[R] in which column is an entry?
Prof Brian Ripley
ripley at stats.ox.ac.uk
Sat Jan 31 18:54:40 CET 2004
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/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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