which(vector1 < vector2)

问题:

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?

回答1:

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.

回答2:

Try this:

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

回答3:

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   

回答4:

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

回答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