Dear All,
How would we do this problem looping over seq(1:2) ?
Thank you,
Ashim
On Thu, Aug 4, 2011 at 10:59 AM, Richard Ma wrote:
> Thank you so much GlenB!
>
> I got it done using your method.
>
> I'm just curious how did you get this idea? Cause for me, this looks so
> tricky
>
> Cheers
Hi C6 (were C1 - 5 already taken in your family?),
I downloaded your data and can replicate your problem. R ceases
responding and terminates. This does not occur with all uses of qr on
a dgCMatrix object. I know nothing about sparse matrices, but if you
believe this should not be occurring, you
Hi,
I've written a package and converted my Rd files into LaTeX using
Rdconv. When I copy and paste these files in to my Sweave document I
get the error message when compiling the Sweave file:
! Undefined control sequence.
l.32 \inputencoding
{utf8}
I also get the error message
Thank you so much GlenB!
I got it done using your method.
I'm just curious how did you get this idea? Cause for me, this looks so
tricky
Cheers,
Richard
-
I'm a PhD student interested in Remote Sensing and R Programming.
--
View this message in context:
http://r.789695.n4.nabble.com/H
You defined temp as a vector:
> temp <- vector(mode="numeric", length = vl)
but you try to extract from it as if it was a 2d object:
colA <- temp[,compareA]
Maybe you meant to use temp1 instead? More meaningful variable names might
help avoid such mistakes.
On Wed, Aug 3, 2011 at 6:14 PM, Matt
I'm fitting a logistic regression model of the form:
outcome ~ covariates + A*B
where A and B are factors -- A has 4 levels, B has 2 levels. The A
and B term each have significant main effects and the interaction term
is significant. I'd like to ask, how does a particular set of A and B
value
Hello
I am trying to draw a basic black and white map of two European countries.
After searching some key words in google and reading many pages I arrived to
the conclusion that persp() could be used to draw that map.
I have prepared three small example files, which are supposed to be the files
Hi everyone,
Suppose I have a list named "lst", see below:
> lst
$sub1
...
$sub1$x
...
$sub1$y
$sub2
...
$sub2$x
...
$sub2$y
…
$sub3
...
...
...
Now, I want to extract the sub-sublist $y from every sublist(sub1, sub2...)
and then storage them to a new list.
I know how to extract them by s
Hi
baptiste auguie wrote:
Dear list,
I have two questions regarding grid.symbols() in the grImport package.
This package allows you to import a vector graphic in R, and
grid.symbols() can be used to plot the resulting glyph at arbitrary
locations in a grid viewport.
I have tried the code in th
All:
Below is my code for creating a basic horizontal, stacked barchart. I'd like
to label the plot in two ways: 1) place the x values in each piece and 2)
place the y values above each piece (angled). I'm currently using lattice, but
I'm open to suggestions using ggplot2.
Questions:
1. C
On Wed, 3 Aug 2011, Peter Langfelder wrote:
Hi all,
in my package I have a function with name plot.cor (this function is
inherited from another legacy package). According to CRAN package
checks reports, the check apparently thinks plot.cor is a method for
the plot generic (I hope I'm using the
'temp' is only a vector and you are trying to reference it as a matrix,
therefore the error message
Sent from my iPad
On Aug 3, 2011, at 18:14, Matt Curcio wrote:
> Greetings all,
> I am getting an error message that is stifling me.
> Any ideas?
>
>> ## Define Directories ##
>> load_from <- "
I want to add a comment related to the calibration plot that was presented in
a previous post (which probably cannot be done optimally in SPSS). The plot
lacks sufficient resolution in the x-axis values that are calibrated. It is
far better to use loess to estimate a smooth nonparametric calibrat
The problem mentioned in the 06 Dec 2010 email below still occurs with
Tinn-R (v.2.3.7.1) when highlighting a string of code, copying to the
clipboard, and trying to send to Rgui via Shift+Ctrl+Q.
I can copy 1 line of code to the clipboard and send it to Rgui with
Shift+CTRL+Q. This fails with 2
Hello R users,
I am trying to give the QR decomposition for a large sparse matrix in
the format of dgCMatrix. When I run qr function for this matrix, the R
session simply stops and exits to the shell.
The matrix is of size 108595x108595, and it has 4866885 non-zeros. I
did the experiment on window
Nomogram is user-friendly, but the equation is also acceptable. It should be
kept in mind that the process of model development really counts.
BTW: You can calculate C-index (AUC) in SPSS. Calibration plot can also be
plotted (may be manually) from the result of SPSS.
*Yao Zhu*
*Department of Uro
Greetings all,
I am getting an error message that is stifling me.
Any ideas?
> ## Define Directories ##
> load_from <- "/home/mcc/Dropbox/abrodsky/kegg_combine_data/"
> save_to <- "/home/mcc/Dropbox/abrodsky/ttest_results/"
>
> ###
> ## Define Columns To Compare ##
> co
> This might be a little late: but how about this (slightly clumsy) function:
>
> putValues <- function(Insert, Destination, Location) {
> Location = as.matrix(Location)
> Location = array(Location,dim(Destination))
> Destination[Location] <- Insert
> return(Destination)
> }
>
> It
A barplot rendered with povray,
http://zoonek2.free.fr/UNIX/48_R/03.html#10
At the other end of the spectrum,
library(txtplot)
x <- factor(c("orange", "orange", "red", "green", "green", "red",
"yellow", "purple", "purple", "orange"))
o <- capture.output(txtbarchart(x))
library(
Hello All,
I am trying to run normalization on the Illumina bead studio output file
version 1.5.1.3 using the lumi package. Sine the original file is a huge
file I am splitting it and reading it as a batch in the lumi package. This
is where the problem begins.. it just seems to be not reading it..
Thanks for all the replies!Unfortunately the solutions only work for
extracting subsets of the data (which was exactly what I was asking for)
and not to replace subsets with other values. I used them, however, to
program a rather akward function to do that. Seems I found one of the
few aspects
Consider I have the following data:
AgeRangeAgeOfPerson PersonNoFriendsAtYear0 FriendsAtYear1
FriendsAtYear2 FriendsAtYear3 FriendsAtYear4 FriendsAtYear5
10 - 12 11 1 0 1 2 2 3 3
10 - 12 12 2 0 1 2 2
Just a very simple follow-up. In the summary table (listed as "summ" below),
the "TR" column I would like to display the total number of rows (i.e.
counts) which I have done via "NROW()" function.
However, in the "RG1" I would only like to count the number of rows with a
'totalread' count >= 1 (i.
Hi all,
in my package I have a function with name plot.cor (this function is
inherited from another legacy package). According to CRAN package
checks reports, the check apparently thinks plot.cor is a method for
the plot generic (I hope I'm using the correct terminology).
checking Rd \usage secti
Some sample data would be useful. If you want to add more lines to a
plot, the use 'lines'
plot(fun1,)
lines(fun2,
lines(fun3,
On Wed, Aug 3, 2011 at 11:21 AM, KnifeBoot wrote:
> Hey,
> Is there any function plotting several "implicit functions" (F(x,y)=0) on
> the same fig.
Did you install R first? R.app is just a GUI around the actual R code
that could run without any assistance in a terminal session. Generally
one installs both R and R.app from the "super-bundle". Since you
provided no details of which .pkg files were chosen we are left
guessing.
(And th
On Aug 3, 2011, at 18:35 , Walter Ludwick wrote:
> Have tried to install R.app several times (6, in fact: versions 2.12, 13 &
> 14, both 32 and 64 bit versions), using packages freshly downloaded from the
> official project page, and failed every time, given exception reports such as
> the fol
It is hard to prove a negative, but to the best of my knowledge lm will not do
what you want. This does not mean there is not a function that will perform
your analyses; the sort of thing you want to do is often accomplished using
non-linear methods.
John
>>> ראובן אברמוביץ 8/3/2011 12:00:04 P
Here's what you _should_ do
1) transpose
2a) as.data.frame
3a) fix the stupid default stringsAsFactor behavior
4a) convert the first 5 columns to numeric
dfrm <- as.data.frame( t( structure(.) ) )
dfrm[, 1:5] <-lapply(dfrm[, 1:5], as.character)
dfrm[, 1:5] <-lapply(dfrm[, 1:5], as.numeric)
Or:
To add to David's comments (nice catch, BTW), I found three
variogram() functions as a result of ??variogram. The one that gets
used is from the package that is highest in the search path (notice
that gstat is 55th (!!)) - that would be the one from the spatial
package. [The other is in the Spatial
Here is one more question,
How could I input an asymmetry in volatility specication in the BEKK
function?
As far as I know, the BEKK estimation function is
mvBEKK.est(eps, order = c(1,1), params = NULL, fixed = NULL, method =
"BFGS", verbose = F)
I totally have no idea to exert an asymmetry int
Dear ALL,
I use BEKK package to estimate Bivariate GARCH model. But when the results
come out, there's no t-stat or p-value of the estimated coeffients. Does
anyone know how to get the significance?
Followings are the codes I input,
>P1=data.frame(x,y)
>y1=mvBEKK.est(P1)
>mvBEKK.diag(y1)
Anyhel
On Aug 3, 2011, at 3:05 PM, Ken wrote:
Sorry about the lack of code, but using Davids example, would:
tapply(itemPrice, INDEX=orderID, FUN=sum)
work?
Doesn't do the cumulative sums or the assignment into column of the
same data.frame. That's why I used ave, because it keeps the sequence
c
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Ken
> Sent: Wednesday, August 03, 2011 12:13 PM
> To: Jeffrey Joh
> Cc:
> Subject: Re: [R] Convert matrix to numeric
>
> How about
> Matrix[1:5,]=as.numeric(Matrix[1:5,])
> -Ken
Hi Jeffrey,
On Wed, Aug 3, 2011 at 3:04 PM, Jeffrey Joh wrote:
>
> I have a matrix that looks like this:
>
>
> structure(c("0.0376673981759913", "0.111066500741386", "1", "1103",
> "18", "OPEN", "DEPR", "0.0404073656092023", "0.115186044704599",
> "1", "719", "18", "OPEN", "DEPR", "0.066534209669
Please use R's search capabilities before posting.
RSiteSearch("Linear Model with Constraints")
appears to give you what you're looking for. Incidentally, with
constraints, the model is no longer linear, I believe.
-- Bert
2011/8/3 ראובן אברמוביץ :
>
> Can I put limits on the lm() command? I
On 03/08/2011 11:21 AM, KnifeBoot wrote:
Hey,
Is there any function plotting several "implicit functions" (F(x,y)=0) on
the same fig. Is there anyone who has an example code of how to do this?
The contour3d function in the misc3d package only work with the functions
with three
How about
Matrix[1:5,]=as.numeric(Matrix[1:5,])
-Ken Hutchison
On Aug 3, 2554 BE, at 3:04 PM, Jeffrey Joh wrote:
>
> I have a matrix that looks like this:
>
>
> structure(c("0.0376673981759913", "0.111066500741386", "1", "1103",
> "18", "OPEN", "DEPR", "0.0404073656092023", "0.11518604470459
Hello,
Perhaps transpose the table attach(as.data.frame(t(data))) and use ColSums()
function with order id as header.
-Ken Hutchison
On Aug 3, 2554 BE, at 1:12 PM, David Winsemius wrote:
>
> On Aug 3, 2011, at 12:20 PM, jim holtman wrote:
>
>> This takes about 2 secs for 1M ro
Have tried to install R.app several times (6, in fact: versions 2.12, 13 & 14,
both 32 and 64 bit versions), using packages freshly downloaded from the
official project page, and failed every time, given exception reports such as
the following (appended below, the 2 reports arising out of my 1st
On 03/08/2011 3:04 PM, Jeffrey Joh wrote:
I have a matrix that looks like this:
structure(c("0.0376673981759913", "0.111066500741386", "1", "1103",
"18", "OPEN", "DEPR", "0.0404073656092023", "0.115186044704599",
"1", "719", "18", "OPEN", "DEPR", "0.0665342096693433", "0.197570061769498",
"1",
Sorry about the lack of code, but using Davids example, would:
tapply(itemPrice, INDEX=orderID, FUN=sum)
work?
-Ken Hutchison
On Aug 3, 2554 BE, at 2:09 PM, David Winsemius wrote:
>
> On Aug 3, 2011, at 2:01 PM, Ken wrote:
>
>> Hello,
>> Perhaps transpose the table attach(as.data.frame(t(dat
Dear List,
I have some difficulties to work with the function lmer from lme4. My
responses are binary form and i want to use forward selection to my 12
covariates but i dont know how can I choose them based on deviance. Can
someone pls give me a example so i can apply. For example my covariates
Hello R Help,
I am attempting to install/build a 64-bit version of R to hopefully resolve
some memory.limit problems for a user who is running a simulation. The
'configure' runs fine and the compilation (make) runs fine until the very last
part (see below). I have libiconv in /usr/local/lib (n
I see a 'variogram' function in both spatial and gstat when I use ??
variogram on my machine that probably does not have even all of those
packages installed. Are you sure they are the same (I looked they
are not) or failing that that the one you expect is being chosen? And
are you eve
Hello R experts,
I am trying to fit an lme model within a function, using a formula that I
passed into the function, and then predict that model from a different
function. Could you please advise me on how to do this? The following code
illustrates the essence of what I'm trying to do. The actu
Since we got this the x-th time now:
Dear Fränzi Korner,
please please please never ever add auto-replies to your account that
also reply to mailing list messages! Thousands of readers of R-help get
your auto reply everey day now!
Best,
Uwe Ligges
On 03.08.2011 12:11, fraenzi.kor...@oiko
Many many thanks, working now.
Best,
BN Mandal
On Wed, Aug 3, 2011 at 10:34 PM, Duncan Murdoch wrote:
> On 03/08/2011 12:47 PM, Baidya Nath Mandal wrote:
>
>> Dear Murdoch,
>>
>> After setting CYGWIN=nodosfilewarning, i re-ran the R CMD check and got
>> following message:
>>
>> * installing *sou
Hey,
Is there any function plotting several "implicit functions" (F(x,y)=0) on
the same fig. Is there anyone who has an example code of how to do this?
The contour3d function in the misc3d package only work with the functions
with three dimensions.
Thanks a lot.
Many thanks
Dear Murdoch,
After setting CYGWIN=nodosfilewarning, i re-ran the R CMD check and got
following message:
* installing *source* package 'mypackage' ...
** libs
ERROR: compilation failed for package 'mypackage'
* removing 'C:/Rpackages/mypackage.Rcheck/mypackage'
The log file contained following.
Hello,
We are trying to use R to simulate a model based on parameters 'a' and 'b'.
This involves the following integration:
model<-function(s,x,a,b)(exp(-s*x*10^-5.5)*(s^(a-1)*(1-s)^(b-1)))
g<- function(x,a,b){
out<-c()
for (i in 1:length(x)){
out[i]<-1- (integrate(model,0,1,x[i]
I have a matrix that looks like this:
structure(c("0.0376673981759913", "0.111066500741386", "1", "1103",
"18", "OPEN", "DEPR", "0.0404073656092023", "0.115186044704599",
"1", "719", "18", "OPEN", "DEPR", "0.0665342096693433", "0.197570061769498",
"1", "1103", "18", "OPEN", "DEPR", "0.1192871
Can I put limits on the lm() command? I only know that you can choose a
liniar model with or without an intercept, but can I put other limits on
the coefficients (for example- the intercept must be bigger than 1) ?
_
Here is my sessionInfo()
> sessionInfo()
R version 2.12.2 (2011-02-25)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
att
On Aug 3, 2011, at 2:01 PM, Ken wrote:
Hello,
Perhaps transpose the table attach(as.data.frame(t(data))) and use
ColSums() function with order id as header.
-Ken Hutchison
Got any code? The OP offered a reproducible example, after all.
--
David.
On Aug 3, 2554 BE, at 1:12
If you add column names to your contrast matrix (treat3) then those names will
be used in the coefficient names.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [
On 2011-08-03 09:40, gbre...@ssc.wisc.edu wrote:
Hello.
I am running the examples provided in the gstat help menus. When I try to
run the following in predict.gstat:
data(meuse)
coordinates(meuse)= ~x+y
v<-variogram(log(zinc)~1, meuse)
I get the following error message:
Error in vector("doub
On Aug 3, 2011, at 12:20 PM, jim holtman wrote:
This takes about 2 secs for 1M rows:
n <- 100
exampledata <- data.frame(orderID = sample(floor(n / 5), n, replace
= TRUE), itemPrice = rpois(n, 10))
require(data.table)
# convert to data.table
ed.dt <- data.table(exampledata)
system.time(
On 03/08/2011 12:47 PM, Baidya Nath Mandal wrote:
Dear Murdoch,
After setting CYGWIN=nodosfilewarning, i re-ran the R CMD check and got
following message:
* installing *source* package 'mypackage' ...
** libs
ERROR: compilation failed for package 'mypackage'
* removing 'C:/Rpackages/mypackage.R
On 3 Aug 2011, at 17:46, Sarah Goslee wrote:
> Hi Federico,
>
> A forward slash isn't a special character:
>
>> strsplit("T/T", "/")
> [[1]]
> [1] "T" "T"
>
> so there's some other problem.
>
> Are you sure that your first column contains strings and not factors?
> What does str(my.data) tell
On 3 Aug 2011, at 17:41, Duncan Murdoch wrote:
>
> It looks as though your my.data[1,1] value is a factor, not a character value.
>
> strsplit(as.character(my.data[1,1]), "/")
Thanks Duncan, this solved it.
Best
Federico
>
> would work, or you could avoid getting factors in the first plac
Hi Federico,
A forward slash isn't a special character:
> strsplit("T/T", "/")
[[1]]
[1] "T" "T"
so there's some other problem.
Are you sure that your first column contains strings and not factors?
What does str(my.data) tell you?
Does
strsplit(as.character(my.data[1,1]), "/")
work?
If you us
On 03/08/2011 12:37 PM, Federico Calboli wrote:
Hi All,
is there a way of using strsplit with a forward slash '/' as the splitting
point?
For data such as:
1 T/TC/C 16/33
2 T/TC/C 33/36
3 T/TC/C 16/34
4 T/TC/C 16/31
5 C/CC/C 28/29
6 T/T
Hello.
I am running the examples provided in the gstat help menus. When I try to
run the following in predict.gstat:
data(meuse)
coordinates(meuse)= ~x+y
v<-variogram(log(zinc)~1, meuse)
I get the following error message:
Error in vector("double", length) : invalid 'length' argument
What's t
Hi All,
is there a way of using strsplit with a forward slash '/' as the splitting
point?
For data such as:
1 T/TC/C 16/33
2 T/TC/C 33/36
3 T/TC/C 16/34
4 T/TC/C 16/31
5 C/CC/C 28/29
6 T/TC/C 16/34
strsplit(my.data[1,1], "/") # and an
This takes about 2 secs for 1M rows:
> n <- 100
> exampledata <- data.frame(orderID = sample(floor(n / 5), n, replace = TRUE),
> itemPrice = rpois(n, 10))
> require(data.table)
> # convert to data.table
> ed.dt <- data.table(exampledata)
> system.time(result <- ed.dt[
+
You can look at the code
coxme:::print.coxme
There you will see that the global test is a chisquare
chi1 <- 2*diff(x$loglik[1:2])
with x$df[1] degrees of freedom.
The fixed effects coefficients are found in x$coefficients$fixed, and
the variances are diag(x$var)[-(1:nfrail)]
1. you wrote to the mailing list rather than to the original poster.
2. you forgot to cite the original post, hence we do not know what you
are referring to.
PLease do read the posting guide to this list!
Uwe Ligges
On 03.08.2011 14:53, mohammad...@gmail.com wrote:
Hello David,
I encounter
There was too many spelling mistakes in my original post so I have
decided to re-submit it. So here is it
Dear R users,
I am trying to determine how many characters can be displayed within
the width of an open grid viewport. Unfortunately, the arithmetic
operation that seems obvious in this case
On 8/3/2011 6:07 AM, wwreith wrote:
So I take it 3D pie charts are out?
At least with ggplot, yes. 2D pie charts are somewhat tricky with
ggplot, even. They can be gone with stacked, normalized bar charts
projected into polar coordinates, if I recall properly.
Not limited to ggplot, there
zcatav gmail.com> writes:
>
> Your suggestion works perfect as i pointed previous message. Now have another
> question about data editing. I try this code:
> X[X[,"c"]==1,"b"]<-X[,"d"]
> and results with error: `[<-.data.frame`(`*tmp*`, X[, "c"] == 1, "b", value
> = c(NA, :
> replacement has
On Aug 3, 2011, at 9:59 AM, ONKELINX, Thierry wrote:
Dear Caroline,
Here is a faster and more elegant solution.
n <- 1
exampledata <- data.frame(orderID = sample(floor(n / 5), n, replace
= TRUE), itemPrice = rpois(n, 10))
library(plyr)
system.time({
+ ddply(exampledata, .(order
Dear R users,
I am trying to determine how many characters can be displayed within the
width of an open grid viewport. Unfortunately, the arithmetic operation that
seems obvious in this case is be permitted with unit objects (see example
below). Although it isa brut force way to get this number (u
On Wed, 2011-08-03 at 11:04 +0300, Antonio Rodriges wrote:
> Hello,
>
> The idea is to grant access of remote users to R running on Linux.
> Users must have ability to run their
> R scripts but avoid corrupting the operating system.
>
> How one can restrict/limit access of remote users to certain
I thought Sarah's reply was great and, alas, should probably be
templated for this list.
Not sure it fits as a fortunes package entry, but I thought it at
least worthy of consideration.
Cheers,
Bert
>> ...
>> I appreciate any suggestions for this problem.
Sarah Goslee replied:
> Suggestions? Ye
Hello All,
I was trying to generate a map of Europe with the following codes:
europe<-map(database="world", fill=FALSE,
plot=TRUE,xlim=c(-25,70),ylim=c(35,71))
However, the "world" database is too old to have right European country
names. Could anyone help?
Thanks,
Tianchan
--
View this messa
Hi all,
I have been using WEKA to do some text classification work and I want to try
out R.
The problem is I cannot load the String to Vector ARFF files created by
WEKA's string parser into Rattle .
Looking at the logs I get something like:
/Error in scan(file, what, nmax, sep, dec, quote, ski
Hi,
I have some difficulties to work with the function lmer from lme4. My
responses are binary form and i want to use forward selection to my 12
covariates but i dont know how can I choose them based on deviance. Can
someone pls give me a example so i can apply. For example my covariates are
gesta
Hi Peter,
Thanks for these information.
I used a column concatenating the listBy data to do this aggregation : (I
don't know if it's the best solution, but it seems to work).
aggregateMultiBy <- function(x, by, FUN){
tableBy = data.frame(by)
tableBy$byKey = ""
for(colBy in
David Winsemius wrote:
>
> On Aug 3, 2011, at 8:09 AM, zcatav wrote:
>
> You need to apply the same logical test/selection on the rows of the
> RHS as you are doing on the LHS.
> Possibly:
>
> X[ X[,"c"]==1, "b"] <- X[ X[,"c"]==1, "d"]
>
>
This solution was suggeste
Gabor Grothendieck wrote:
>
> On Wed, Aug 3, 2011 at 8:09 AM, zcatav wrote:
>> Your suggestion works perfect as i pointed previous message. Now have
>> another
>> question about data editing. I try this code:
>> X[X[,"c"]==1,"b"]<-X[,"d"]
>> and results with error: `[<-.data.fr
Hello I am using the "step" function in order to do backward selection for a
linear model of more than 200 variables but it doesn't work correctly.
I think, there is a problem, if the matrix has same or more columns than
rows.
And if the matrix has too much columns the step-function doesn't work
b
On Wed, 3 Aug 2011, peter dalgaard wrote:
On Aug 3, 2011, at 12:30 , Jannis wrote:
Dear List,
i would like to mimic the behaviour or the following indexing with a do.call
construct to be able to supply the arguments to `[` as a list:
test = matrix[1:4,2]
result = test[2,]
My try, h
On Aug 3, 2011, at 9:25 AM, Caroline Faisst wrote:
Hello there,
Im computing the total value of an order from the price of the
order items
using a for loop and the ifelse function.
Ouch. Schools really should stop teaching SAS and BASIC as a first
language.
I do this on a large
On Aug 3, 2011, at 12:30 , Jannis wrote:
> Dear List,
>
>
>
> i would like to mimic the behaviour or the following indexing with a do.call
> construct to be able to supply the arguments to `[` as a list:
>
>
> test = matrix[1:4,2]
>
> result = test[2,]
>
>
> My try, however, did not wo
Hi there,
I had a problem when I hoped to get confidence intervals for the
parameters I got using mle() of stats4 package. This problem would not
appear if ``fixed'' option was not used. The following mini-example will
demo the problem:
x <- c(100, 56, 32, 18, 10, 1)
r <- c(18, 17, 10, 6, 4,
On 2011-08-03 00:24, Thaler,Thorn,LAUSANNE,Applied Mathematics wrote:
Does
xyplot(y ~ seq_along(y), xlab = "Index")
do what you want?
Not exactly, because it does not work once multipanel conditioning comes
into play:
xyplot(y~seq_along(y)|factor(rep(1:2, each=5)), xlab = "Index")
The p
Hi,
On Wed, Aug 3, 2011 at 10:06 AM, Vishal Thapar wrote:
> Hi,
>
> I apologize for posting this here, I am also trying to post this on machine
> learning emailing lists.
> I have a set (18K) of sequences (22 nt long) and I have their counts at 4
> different stages. The difference in counts from
Thank you very much Peter,
I'm going to dig deeper into the code of the functions you've listed.
On Wed, Aug 3, 2011 at 6:57 AM, Peter Ehlers wrote:
> On 2011-08-02 11:51, Sébastien Bihorel wrote:
>
>> Hi,
>>
>> This might be a simple problem but I don't know how to calculate a random
>> varia
Hi,
I apologize for posting this here, I am also trying to post this on machine
learning emailing lists.
I have a set (18K) of sequences (22 nt long) and I have their counts at 4
different stages. The difference in counts from one stage to the next
represents how well the sequence performed in th
Dear Caroline,
Here is a faster and more elegant solution.
> n <- 1
> exampledata <- data.frame(orderID = sample(floor(n / 5), n, replace = TRUE),
> itemPrice = rpois(n, 10))
> library(plyr)
> system.time({
+ ddply(exampledata, .(orderID), function(x){
+ data.frame(itemPr
Dear R-users,
I am comparing differences in variance, skew, and kurtosis between two groups.
For variance the comparison is easy: just
var.test(group1, group2)
I am using agostino.test() for skew, and anscombe.test() for kurtosis.
However, I can't find an equivalent of the F.test or Mood.tes
Hello there,
Im computing the total value of an order from the price of the order items
using a for loop and the ifelse function. I do this on a large dataframe
(close to 1m lines). The computation of this function is painfully slow: in
1min only about 90 rows are calculated.
The computati
Hello David,
I encountered the same problem of yours.
What did you do to resolve it?
Thanks for your reply
Mohammad
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On Wed, Aug 3, 2011 at 9:04 AM, Antonio Rodriges wrote:
> Hello,
>
> The idea is to grant access of remote users to R running on Linux.
> Users must have ability to run their
> R scripts but avoid corrupting the operating system.
Ordinary users can't corrupt the operating system on Linux[1]. The
On Aug 3, 2011, at 8:09 AM, zcatav wrote:
Your suggestion works perfect as i pointed previous message. Now
have another
question about data editing. I try this code:
X[X[,"c"]==1,"b"]<-X[,"d"]
and results with error: `[<-.data.frame`(`*tmp*`, X[, "c"] == 1,
"b", value
= c(NA, :
replaceme
So I take it 3D pie charts are out?
P.S. It is not about hiding anything. It is about consulting and being told
by your client to make 3D pie charts and change this font or that color to
make the graphs more apealing. Given that I am the one trying to open the
door to using R where I work it woul
Andrew Winterman wrote:
>
>
> I'm trying to use the xlsx package to read a series of excel spreadsheets
> into R, but my code is failing at the first step.
>
> I setwd into my the directory with the spreadsheets, and, as a test ask
> for
> the first one:
>
>read.xlsx(file = "Argentina Fina
Antonio Rodriges wrote:
>
>
> The idea is to grant access of remote users to R running on Linux. Users
> must have ability to run their
> R scripts but avoid corrupting the operating system.
>
>
Check RStudio.org
Dieter
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The nomogram you included was produced by the Design package, the precursor
to the rms package. You will have to take the time to intensively read the
rms package documentation. Note that how you developed the model (e.g.,
allowing for non-linearity in log PSA, not using stepwise regression which
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