We can loop (apply
) through the rows (MARGIN=1
) of the logical matrix (df=='Yes'
), convert to 'numeric' index (which
), get the names
and paste
it together with a wrapper toString
which is paste(., collapse=', ')
. We may also need a if/else
logical condition to check if there are any
'Yes' values in a row. If not, it should return NA
.
df$Expected_Result <- apply(df=='Yes', 1, function(x) {
if(any(x)) {
toString(names(which(x)))
}
else NA
})
Or another option would to get the row/column
index with which
by specifying the arr.ind=TRUE
. Grouped by the row
of 'indx' (indx[,1]
), we paste
the column names of 'df' ('val'). If there are some rows missing i.e. without any 'Yes' element, then use ifelse
to create NA
for the missing row.
indx <- which(df=='Yes', arr.ind=TRUE)
val <- tapply(names(df)[indx[,2]], indx[,1], FUN=toString)
df$Expected_Result <- ifelse(seq_len(nrow(df)) %in% names(val), val, NA)
data
df <- structure(list(c1 = c("Yes", "Yes", "No", "Yes", "Yes", "Yes",
"No"), c2 = c("No", "Yes", "Yes", "No", "No", "No", "No"), c3 = c("Yes",
"No", "No", "No", "Yes", "No", "Yes"), c4 = c("No", "No", "No",
"No", "No", "No", "No")), .Names = c("c1", "c2", "c3", "c4"),
class = "data.frame", row.names = c(NA, -7L))