wondering if there is a useR group in cincy area
(OH in USA) or not. If not, how many on this list in cincy area would
like to have a useR group in cincy area?
Thank you so much!
__
R-help@r-project.org mailing list
http
Exactly what are you doing and how are you 'sending' in the commands?
Are you doing a cut/paste to the RGUI? If so, have you tried
'source'? I use RGUI all the time and have no performance/memory
problems with executing scripts; I always 'source' in most of the
scripts.
Some more detail on exact
See if this example helps; show how to either plot the row or columns
of a data frame:
> test <- data.frame(C1=runif(10), C2=runif(10), C3=runif(10))
> test
C1C2C3
1 0.91287592 0.3390729 0.4346595
2 0.29360337 0.8394404 0.7125147
3 0.45906573 0.3466835 0.344
4 0.
OK, I guess I'm getting better at the data part of R. I wrote a
program outside of R this morning to dump a bunch of experimental
data. It's a sort of ragged array - about 700 rows and 400 columns,
but the amount of data in each column varies based on the length of
the experiment. The real data end
Thanks, embarrased that I didn't think of that myself :)
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, & Mobile & VoiceM
On Jul 4, 2009, at 6:15 PM, sjkimble wrote:
R Help:
Another question:
My data are:
x.crd y.crd A p
1 361762 1034052 3.414 0
2 361763 1034071 94.169 0
3 370325 1027277 127.642 1
4 370416 1027166 127.961 0
5 370471 10271481804.846 1
6 369050 1031312
On 04/07/2009 6:09 PM, Henrique Dallazuanna wrote:
You cas use isTRUE:
isTRUE(grep("a", "") > 0)
Not really. If there are multiple hits, that returns FALSE.
You want to use length(grep(...)) to detect any hits.
Duncan Murdoch
On Sat, Jul 4, 2009 at 2:56 PM, Mark Kimpel wrote:
I
Thanks a lot! I got it! I haven't set the level before.Hao
On Sat, Jul 4, 2009 at 6:33 PM, Duncan Murdoch wrote:
> On 04/07/2009 3:31 PM, Hao Jiang wrote:
>
>> Hi Duncan,Thanks!
>>
>> But I still get a little confused about outer() func. Would give me a
>> simple
>> example to contour it? Just l
R Help:
Another question:
My data are:
x.crd y.crd A p
1 361762 1034052 3.414 0
2 361763 1034071 94.169 0
3 370325 1027277 127.642 1
4 370416 1027166 127.961 0
5 370471 10271481804.846 1
6 369050 10313121790.493 0
7 370799 1026557 103.994 0
On 04/07/2009 3:31 PM, Hao Jiang wrote:
Hi Duncan,Thanks!
But I still get a little confused about outer() func. Would give me a simple
example to contour it? Just like the formula x^2 + y^2 + x + y -5 = 0.
(Sorry I am a newbie to R, found really hard to use the R manual)
> x <- seq(-6,6,len=10
You cas use isTRUE:
isTRUE(grep("a", "") > 0)
On Sat, Jul 4, 2009 at 2:56 PM, Mark Kimpel wrote:
> I am using grep to locate colnames to automate a report build and have
> run into a problem when a colname is not found. The use of integer(0)
> in a conditional statement seems to be a no no
On Sat, Jul 4, 2009 at 7:56 PM, Mark Kimpel wrote:
> I am using grep to locate colnames to automate a report build and have
> run into a problem when a colname is not found. The use of integer(0)
> in a conditional statement seems to be a no no as it has length 0.
> Below is a self-contained trivia
Hello everyone,
This is my first post on this forum and I hope I communicate my problem
correctly.
I have been using R for sometime and have always used RGUI to send commands
to R. However recently I had to work on large data sets and memory was an
issue. I have started using R terminal and the
Hi,
I have a problem in R - I can't load my database. When I use Gentoo
everything is ok but in Ubuntu I get a message:
Error in make.names(col.names, unique = TRUE) :
invalid multibyte string 5
--
View this message in context:
http://www.nabble.com/I-can%27t-load-my-database-tp24337284p2433
On Sat, Jul 4, 2009 at 12:24 PM, Mark Knecht wrote:
> On Sat, Jul 4, 2009 at 12:10 PM, Allan Engelhardt wrote:
>> Try help("strptime"). Example
>>
>> strptime(1080103L + 19e6L, "%Y%m%d")
>> # [1] "2008-01-03"
>>
>> (This assumes your input is an integer but you can just drop the L if you
>> want)
Hi Duncan,Thanks!
But I still get a little confused about outer() func. Would give me a simple
example to contour it? Just like the formula x^2 + y^2 + x + y -5 = 0.
(Sorry I am a newbie to R, found really hard to use the R manual)
Thanks,
Hao
On Sat, Jul 4, 2009 at 7:10 AM, Duncan Murdoch wrot
On Sat, Jul 4, 2009 at 12:10 PM, Allan Engelhardt wrote:
> Try help("strptime"). Example
>
> strptime(1080103L + 19e6L, "%Y%m%d")
> # [1] "2008-01-03"
>
> (This assumes your input is an integer but you can just drop the L if you
> want)
>
>
> On 04/07/09 19:37, Mark Knecht wrote:
>
> Hi,
>Is t
Try help("strptime"). Example
strptime(1080103L + 19e6L, "%Y%m%d")
# [1] "2008-01-03"
(This assumes your input is an integer but you can just drop the L if
you want)
On 04/07/09 19:37, Mark Knecht wrote:
> Hi,
> Is there a function that will convert this sort of date code which
> looks li
On Jul 3, 2009, at 1:29 PM, Steven Rytina wrote:
The rules for conformability of objects required for various
operators remain a mystery as do some related problems like the
rules for recycling in creating arrays etc. Would someone be able to
direct me to where such rules are stated?
The break command can only break the innermost loop (the b loop in your
case). Rewrite the loop control logic. Here is one example
end.a <- FALSE;
for (a in 1:10) {
print(a);
for (b in 1:20) {
if (a > 5 && b > 5 ) {
end.a <- TRUE;
break; # Breaks the b loop
}
};
On 04/07/09 18:56, Mark Kimpel wrote:
I am using grep to locate colnames to automate a report build and have
run into a problem when a colname is not found. The use of integer(0)
in a conditional statement seems to be a no no as it has length 0.
Below is a self-contained trivial example. I woul
Hi,
Is there a function that will convert this sort of date code which
looks like "years from 1900 + month_number + day_number" with no
spaces? As an example Jan. 3rd 2008 would be written as 1080103.
Thanks,
Mark
__
R-help@r-project.org mailing list
I am using grep to locate colnames to automate a report build and have
run into a problem when a colname is not found. The use of integer(0)
in a conditional statement seems to be a no no as it has length 0.
Below is a self-contained trivial example. I would like to get
something like "NA" or -1 fo
Hi all,
Thanks very much for the prompt responses! All the suggestions worked :)
Paul
From: Martin Morgan
Cc: r-h...@stat.math.ethz.ch
Sent: Saturday, July 4, 2009 1:01:16 PM
Subject: Re: [R] newby question
Hi Paul --
Paul Evans wrote:
> Hi,
>
> I work w
On Thu, Jul 2, 2009 at 1:52 AM, Thomas Mang wrote:
> Hi,
> I just came across the following issue regarding mixed effects models:
> In a longitudinal study individuals (variable ind) are observed for some
> response variable. One explanatory variable, f, entering the model as fixed
> effect, is a
Hi Paul --
Paul Evans wrote:
> Hi,
>
> I work with bio-conductor, but this is probably a basic R question.
>
> I want to emulate the GOBPOFFSPRING$"GO:0008150" command:
>
>
>> allBP <- GOBPOFFSPRING$"GO:0008150"
>> class(allBP)
> [1] "character"
>> length(allBP)
> [1] 16066
>
> I want to crea
You probably want:
str <- "GO:0008150"
ifunc <- function(str){
allBP <- GOBPOFFSPRINGS[[str]]
return(allBP)
}
On Sat, Jul 4, 2009 at 12:38 PM, Paul Evans wrote:
> Hi,
>
> I work with bio-conductor, but this is probably a basic R question.
>
> I want to emulate the GOBPOFFSPRING$"GO:0008150" comma
Try this:
ifunc <- function(str){
allBP <- GOBPOFFSPRING[[str]]
return(allBP)}
On Sat, Jul 4, 2009 at 1:38 PM, Paul Evans wrote:
> Hi,
>
> I work with bio-conductor, but this is probably a basic R question.
>
> I want to emulate the GOBPOFFSPRING$"GO:0008150" command:
>
>
> > allBP <- GO
I have R on a couple of USB drives for Windows XP, I simply download the .exe
file to my hard drive and install the program on the USB drive. It runs off
the USB just as it would if installed on a hard drive.
--- On Fri, 7/3/09, Raphael Saldanha wrote:
> From: Raphael Saldanha
> Subject: [R
Hi,
I work with bio-conductor, but this is probably a basic R question.
I want to emulate the GOBPOFFSPRING$"GO:0008150" command:
> allBP <- GOBPOFFSPRING$"GO:0008150"
> class(allBP)
[1] "character"
> length(allBP)
[1] 16066
>
I want to create a function so that I can execute the command by p
Try this:
foo <- function(x, nrow, ncol){
m <- matrix(0, nrow = nrow, ncol = ncol)
m[cbind(unlist(lapply(0:(ncol - 1), `+`, seq(x))),rep(1:ncol, each =
length(x)))] <- x
m
}
foo(c(2, 1, 1, 1), nrow = 5, ncol = 2)
On Sat, Jul 4, 2009 at 1:26 PM, William Simpson wrote:
> Doesn't wor
Tal Galili wrote:
Hello Frank,
Thank you for the extension and remarks.
The basic weakness of stepwise regression VS going through all-subsets
is very much agreed upon. Although from what I gather there is one case
where all subsets will be a problem to implement, that is for very LARGE
dat
Doesn't work:
> Make(x=c(2,1,1,1),nR=5,nC=2)
[,1] [,2]
[1,]20
[2,]12
[3,]01
[4,]00
[5,]00
should be
[,1] [,2]
[1,]20
[2,]12
[3,]11
[4,]11
[5,]01
On Sat, Jul 4, 2009 at 5:04 PM, Jorge Ivan
Velez wrote:
> D
Here are two things to try.
First check the data. There may be a factor that does not have
variation in the sample. For example,
if you had a predictor such as 'present'/'absent', in the current
sample, all of them
may be 'present'.
Second, you can put a 'try' statement in your function.
try(
On Jul 4, 2009, at 11:59 AM, William Simpson wrote:
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp<-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), n
Dear William,
Here is one way using Henrique's solution:
Make <- function(x, nR, nC){
m <- matrix(0, nrow = nR, ncol = nC)
diag(m) <- x[1]
diag(m[-1,]) <- x[2]
m
}
Make(x = c(1,2), nR = 5, nC = 4)
HTH,
Jorge
On Sat, Jul 4, 2009 at 11:59 AM, William Simpson <
william.a
On Sat, 4 Jul 2009, William Simpson wrote:
Can anybody please tell me a good way to do the following?
Not sure how you want cases like x = 1:3, nrow=2 , ncol=7 to be handled,
but for the example you give, this works:
mat <- matrix(0,nr=nrow,nc=ncol)
indx <- outer(
seq( from=
Thanks everyone for the help.
I should have said that I want to do this generally, not as a one-off.
So I want a function to do it. Like this
tp<-function(x, nr, nc)
{
matrix( c(x,rep(0, nr-length(x)+1)), nrow=nr, ncol=nc)
}
tp(x=c(1,2), nr=5, nc=4)
This one looks good -- the warning message i
On Jul 4, 2009, at 11:36 AM, sdzhangping wrote:
Dear R users:
In my recent works, I compared the cumulative incidences among
three different treatment groups. The cuminc function (cmprsk
package ) yielded a graph (refer to figure 1)
No figure attached. Perhaps it did not conform to th
On Jul 4, 2009, at 11:17 AM, William Simpson wrote:
Can anybody please tell me a good way to do the following?
Given a vector, number of rows and number of columns, return a matrix
as follows. Easiest to give an example:
x=c(1,2), nrow=5, ncol=4
You ought to separate those assignments with
Try this:
m <- matrix(0, nrow = 5, ncol = 4)
diag(m) <- x[1]
diag(m[-1,]) <- x[2]
On Sat, Jul 4, 2009 at 12:17 PM, William Simpson <
william.a.simp...@gmail.com> wrote:
> Can anybody please tell me a good way to do the following?
>
> Given a vector, number of rows and number of columns, return a
Dear R users:
In my recent works, I compared the cumulative incidences among three
different treatment groups. The cuminc function (cmprsk package ) yielded a
graph (refer to figure 1) and a p value (p = 0.0007). I don’t know how to
interpret the meaning of the p value ( one p value and thr
I looked at the question and Gabor's reply and could not figure out
why you were not simply replacing the function list() for head() in
the second line of his example. Or perhaps you were, and the result
was not what you wanted, in which case it would be of you not being
clear about the des
Try:
?package.skeleton
?prompt
On Sat, Jul 4, 2009 at 10:59 AM, Jason Rupert wrote:
>
> By any chance is there a skeleton package to use as a template to develop an
> R package?
>
> I downloaded "Writing R Extensions", which was evidently updated pretty
> recently, but I did not see any referen
Can anybody please tell me a good way to do the following?
Given a vector, number of rows and number of columns, return a matrix
as follows. Easiest to give an example:
x=c(1,2), nrow=5, ncol=4
return the matrix:
1 0 0 0
2 1 0 0
0 2 1 0
0 0 2 1
0 0 0 2
Thanks very much for any h
Hello Frank,
Thank you for the extension and remarks.
The basic weakness of stepwise regression VS going through all-subsets is
very much agreed upon. Although from what I gather there is one case where
all subsets will be a problem to implement, that is for very LARGE datasets
- especially in th
By any chance is there a skeleton package to use as a template to develop an R
package?
I downloaded "Writing R Extensions", which was evidently updated pretty
recently, but I did not see any references (and of course I may have totally
missed it) to a package template to use as a go by.
Do
Try some reproducible example and why you want this list. I don't
understand why you are trying to do what you want to do.
On Sat, Jul 4, 2009 at 2:51 AM, Bogaso wrote:
>
> No reply yet. Is my question not clear? Please let me know.
>
> Thanks
>
>
> Bogaso wrote:
>>
>> Thanks Gabor for this reply
Ravi Varadhan wrote:
But you are still left with the problem of choosing the regularization
parameter, i.e. how much to shrink the coefficients? In other words, there is
no free ride.
Ravi.
Ravi,
Choosing a penalty factor is extremely easy compared to variable
selection. There is a uniqu
On Jul 3, 2009, at 12:10 PM, John Lipkins wrote:
Dear All,
I want to create a table for several variables. As example. I have a
dataframe with following data:
Gendertransport driving
1 0 1
0 1 0
1
On Jul 3, 2009, at 1:10 PM, John Lipkins wrote:
Dear All,
I want to create a table for several variables. As example. I have a
dataframe with following data:
Gendertransport driving
1 0 1
0 1 0
1
But you are still left with the problem of choosing the regularization
parameter, i.e. how much to shrink the coefficients? In other words, there is
no free ride.
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of G
On Jul 4, 2009, at 6:21 AM, Vincent Vinh-Hung wrote:
Dear List:
I have a question related to a previous discussion
How to sum one column in a data frame keyed on other columns
https://stat.ethz.ch/pipermail/r-help/2006-December/122141.html
(George Nachman, Bill Venables)
The original query wa
Tal Galili wrote:
Hi Ben,
I just wished to give a small remark about your claim:
"it's best not to consider hypothesis testing (statistical significance) and
AIC in the same analysis."
Since in the case of forward selection for orthogonal matrix's, it can be
shown that AIC is like using a P to e
mean(x[x[,'region'] == 'south', 'profit'])
or if dataframe
mean(x$profit[x$region == 'south'])
or for all regions
tapply(x$profit, x$region, mean)
On Fri, Jul 3, 2009 at 6:55 PM, JoK LoQ wrote:
>
> Just a quickly beginner's question.
>
> I wanna find the mean only from the values from a column
Hao Jiang wrote:
Hi,
I want to plot a polynomial in the form like ax^2 + bxy + cy^2 + dx + ey + f
=0 without solving it(since I may have 3 or 4 dimensional polynomial and
it's really hard to solve). Is there any way to plot this kind of
polynomial?
Thanks a lot!
There are lots of ways. A c
Tal Galili wrote:
> Hi Ben,
> I just wished to give a small remark about your claim:
> "it's best not to consider hypothesis testing (statistical significance) and
> AIC in the same analysis."
>
> Since in the case of forward selection for orthogonal matrix's, it can be
> shown that AIC is like
Dear List:
I have a question related to a previous discussion
How to sum one column in a data frame keyed on other columns
https://stat.ethz.ch/pipermail/r-help/2006-December/122141.html
(George Nachman, Bill Venables)
The original query was to calculate the sum of visits for each unique
tuple of
francogrex wrote:
Data from Fisher's paper: Confidence Limits for a Cross-Product Ratio.
y
col1 col2
[1,] 103
[2,]2 15
fisher.test(y)
Fisher's Exact Test for Count Data
data: y
p-value = 0.0005367
alternative hypothesis: true odds ratio is not equal to 1
95 perce
Steven Rytina wrote:
The rules for conformability of objects required for various operators
remain a mystery as do some related problems like the rules for recycling in
creating arrays etc. Would someone be able to direct me to where such rules are
stated?
In a related vein, there ar
The reason I had asked how and what method was used to calculated the
confidence intervals of the odds ratio in the fisher.test is because I don't
think it's the exact method based on permutations. It seems that the exact
method is not implemented in R or anywhere else:
"In a tactic originally sug
Hannu Kahra wrote:
Hi,
I have a graph with seven parallel horizontal lines. Is it possible to
shade the area between two adjacent lines?
Hi Hannu,
Easy if you know the endpoints of the lines.
plot(rep(1,2),type="l",ylim=c(0,8))
for(i in 1:6) {
lines(rep(i+1,2))
polygon(c(1,1,2,2),c(i,i+1,i
Hi,
You could use ?rect
x = 1:5
y=1:5
plot(x,y,t="n",xaxs="i")
abline(h=y)
xlims = range(x)
y.rect = matrix(rep(y,each=2)[-c(1, 2*length(y))], ncol=2, byrow=T)
x.rect = matrix(rep(xlims), ncol=2,nrow=nrow(y.rect), byrow=T)
cols = 1:4
rect(x.rect[,1], y.rect[,1], x.rect[,2], y.rect[,2],col=
having been bitten by this behavior many times, could I register a
plea for
allowing a col argument to plot.stepfun that would deal with both the
horizontal
and vertical segments -- I rather doubt that it is often desirable to
have different
colors for these. Note that verticals = TRUE is t
The rules for conformability of objects required for various operators
remain a mystery as do some related problems like the rules for recycling in
creating arrays etc. Would someone be able to direct me to where such rules are
stated?
In a related vein, there are all manner of operati
Hello,
Please could someone explain
break and next??
Is there a way to break a loop that is not the innermost loop??
for example:
#=
dim A = 1000 x 3
dim B = 2000 x 3
for (a in 1:1000){
for (b in 1:2000){
expr = intersect(A[a,1]:A[a,2],B[b,1]:B[b,2])
Just a quickly beginner's question.
I wanna find the mean only from the values from a column related to specific
values from another one. Like, theres a 'region' column, i want the mean of
the value on 'profit' column only from "south" sells from 'region' column
--
View this message in context:
Dear All,
I want to create a table for several variables. As example. I have a
dataframe with following data:
Gendertransport driving
1 0 1
0 1 0
1 0 1
Now I want to crea
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