问题描述
which.max和which.min将返回最大值或最小值的最小索引.
which.max and which.min will return the smallest index of the max or min value if there are ties.
是否可以解决这个问题，以便最大程度地返回索引而不会影响函数的效率?
Is there a way around this so that the largest index is returned without affecting the efficiency of the function?
max.col具有此确切功能，但我要处理的是向量而不是矩阵.
max.col has this exact functionality, but I am dealing with a vector not a matrix.
推荐答案
您可以这样做:
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
或者，您也可以对以下矢量使用max.col函数:
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
一些基准测试:
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
因此所有方法似乎都比原始方法慢(这会导致错误的结果)，但是z <- which(x == max(x));z[length(z)]似乎是这些方法中最快的选择.
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.
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