similar to: Matrix diagonal help

Hi, I have two 3 D arrays. Both are of this form array_1<- array[n,n,k] array_2<-array[m,m,k] Lets say n=83 and m=80 Since n>m. I would like to add rows and columns to array_2 to make them equal. I want to keep the size of the third dimension fixed i.e.. k. i.e. if (nrow(array_1)>nrow(array_2)) { array_2[m:n,m:n,]<- 10^6 } But this doesn't work. I tried abind and rbind but

Hello! I wrote a function that returns a data frame. Nowhere in the function do I say print(my.data.frame), but when I run the function - the data frame is printed on the console. Is there any way to suppress it? Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hello! s<-"start"; e<-"end" middle<-as.character(c(1,2,3)) I would like to get the following result: "start 123 end" or "start 1 2 3 end" or "start 1,2,3 end" How can I avoide this (undesired) result: paste(s,middle,e,sep=" ") Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

Hello! I am wondering if it's possible to run - in sequence - code that is stored in several R scripts. For example: Script in the file "code1.r" contains the code: a = 3; b = 5; c = a + b Script in the file "code2.r" contains the code: d = 10; e = d - c Script in the file "code3.r" contains the code: result=e/a I understand that I could write those 3 scripts

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

I know that I can use as.yearmon in the package "zoo" to find the year and the month of a date. I can use as. yearqtr to find the year and the quarter. But how can one find just the year of a date? Thanks a lot! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

I am sorry, I'd like to split my column ("names") such that all the beginning of a string ("X..") is gone and only the rest of the text is left. x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd")) x$names<-as.character(x$names) (x) str(x) Can't figure out how to apply strsplit in this situation - without using a

Hello! If I have in my data frame MyFrame a variable saved as a Date and want to translate it into years, I currently do it like this using "zoo": library(zoo) as.year <- function(x) as.numeric(floor(as.yearmon(x))) myFrame$year<-as.year(myFrame$date) Is there a function that would do it directly - like "as.yearmon" - but for years? Thank you! -- Dimitri

Dear R-ers, I have a large data frame (several thousands of rows and about 2.5 thousand columns). One variable ("group") is a grouping variable with over 30 levels. And I have a lot of NAs. For each variable, I need to divide each value by variable mean - by subgroup. I have the code but it's way too slow - takes me about 1.5 hours. Below is a data example and my code that is too

Dear R'rs, is there a function that checks if a given vector contains a certain value. E.g., x<-c(1,2,3,4). How can I get a TRUE or FALSE for whether x contains a 2? -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Hello. Until today I've been using R2.9 and since today R2.10 (on a PC). In both of them it takes about 20 sec for the prompt to appear IN R console after I start R. And every time it says: "Previous saved work space restored" - even if I have not saved any workspace or, in case of R2.10 - even though I have not used it once. In the older versions - R would start within 2-3 sec. Is

Dear developers of R-project, I have such data height of male persons 181, 178, 182, 160, 187, 193, 184, 184, 175, 178, 184, 184, 174, 185, 175 height of female persons 164, 165, 160, 174, 167, 161, 164, 165, 169, 175, 165, 155, 172, 164, 172, 166, 160, 159, 158, 173 I want to determine if height is dependent on gender, and I need to find, as I understand, point biserial correlation. I had tried

Dear R-ers, I have a data frame data with predictors x1 through x5 and the response variable y. I am running a simple regression: reg<-lm(y~x1, data=data) I would like to loop through all predictors. Something like: predictors<-c("x1","x2",... "x10) for(i in predictors){ reg<-lm(y~i) etc. } But it's not working. I am getting an error: Error in

Dear all, I have a function that predicts DV based on one predictor pred: pred<-c(0,3000000,7800000,15600000,23400000,131200000) DV<-c(0,500,1000,1400,1700,1900) ## I define Function 1 that computes the predicted value based on pred values and parameters a and b: calc_DV_pred <- function(a,b) { DV_pred <- rep(0,(length(pred))) for(i in 1:length(DV_pred)){ DV_pred[i] <- a *

Hello, dear R-ers! I have a data frame: x<-data.frame(a=c(4,2,4,1,3,4),b=c(1,3,4,1,5,0),c=c(NA,2,5,3,4,NA),d=rep(NA,6),e=rep(NA,6)) x When x$a==1, I would like to replace NAs in columns d and e with 8 and 9, respectively When x$a != 1, I would like to replace NAs in columns d and e 101 and 1022, respectively. However, I only want to do it for rows 2:5 - while ignoring what's happening in

Hello! I have 2 data frames like this (well, actually, I have 200 of them): df1<-data.frame(location=c("loc 1","loc 2","loc 3"),date=c("1/1/2010","1/1/2010","1/1/2010"), a=1:3,b=11:13,c=111:113) df2<-data.frame(location=c("loc 1","loc 2","loc

Dear everyone, I am trying to run rgenoud on several chips simultaneusly. I used the instructions provided on Jasjeet Sekhon's Homepage (http://sekhon.berkeley.edu/rgenoud/multiple_cpus.html). However, I have the newer version of R (R 2.12) installed - for a 64-bit machine. So, when I tried to install the special version of "snow" from a zip file provided by Jasjeet on his page, R

I have just installed R 12. I have Windows 7, 64-bit verison. I currently have IE as my default browser. The internet connection is very good. Whenever I try to run a help command (?lm, for example), I get this error: Error in shell.exec(url) : access to 'http://127.0.0.1:20271/library/stats/html/lm.html' denied I first got this message when Google Chrome was my default browser. For some

Sorry for no code - but it's a more of a general question. I have read in a data frame ("|"-delimited, .txt). daily<-read.table(file="filename.txt",sep="|",header=T) One of the variables is a factor with 110 levels: >str(daily$dma_id) Factor w/ 110 levels "500","501","503",... 108 levels of this factor happen to be numbers