Hi r-help-boun...@r-project.org napsal dne 25.10.2010 20:41:55:
> Adrienne, there's one glitch when I implement your solution below. When the > loop encounters a case with no data at all (that is, all 140 item responses > are missing), it aborts and prints this error message: " ERROR: argument is > of length zero". > > I wonder if there's a logical condition I could add that would enable R to > skip these empty cases and continue executing on the next case that contains data. > > Thanks, Dave > > David S. Herzberg, Ph.D. > Vice President, Research and Development > Western Psychological Services > 12031 Wilshire Blvd. > Los Angeles, CA 90025-1251 > Phone: (310)478-2061 x144 > FAX: (310)478-7838 > email: dav...@wpspublish.com > > > > From: wootten.adrie...@gmail.com [mailto:wootten.adrie...@gmail.com] On Behalf > Of Adrienne Wootten > Sent: Friday, October 22, 2010 9:09 AM > To: David Herzberg > Cc: r-help@r-project.org > Subject: Re: [R] Conditional looping over a set of variables in R > > David, > > here I'm referring to your data as testmat, a matrix of 140 columns and 1500 > rows, but the same or similar notation can be applied to data frames in R. If > I understand correctly, you are looking for the first response (column) where > you got a value of 1. I'm assuming also that since your missing values are > characters then your two numeric values are also characters. keeping all this > in mind, try something like this. If you really only want to know which column in each row has first occurrence of 1 (or any other value) you can get rid of looping and use other R capabilities. > set.seed(111) > mat<-matrix(sample(1:3, 20, replace=T),5,4) > mat [,1] [,2] [,3] [,4] [1,] 2 2 2 2 [2,] 3 1 2 1 [3,] 2 2 1 3 [4,] 2 2 1 1 [5,] 2 1 1 2 > mat.w<-which(mat==1, arr.ind=T) > tapply(mat.w[,2], mat.w[,1], min) 2 3 4 5 2 3 3 2 > mat[2, ]<-NA > mat [,1] [,2] [,3] [,4] [1,] 2 2 2 2 [2,] NA NA NA NA [3,] 2 2 1 3 [4,] 2 2 1 1 [5,] 2 1 1 2 and this approach smoothly works with NA values too > mat.w<-which(mat==1, arr.ind=T) > tapply(mat.w[,2], mat.w[,1], min) 3 4 5 3 3 2 You can then use modify such output as you have info about columns and rows. I am sure there are other maybe better options, e.g. lll<-as.list(as.data.frame(t(mat))) > unlist(lapply(lll, function(x) min(which(x==1)))) V1 V2 V3 V4 V5 Inf Inf 3 3 2 Regards Petr > > first = c() # your extra variable which will eventually contain the first > correct response for each case > > for(i in 1:nrow(testmat)){ > > c = 1 > > while( c<=ncol(testmat) | testmat[i,c] != "1" ){ > > if( testmat[i,c] == "1"){ > > first[i] = c > break # will exit the while loop once it finds the first correct answer, and > then jump to the next case > > } else { > > c=c+1 # procede to the next column if not > > } > > } > > } > > > Hope this helps you out a bit. > > Adrienne Wootten > NCSU > > On Fri, Oct 22, 2010 at 11:33 AM, David Herzberg <dav...@wpspublish.com< > mailto:dav...@wpspublish.com>> wrote: > Here's the problem I'm trying to solve in R: I have a data frame that consists > of about 1500 cases (rows) of data from kids who took a test of listening > comprehension. The columns are their scores (1 = correct, 0 = incorrect, . = > missing) on 140 test items. The items are numbered sequentially and are > ordered by increasing difficulty as you go from left to right across the > columns. I want R to go through the data and find the first correct response > for each case. Because of basal and ceiling rules, many cases have missing > data on many items before the first correct response appears. > > For each case, I want R to evaluate the item responses sequentially starting > with item 1. If the score is 0 or missing, proceed to the next item and > evaluate it. If the score is 1, stop the operation for that case, record the > item number of that first correct response in a new variable, proceed to the > next case, and restart the operation. > > In SPSS, this operation would be carried out with LOOP, VECTOR, and DO IF, as > follows (assuming the data set is already loaded): > > * DECLARE A NEW VARIABLE TO HOLD THE ITEM NUMBER OF THE FIRST CORRECT > RESPONSE, SET IT EQUAL TO 0. > numeric LCfirst1. > comp LCfirst1 = 0 > > * DECLARE A VECTOR TO HOLD THE 140 ITEM RESPONSE VARIABLES. > vector x=LC1a_score to LC140a_score. > > * SET UP A LOOP THAT WILL RUN FROM 1 TO 140, AS LONG AS LCfirst1 = 0. "#i" IS > AN INDEX VARIABLE THAT INCREASES BY 1 EACH TIME THE LOOP RUNS. > loop #i=1 to 140 if (LCfirst1 = 0). > > * SET UP A CONDITIONAL TRANSFORMATION THAT IS EVALUATED FOR EACH ELEMENT OF > THE VECTOR. THUS, WHEN #i = 1, THE EXPRESSION EVALUATES THE FIRST ELEMENT OF > THE VECTOR (THAT IS, THE FIRST OF THE 140 ITEM RESPONSES). AS THE LOOP RUNS > AND #i INCREASES, SUBSEQUENT VECTOR ELELMENTS ARE EVALUATED. THE do if > STATEMENT RETAINS CONTROL AND KEEPS LOOPING THROUGH THE VECTOR UNTIL A '1' IS > ENCOUNTERED. > + do if x(#i) = 1. > > * WHEN A '1' IS ENCOUNTERED, CONTROL PASSES TO THE NEXT STATEMENT, WHICH > RECODES THE VALUE OF THAT VECTOR ELEMENT TO '99'. > + comp x(#i) = 99. > > * AND THEN CONTROL PASSES TO THE NEXT STATEMENT, WHICH RECODES THE VALUE OF > LCfirst1 TO THE CURRENT INDEX VALUE, THUS CAPTURING THE ITEM NUMBER OF THE > FIRST CORRECT RESPONSE FOR THAT CASE. CHANGING THE VALUE OF LCfirst1 ALSO > CAUSE S THE LOOP TO STOP EXECUTING FOR THAT CASE, AND THE PROGRAM MOVES TO THE > NEXT CASE AND RESTARTS THE LOOP. > + comp LCfirst1 = #i. > + end if. > end loop. > exe. > > After several hours of trying to translate this procedure to R, I'm stumped. I > played around with creating a list to hold the item responses variables > (analogous to 'vector' in SPSS), but when I tried to use the list in an R > procedure, I kept getting a warning along the lines of 'the list contains > 1 > element, only the first element will be used'. So perhaps a list is not the > appropriate class to 'hold' these variables? > > It seems that some nested arrangement of 'for' 'while' and/or 'lapply' will > allow me to recreate the operation described above? How do I set up the > indexing operation analogous to 'loop #i' in SPSS? > > Any help is appreciated, and I'm happy to provide more information if needed. > > David S. Herzberg, Ph.D. > Vice President, Research and Development > Western Psychological Services > 12031 Wilshire Blvd. > Los Angeles, CA 90025-1251 > Phone: (310)478-2061 x144 > FAX: (310)478-7838 > email: dav...@wpspublish.com<mailto:dav...@wpspublish.com> > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org<mailto:R-help@r-project.org> mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
