similar to: why "object 'x' not found"?

Dear Listers, can anyone help me, please. Since several days i try to figure out, how to assign values, vectors, functions etc to variables with dynamically generated names inside of functions. Sometimes I succeed, but the success is rather arbitrary, it seems. up to now i don't fully understand, why things like get, assign, <<- etc do sometimes work, and sometimes not. here's one

Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a b c d TRUE TRUE TRUE TRUE FALSE FALSE FALSE TRUE FALSE TRUE FALSE FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a b c

Hi, Suppose, if I create 15 files in my working directory. set.seed(48) lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)}) ?D <-dir() D1 <- D[order(as.numeric(gsub("\\D+","",D)))] D1 ?res <- t(sapply(D1,function(x) {x1<-

dear dirk, i am following up on our thread on r-help-mailinglist. you advised me to bring it over here. when installing packages from within R, they usually go to a directory defined in R_LIBS_USER, which i have it set in .Renviron. when installing packages via sudo apt install r-cran-.., they go to /usr/lib/R/site-library. is there a possibility to have packages installed to the R_LIBS-USER

Dear R listers, I am trying to create a new variable that uniquely identifies groups of observations in a dataset. So far I couldn't figure out how to do this in R. In Stata I would simply type: egen newvar = group(dim1, dim2, dim3) Please, find below a quick example to show what I am dealing with: I have a dataset with 4 variables: var <- runif(50) ## a variable that I want to group

Ein eingebundener Text mit undefiniertem Zeichensatz wurde abgetrennt. Name: nicht verf?gbar URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20130513/fe848ce7/attachment.pl>

Hi, I am trying to debug the following code: for (i in 1:10){ t <- paste("d0$V",i,sep="") t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0) } and I would like to see what code is actually processing R, how can I do that? More to the point, I am trying to update my variables d0$V1 to d0$V10 according to the presence or absence of some text (contained in the file d1)

Thanks Don, for (i in 1:10){ nm <- paste0("V", i) d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0) } is exaclty what I needed. Best regards, Luca 2018-04-25 23:03 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>: > Your code doesn't make sense to me in a couple of ways. > > Inside the loop, the first line assigns a value to an

Hi all, I am attempting to create a new variable based on values of other variables. The variable is called pharm. It basically takes the numeric code of 1 as yes and 0 to be No from the variable B20_C1 (a question on a survey). However, I would also like to have a level for non-respondents and these are captured in the variable nr.B20C, which is a 1 when there is a non-response on the whole group

I forgot to explain why my suggestion. The logical condition returns FALSE/TRUE that in R are coded as 0/1. So all you have to do is coerce to integer. This works because the ifelse will return a 1 or a 0 depending on the condition. Meaning exactly the same values. And is more efficient since ifelse creates both vectors, the true part and the false part, and then indexes those vectors in

Your code doesn't make sense to me in a couple of ways. Inside the loop, the first line assigns a value to an object named "t". Then, the second line does the same thing, assigns a value to an object named "t". The value of the object named "t" after the second line will be the output of the ifelse() expression, whatever that is. This has the effect of making

Luca, If speed is important, you might improve performance by making d0 into a true matrix, rather than a data frame (assuming d0 is indeed a data frame at this point). Although data frames may look like matrices, they aren?t, and they have some overhead that matrices don?t. I don?t think you would be able to use the [[nm]] syntax with a matrix, but [ , nm] should work, provided the matrix has

Hello, instead of ifelse, the following is exactly the same and much more efficient. d0[[nm]] <- as.integer(regexpr(d1[i,1], d0$X0) > 0) Hope this helps, Rui Barradas On 4/28/2018 8:45 PM, Luca Meyer wrote: > Thanks Don, > > for (i in 1:10){ > nm <- paste0("V", i) > d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0) > }

Dear List, I want to compare the presence and absence of bird species based on the sites in a matrix. The matrix has 5 rows for Island A, B, C, D, and E. It has 100 columns for bird species D001-D100. In each cell of the matrix, the presence-absence of bird species will be recorded as 1 or 0. (For example, if species D001 is found on Island D, the matrix cell of species D001 and Island D

ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0,

I am new to R. I just tried to recreate in R (using sem package and the identical input data) a solution for a simple measurment model I have found before in LISREL. LISREL had no problems and converged in just 3 iterations. In sem, I got no solution, just the warning message: "Could not compute QR decomposition of Hessian. Optimization probably did not converge. in: sem.default(ram =

Hi Rui Thank you for your suggestion, I have tested the code suggested by you against that supplied by Don in terms of timing and results are very much aligned: to populate a 5954x899 0/1 matrix on my machine your procedure took 79 secs, while the one with ifelse employed 80 secs, hence unfortunately not really any significant time saved there. Nevertheless thank you for your contribution.

Dear all, Without a loop, I would like transform 3 numeric vectors empty of 0/1 of same length Vec1 : transform 1 to A and 0 to "" Vec2 : transform 1 to B and 0 to "" Vec3 : transform 1 to C and 0 to "" to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC, BC, AC, AB,...) Any idea ? Thank you for your help -- Michel ARNAUD

Dear R-users, I have dataset with column A and B with1000 values, For each of column C value (C = seq(1,1000,1), I want to repeat A and B values and calculate R = A+B*C for each row. I want to get output as A B C R 1 10 1 11 2 30 1 32 3 50 1 53 1000 1000 1

Hello, I finally run my measurement model in sem - successfully. Now, I am trying to print out the path diagram that is based on the results - but for some reason it's not working. Below is my script - but the problem is probably in my very last line: # ANALYSIS OF ANXIETY, DEPRESSION, AND FEAR - LISREL P.31 library(sem) # Creating the ANXIETY, DEPRESSION, AND FEAR intercorrelation matrix