问题描述
首先让我们举一个小例子，它以R计算:
Let's make a small example first, that computes in R:
x<- c(1,3,1,4,2)
max(which(x<2))
[1] 3
现在，我不仅要对一个值2进行此操作，还要同时对许多值进行此操作.它应该给我这样的东西:
Now, I would like to do this not just for one value 2, but for many values simultaneously. It should give me something like that:
max(which(x<c(1,2,3,4,5,6)))
[1] NA 3 5 5 5 5
我当然可以运行for循环，但这很慢:
Of course I could run a for loop, but that is very slow:
for(i in c(1,2,3,4,5,6)){
test[i]<-max(which(x<i))
}
有快速的方法吗?
推荐答案
查找在x中看到的每个值的最大索引:
Find the max index of each value seen in x:
xvals <- unique(x)
xmaxindx <- length(x) - match(xvals,rev(x)) + 1L
重新排列
xvals <- xvals[order(xmaxindx,decreasing=TRUE)]
xmaxindx <- xmaxindx[order(xmaxindx,decreasing=TRUE)]
# 2 4 1 3
# 5 4 3 2
从这些中选择:
xmaxindx[vapply(1:6,function(z){
ok <- xvals < z
if(length(ok)) which(ok)[1] else NA_integer_
},integer(1))]
# <NA> 1 2 2 2 2
# NA 3 5 5 5 5
它方便地报告值(第一行)和索引(第二行).
It handily reports the values (in the first row) along with the indices (second row).
sapply方法更简单，而且可能不会更慢:
The sapply way is simpler and probably not slower:
xmaxindx[sapply(1:6,function(z) which(xvals < z)[1])]
基准.OP的情况没有得到充分描述，但是无论如何，这里有一些基准:
Benchmarks. The OP's case is not fully described, but here are some benchmarks anyway:
# setup
nicola <- function() max.col(outer(y,x,">"),ties.method="last")*NA^(y<=min(x))
frank <- function(){
xvals <- unique(x)
xmaxindx <- length(x) - match(xvals,rev(x)) + 1L
xvals <- xvals[order(xmaxindx,decreasing=TRUE)]
xmaxindx <- xmaxindx[order(xmaxindx,decreasing=TRUE)]
xmaxindx[vapply(y,function(z){
ok <- xvals < z
if(length(ok)) which(ok)[1] else NA_integer_
},integer(1))]
}
beauvel <- function()
Vectorize(function(u) ifelse(length(which(x<u))==0,NA,max(which(x<u))))(y)
davida <- function() vapply(y, function(i) c(max(which(x < i)),NA)[1], double(1))
hallo <- function(){
test <- vector("integer",length(y))
for(i in y){
test[i]<-max(which(x<i))
}
test
}
josho <- function(){
xo <- sort(unique(x))
xi <- cummax(1L + length(x) - match(xo, rev(x)))
xi[cut(y, c(xo, Inf))]
}
require(microbenchmark)
(@ MrHallo和@DavidArenburg的警告方式与我现在编写它们的方式类似，但是可以解决.)以下是一些结果:
(@MrHallo's and @DavidArenburg's throw a bunch of warnings the way I have them written now, but that could be fixed.) Here are some results:
> x <- sample(1:4,1e6,replace=TRUE)
> y <- 1:6
> microbenchmark(nicola(),frank(),beauvel(),davida(),hallo(),josho(),times=10)
Unit: milliseconds
expr min lq mean median uq max neval
nicola() 76.17992 78.01171 99.75596 98.43919 120.81776 127.63058 10
frank() 25.27245 25.44666 36.41508 28.44055 45.32306 73.66652 10
beauvel() 47.70081 59.47828 67.44918 68.93808 74.12869 95.20936 10
davida() 26.52582 26.55827 33.93855 30.00990 35.55436 57.24119 10
hallo() 26.58186 26.63984 32.68850 28.68163 33.54364 50.49190 10
josho() 25.69634 26.28724 37.95341 30.50828 47.90526 68.30376 10
There were 20 warnings (use warnings() to see them)
>
>
> x <- sample(1:80,1e6,replace=TRUE)
> y <- 1:60
> microbenchmark(nicola(),frank(),beauvel(),davida(),hallo(),josho(),times=10)
Unit: milliseconds
expr min lq mean median uq max neval
nicola() 2341.96795 2395.68816 2446.60612 2481.14602 2496.77128 2504.8117 10
frank() 25.67026 25.81119 42.80353 30.41979 53.19950 123.7467 10
beauvel() 665.26904 686.63822 728.48755 734.04857 753.69499 784.7280 10
davida() 326.79072 359.22803 390.66077 397.50163 420.66266 456.8318 10
hallo() 330.10586 349.40995 380.33538 389.71356 397.76407 443.0808 10
josho() 26.06863 30.76836 35.04775 31.05701 38.84259 57.3946 10
There were 20 warnings (use warnings() to see them)
>
>
> x <- sample(sample(1e5,1e1),1e6,replace=TRUE)
> y <- sample(1e5,1e4)
> microbenchmark(frank(),josho(),times=10)
Unit: milliseconds
expr min lq mean median uq max neval
frank() 69.41371 74.53816 94.41251 89.53743 107.6402 134.01839 10
josho() 35.70584 37.37200 56.42519 54.13120 63.3452 90.42475 10
当然，对于OP的真实情况，比较结果可能会有所不同.
Of course, comparisons might come out differently for the OP's true case.
这篇关于其中(vector1< vector2)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！