similar to: Pivoting a data.frame

I have observed the following differences in the use of min/max functions between R and Splus (4.5) R (0.64.1): min(c(NA,NA,NA),na.rm=T) returns the following Warning in min(..., na.rm = na.rm) : no finite arguments to min/max; returning extreme. [1] 2147483647 S-Plus 4.5( Win32 edition) min(c(NA,NA,NA),na.rm=T) returns [1] NA The S-PLUS behaviour seems to be the right one Thanks,

Hi, I wonder whether the mechanism of parsing parameters to functions has changed between 0.63.1 and 0.63.3? The following code yeils different results in R 0.63.1 (Version 0.63.1 (Dec 5, 1998)) and R 0.63.3. cave<-function(x,a,b) { return(c(mean(x[a],na.rm=T),mean(x[b],na.rm=T))) } datx <- data.frame(rbind(c(1,2,3,4),c(4,5,6,7)))

Hi, I wonder whether the mechanism of parsing parameters to functions has changed between 0.63.1 and 0.63.3? The following code yeils different results in R 0.63.1 (Version 0.63.1 (Dec 5, 1998)) and R 0.63.3. cave<-function(x,a,b) { return(c(mean(x[a],na.rm=T),mean(x[b],na.rm=T))) } datx <- data.frame(rbind(c(1,2,3,4),c(4,5,6,7)))

Dear HelpeRs, I would be grateful for anybody who might help to produce the following plot (the code for matlab is below) using the "rgl" package of R? [t,r] = meshgrid(linspace(0,2*pi,361),linspace(-4,4,101)); [x,y] = pol2cart(t,r); P = peaks(x,y); figure('color','white'); polarplot3d(P,'colordata',gradient(P)); view([-18 72]); You may see the code and plot

Hello HelpeRs, The last print statement in the code segment below results in : Error in data$x[data$x == -Inf] <- range$x.range[1] : replacement has length zero R version 2.14.1 Patched (2011-12-23 r57982) ggplot2: version 0.90 OS : Linux (64bit) Any thoughts? Many Thanks, A. ## code segment starts here df <- data.frame( trt = factor(c(1, 1, 2, 2)), resp = c(1, 5,

Hi Guys, I'm having problems getting the new idmap_adex module to work. When using the idmap_adex plugin I get the following: # wbinfo -n administrator S-1-5-21-XXXXXXXXX-XXXXXXXXXX-XXXXXXXXXX-500 User (1) # wbinfo -i administrator Could not get info for user administrator As expected attempting to lookup user & group info via commands which use libnss also fail. The

Thankyou for taking the time to read this ... We have large PC userbase and a not-so-large (but equally loud) Unix user base. We have a requirement to "PopUp" information windows on ALL our user's screens with an absolute minimum of manual intervention (ie type the message once and then forget it). We can manage the Unix X servers but can't seem to find a UNIX application that

Dear I am very beginners for samba. I want to install samba on Debian. I want maintain file server. We have 50 PCs that are running on WinXP I want connect these PCs to PC it run on Debian . Over IP Range is 192.168.1.0 255.255.255.0 What are the main step to install samba on debian . DNS server is must or not? Can you help me thanks

Hello everyone, I have a page that populates/updates make, model and year select lists via Ajax.Updater and I think that an EventObserver is causing a conflict. Here''s what I''m trying to do. I''m using this page as a create and edit page for a car database. If I''m calling the page in edit mode, I''ll want the make, model and year set (using the

Hello everyone, This is related to my post a few days ago regarding multi select lists. I''ve rewritten the code and now can''t seem to get the Ajax.Updater in getModels() to "execute", and the element never gets populated. I know it''s something obvious and hope someone could point it out to me. Thanks for any help. Here''s my code: <script

Hi I would like to know how to pivot a table that will sum the number of plants (a or b) for each location (x,y,z) I have read on the listserve similar questions but which involve summing up numbers rather than factors. I have also read about the R package reshape on the listserve but wanted to know if there's a way to do it easily without the package Thanks for the help! location plant x

Hi Everyone, I have modified my version of R-1.4.1 to include choleski with pivoting (like in Splus). I thought R-core might consider including this in the next version of R, so I give below the steps required to facilitate this. 1. Copied Linpack routine "dchdc.f" into src/appl 2. Inserted line F77_SUBROUTINE(dchdc) in src/appl/ROUTINES 3. Inserted "dchdc.f" into

It works! But Once I have the square root of this matrix, how do I convert it to a real (not imaginary) matrix which has the same property? Is that possible? Best, Simon >----Messaggio originale---- >Da: p.dalgaard at biostat.ku.dk >Data: 21-nov-2009 18.56 >A: "Charles C. Berry"<cberry at tajo.ucsd.edu> >Cc: "simona.racioppi at

Dear I am very beginners for samba. I want to install samba on Debian. I want maintain file server. We have 50 PCs that are running on WinXP I want connect these PCs to one server PC it run on Debian . Over IP Range is 192.168.1.0 255.255.255.0 What are the main step to install samba on debian . DNS server is must or not? Can you help me thanks

Thanks for your message! Actually it works quite well for me too. If I then take the trace of the final result below, I end up with a number made up of both a real and an imaginary part. This does not probably mean much if the trace of the matrix below givens me info about the degrees of freedom of a model... Simona >----Messaggio originale---- >Da: RVaradhan at jhmi.edu >Data:

Hi all, I have a table in database that is very long and when simplified it has only two columns in it (id, text). id is the row, and text is the column. Technically the text is a term and and id is the document. If simplifying this and assuming there is only one occurrence of the term per the document. I shall be able to convert this into a binary matrix. Table looks like this... *ID** **Text*

Dear Peter, thank you very much for your answer. My problem is that I need to calculate the following quantity: solve(chol(A)%*%Y) Y is a 3*3 diagonal matrix and A is a 3*3 matrix. Unfortunately one eigenvalue of A is negative. I can anyway take the square root of A but when I multiply it by Y, the imaginary part of the square root of A is dropped, and I do not get the right answer. I tried

Dear R experts, I am trying to solve this problem, related to the possibility of changing the shape of a data frame using a "pivoting-like" function. I have a dataframe df of observations as follows: ID VALIDITY YEAR PROPERTY PROPERTY VALUE A1 2007 P1 V1 A1 2007 P2 V2 A1 2007 P3 V3 A1 2008 P1 V10 A1 2008 P2 V20 A2 2007 P5 V50 A2 2008 P6 V20 A3 2007

Hello, I have been struggling to do a plot in ggplot(2) that's of lattice equivalent. The following code shows the lattice plot.

Hi, I'd like to use R to do what excel pivot tables do, and plot results. I've never used R before, and I've managed to do something, but it's quite a lot of code to do something simple. I can't help but think I'm not "Doing it the R way". I could be using R for the wrong thing, in which case, please tell me off. I was hoping something like plot(by(t,