On Aug 29, 2011, at 8:17 PM, David Winsemius wrote:
On Aug 29, 2011, at 8:02 PM, array chip wrote:
Hi, I had a weird results from using apply(). Here is an simple
example:
y<-data.frame(list(a=c(1,NA),b=c('2k','0')))
y
a b
1 1 2k
2 NA 0
apply(y,1,function(x){x<-unlist(x);
That is quite unnecessary since apply coerces the rows into vectors
whether you want it to or not,
if (!is.na(x[2]) & x[2]=='2k' & !is.na(x[1]) & x[1]=='1') 1 else
0} )
So everything get coerced to character and I suspect NA -> "NA" or
NA_character and so !is.na() gives the wrong result. Try printing
str(x) inside that apply loop.
I was wrong abut my diagnosis but my methods were sound.
This is str from the first row:
Named chr [1:2] " 1" "2k"
So probably the default stringsAsFactors rule created "b" as a factor
and something else made that integer `1` turn into " 1" .
--
DA\avid.
--
David.
This should print "1 0" as output, as demonstrated by:
apply(y[1,],1,function(x){x<-unlist(x); if (!is.na(x[2]) &
x[2]=='2k' & !is.na(x[1]) & x[1]=='1') 1 else 0} )
1
1
apply(y[2,],1,function(x){x<-unlist(x); if (!is.na(x[2]) &
x[2]=='2k' & !is.na(x[1]) & x[1]=='1') 1 else 0} )
2
0
But it actually prints:
apply(y,1,function(x){x<-unlist(x); if (!is.na(x[2]) & x[2]=='2k'
& !is.na(x[1]) & x[1]=='1') 1 else 0} )
[1] 0 0
Anyone has any suggestion?
Thanks
John
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David Winsemius, MD
West Hartford, CT
______________________________________________
[email protected] 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.
David Winsemius, MD
West Hartford, CT
______________________________________________
[email protected] 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.
