Error in for loop

Question

k<-21;

for(i in 5:k)
{
pharma[,i][pharma[,i]=="#N/A"]<- NA
pharma[,i][pharma[,i]=="NM"]<- NA
num<-sum(is.na(pharma[,i]))
n=1-num/length(pharma[,i])

if(n<0.8) {
rm(pharma[,i])
Else n=0
}
}

Basically trying to replace columns with NA and removing those where there are too many NA.

Can you expand the question to explain where the error is arising? — James, Apr 05 '13 at 12:16
Read this! http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example — Richie Cotton, Apr 05 '13 at 13:06
Folks, sorry was busy at work. But nailed the error. below are the codes: for(i in c(5:18)){ pharma[,i][pharma[,i]== "#N/A"]<- NA pharma[,i][pharma[,i]== "NM"]<- NA num<-sum(is.na(pharma[,i])) n<-1-num/length(pharma[,i]) if(n >0.95) pharma1<-merge(m,pharma[,c(1,4,i)],all=TRUE) m<-pharma1 } — vj.vijay, Apr 10 '13 at 11:10

score 2 · Answer 1 · answered Apr 05 '13 at 12:20

2

You did not tell us the error the code creates. But several observations:

R is case sensitive. Else is not the same as else and Else is not correct
You do not close the if statement before the else.
There is no need to loop over the columns explicitly

answered Apr 05 '13 at 12:20

Paul Hiemstra

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Thanks Paul, found a way through, below is the new code: for(i in c(5:18)){ pharma[,i][pharma[,i]== "#N/A"]<- NA pharma[,i][pharma[,i]== "NM"]<- NA num<-sum(is.na(pharma[,i])) n<-1-num/length(pharma[,i]) if(n >0.95) pharma1<-merge(m,pharma[,c(1,4,i)],all=TRUE) m<-pharma1 } – vj.vijay Apr 10 '13 at 11:14
1

If this answers your question, I would add it as a separate answer, describe what you did to fix the situation, post the code, and accept it as the correct answer. – Paul Hiemstra Apr 10 '13 at 11:22

score 1 · Answer 2 · answered Apr 05 '13 at 12:29

You probably want something like

## extract the columns to manipulate
pp <- pharma[,5:21]
## set relevant values to NA
pp <- lapply(pp,function(x) x[x %in% c("#N/A","NM")] <- NA)
## estimate fraction NA and test
badcols <- colMeans(is.na(pp))>0.2
## remove bad columns
pp <- pp[,!badcols]
## put the manipulated stuff back together with the original structure
pharma <- cbind(pharma[,1:4],pp)

but it's hard to tell exactly without a reproducible example.

Error in for loop

2 Answers2