Displaying 1 result from an estimated 1 matches for "first_lett".
Did you mean:
first_letter
2005 Jul 04
1
compare two lists with differents levels
...e=r.ID and r.ID=pr.release and pr.product=p.ID and r.ID='",param2,"';",sep=""))
data_NA<-matrix(data=0,nrow=length(release$Title),ncol=length(formv$TitleCrit),byrow=TRUE, dimnames=list(as.factor(release$Title),paste(as.factor(formv$TitleCrit),"(",formv$first_letter,")",sep="")))
for (i in 1:length(formv$TitleCrit))
{
for(k in 1: length(release1$TitleCrit))
{
if(formv$TitleCrit[i]==release1$TitleCrit[k])
{
data_NA[1,formv$TitleCrit[[i]]]<-1
}
else{data_NA[1,formv$TitleCrit[[i]]]&...