[R] in which column is an entry?

[Prof Brian Ripley]

On Sat, 31 Jan 2004, Christian Schulz wrote:

> 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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-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
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