which.max and which.min will return the smallest index of the max or min value if there are ties.
Is there a way around this so that the largest index is returned without affecting the efficiency of the function?
max.col has this exact functionality, but I am dealing with a vector not a matrix.
Answer
You could do like this:
x<-c(1,2,1,4,3,4)
#identical to which.max, except returns all indices with max
which(x==max(x))
[1] 4 6
z<-which(x==max(x))
z[length(z)]
[1] 6
#or with tail
tail(which(x==max(x)),1)
[1] 6
edit:
Or, you could also use max.col function for vectors like this:
max.col(t(x),"last")
[1] 6
#or
max.col(matrix(x,nrow=1),"last")
[1] 6
edit: Some benchmarking:
x<-sample(1:1000,size=10000,replace=TRUE)
library(microbenchmark)
microbenchmark(which.max(x),{z<-which(x==max(x));z[length(z)]},
tail(which(x==max(x)),1),max.col(matrix(x,nrow=1),"last"),
max.col(t(x),"last"),which.max(rev(x)),times=1000)
Unit: microseconds
expr min lq median uq max neval
which.max(x) 29.390 30.323 30.323 31.256 17550.276 1000
{ z <- which(x == max(x)) z[length(z)] } 40.586 42.452 42.919 44.318 631.178 1000
tail(which(x == max(x)), 1) 57.380 60.646 61.579 64.844 596.657 1000
max.col(matrix(x, nrow = 1), "last") 134.353 138.085 139.485 144.383 710.949 1000
max.col(t(x), "last") 116.159 119.425 121.291 125.956 729.610 1000
which.max(rev(x)) 89.569 91.435 92.368 96.566 746.404 1000
So all methods seem to be slower than the original (which gives wrong result), but z <- which(x == max(x));z[length(z)] seems to be fastest option of these.