which(vector1 < vector2) - r

Question

Let's make a small example first, that computes in R:
x<- c(1,3,1,4,2)
max(which(x<2))
[1] 3
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
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))
}
Is there a fast way to do this?

Answer 1

Try this:
vapply(1:6, function(i) max(which(x < i)), double(1))

Answer 2

A fully vectorized approach:
x <- c(1,3,1,4,2)
y <- c(1,2,3,4,5,6)
f <- function(x, y) {
xo <- sort(unique(x))
xi <- cummax(1 + length(x) - match(xo, rev(x)))
xi[cut(y, c(xo, Inf))]
}
f(x,y)
# [1] NA 3 5 5 5 5
The advantages of full vectorization really start to kick in when both x and y are relatively long and each contains many distinct values:
x <- sample(1:1e4)
y <- 1:1e4
microbenchmark(nicola(), frank(), beauvel(), davida(), hallo(), josho(),times=5)
Unit: milliseconds
expr min lq mean median uq max neval cld
nicola() 4927.45918 4980.67901 5031.84199 4991.38240 5052.6861 5207.00330 5 d
frank() 513.05769 513.33547 552.29335 517.65783 540.9536 676.46221 5 b
beauvel() 1091.93823 1114.84647 1167.10033 1121.58251 1161.3828 1345.75158 5 c
davida() 562.71123 575.75352 585.83873 590.90048 597.0284 602.80002 5 b
hallo() 559.11618 574.60667 614.62914 624.19570 641.9639 673.26328 5 b
josho() 36.22829 36.57181 37.37892 37.52677 37.6373 38.93044 5 a

Answer 3

Are you looking for this?
y<-1:6
max.col(outer(y,x,">"),ties.method="last")*NA^(y<=min(x))
#[1] NA 3 5 5 5 5

Answer 4

Find the max index of each value seen in x:
xvals <- unique(x)
xmaxindx <- length(x) - match(xvals,rev(x)) + 1L
Rearrange
xvals <- xvals[order(xmaxindx,decreasing=TRUE)]
xmaxindx <- xmaxindx[order(xmaxindx,decreasing=TRUE)]
# 2 4 1 3
# 5 4 3 2
Select from those:
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
It handily reports the values (in the first row) along with the indices (second row).
The sapply way is simpler and probably not slower:
xmaxindx[sapply(1:6,function(z) which(xvals < z)[1])]
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'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
Of course, comparisons might come out differently for the OP's true case.

Answer 5

You can use Vectorize:
func = Vectorize(function(u) ifelse(length(which(x<u))==0,NA,max(which(x<u))))
> func(1:6)
#[1] NA 3 5 5 5 5