in which column is an entry?

Yes, many thanks i have really to  avoid think in loops  :-)
Unfortunately Chuck's solution is a loop over rows, disguised by the use 
of apply.

Let us assume that the dataframe has all numeric entries and coerce to a 
matrix (as apply will, BTW).

tmp <- as.matrix(df)
tmp[is.na(tmp)] <- -1   # get rid of the NAs
tmp <- tmp >= 0         # a logical matrix
tmp <- cbind(tmp, TRUE) # add a fence column

I am happy to loop over 43 columns, though, so

for(i in 2:44) tmp[, i] <- tmp[, i] | tmp[, i-1]
for(i in 44:2) tmp[, i] <- tmp[, i] & !tmp[, i-1]
rtmp <- t(tmp)
z <- row(rtmp)[rtmp]
z[z==44] <- NA
z

is what you want.  It's a lot faster (about 12x).
christian

Am Samstag, 31. Januar 2004 16:57 schrieb Chuck Cleland:
Christian Schulz wrote:
df  is a data.frame with 43 colums and 29877 rows with lot of NA.
I want the column number for all respondendts in one column
where is the first entry >=0 as columnnumber.

my first step:
 time <- function(df)
+     {        for (i in 1:length(df [,1])) {
+     which(df[i,1]:df[i,43] >= 0)
+     }
+            }

t1 <-  time(YS)
Error in df[i, 1]:df[i, 43] : NA/NaN argument
   I am not sure, but I think you might want something like this:

t1 <- apply(df, 1, function(x){
             ifelse(all(is.na(x)) | all(na.omit(x) < 0),
             NA, which(x >= 0))})

hope this helps,

Chuck Cleland

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in which column is an entry?

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