On Mar 17, 2019, at 8:17 PM, Evan Cooch <evan.cooch at gmail.com> wrote:
Solved --
hold=apply(test,1,which.max)
hold[apply(test,1,isUnique)==FALSE] <- 'T'
Now, all I need to do is figure out how to get <- 'T' from turning everything in the matrix to a string.
On 3/17/2019 8:00 PM, Evan Cooch wrote:
Got relatively close - below:
On 3/17/2019 7:39 PM, Evan Cooch wrote:
Suppose I have the following sort of structure:
test <- matrix(c(2,1,1,2,2,2),3,2,byrow=T)
What I need to be able to do is (i) find the maximum value for each row, (ii) find the column containing the max, but (iii) if the maximum value is a tie (in this case, all numbers of the row are the same value), then I want which.max (presumably, a tweaked version of what which.max does) to reurn a T for the row where all values are the same.
Parts (i) and (ii) seem easy enough:
apply(test,1,max) --- gives me the maximum values
apply(test,1,which.max) --- gives me the column
But, standard which.max doesn't handles ties/duplicates in a way that serves my need. It defaults to returning the first column containing the maximum value.
What I'd like to end up with is, ultimately, something where apply(test,1,which.max) yields 1,2,T (rather than 1,2,1).
So, a function which does what which.max currently does if the elements of the row differ, but which returns a T (or some such) if in fact the row values are all the same.
I've tried a bunch of things, to know avail. Closest I got was to use a function to test for whether or not a vector
isUnique <- function(vector){
return(!any(duplicated(vector)))
}
which returns TRUE if values of vector all unique. So
apply(test,1,isUnique)
returns
[1] TRUE TRUE FALSE
but I'm stuck beyond this.
The following gets me pretty close,
test_new <- test
test_new[which(apply(test,1,isUnique)==FALSE),] <- 'T'
but is clunky.