Kate — here is a transparent solution (tested but without NA treatment).
Doubtless there are cleverer faster ones, which later posters will present.
HTH
# example with four columns and 20 rows
nrows <- 20
A <- sample(c(1:100), nrows, replace=T)
B <- sample(c(1:100), nrows, replace=T)
C <- sample(c(1:100), nrows, replace=T)
D <- sample(c(1:100), nrows, replace=T)
locs <-
c(c(1:nrows)[A==max(A)],c(1:nrows)[B==max(B)],c(1:nrows)[C==max(C)],c(1:nrows)[D==max(D)])
mat1 <- matrix(rep(0,4*nrows),nrows,4)
for (i in 1:4)
mat1[,i][locs[i]] <- 1
SUM <- rowSums(mat1)
> On Feb 26, 2015, at 12:23 PM, Kate Ignatius <kate.ignat...@gmail.com> wrote:
>
> Hi,
>
> Supposed I had a data frame like so:
>
> A B C D
> 0 1 0 7
> 0 2 0 7
> 0 3 0 7
> 0 4 0 7
> 0 1 0 0
> 0 0 0 0
> 0 0 0 0
> 0 0 0 0
> 0 0 1 5
> 0 5 1 5
> 0 4 1 5
> 0 8 4 7
> 0 0 3 0
> 0 0 3 4
> 0 0 3 4
> 0 0 0 5
> 0 2 0 6
> 0 0 4 0
> 0 0 4 0
> 0 0 4 0
>
> For each row, I want to count how many max column values appear to
> adventurely get the following outcome, while ignoring zeros and N/As:
>
> A B C D Sum
> 0 1 0 7 1
> 0 2 0 7 1
> 0 3 0 7 1
> 0 4 0 7 1
> 0 1 0 0 0
> 0 0 0 0 0
> 0 0 0 0 0
> 0 0 0 0 0
> 0 0 1 5 0
> 0 5 1 5 0
> 0 4 1 5 0
> 0 8 4 7 3
> 0 0 3 0 0
> 0 0 3 4 0
> 0 0 3 4 0
> 0 0 0 5 0
> 0 2 0 6 0
> 0 0 4 0 1
> 0 0 4 0 1
> 0 0 4 0 1
>
> I've used the following code but it doesn't seem to work (my sum
> column column is all 1s):
>
> (apply(df,1, function(x) (sum(x %in% c(pmax(x))))))
>
> Is this code too simple?
>
> Thanks!
>
> K.
>
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