How to vectorize this operation on every row of a matrix

Question

I have a matrix filled with TRUE/FALSE values and I am trying to find the index position of the first TRUE value on each row (or retur

Answer 1

Not sure this is any better, but this is one solution:

> x2 <- t(t(matrix(as.numeric(x), nrow=10)) * 1:3)
> x2[x2 == 0] <- Inf
> rowMins(x2)
 [1] 2 1 1 2 1 1 2 1 1 2

Edit: Here's a better solution using base R:

> x2 <- (x2 <- which(x, arr=TRUE))[order(x2[,1]),]
> x2[as.logical(c(1,diff(x2[,1]) != 0)),2]
 [1] 2 1 1 2 1 1 2 1 1 2

Answer 2

A couple of years later, I want to add two alternative approaches.

1) With max.col:

> max.col(x, "first")
 [1] 2 1 1 2 1 1 2 1 1 2

2) With aggregate:

> aggregate(col ~ row, data = which(x, arr.ind = TRUE), FUN = min)$col
 [1] 2 1 1 2 1 1 2 1 1 2

---

As performance is an issue, let's test the different methods on a larger dataset. First create a function for each method:

abiel <- function(n){apply(n, 1, function(y) which(y)[1])}
maxcol <- function(n){max.col(n, "first")}
aggr.min <- function(n){aggregate(col ~ row, data = which(n, arr.ind = TRUE), FUN = min)$col}
shane.bR <- function(n){x2 <- (x2 <- which(n, arr=TRUE))[order(x2[,1]),]; x2[as.logical(c(1,diff(x2[,1]) != 0)),2]}
joris <- function(n){z <- which(t(n))-1;((z%%ncol(n))+1)[match(1:nrow(n), (z%/%ncol(n))+1)]}

Second, create a larger dataset:

xl <- matrix(sample(c(F,T),9e5,replace=TRUE), nrow=1e5)

Third, run the benchmark:

library(microbenchmark)
microbenchmark(abiel(xl), maxcol(xl), aggr.min(xl), shane.bR(xl), joris(xl),
               unit = 'relative')

which results in:

Unit: relative
         expr        min         lq       mean     median         uq       max neval   cld
    abiel(xl)  55.102815  33.458994  15.781460  33.243576  33.196486  2.911675   100    d 
   maxcol(xl)   1.000000   1.000000   1.000000   1.000000   1.000000  1.000000   100 a    
 aggr.min(xl) 439.863935 262.595535 118.436328 263.387427 256.815607 16.709754   100     e
 shane.bR(xl)  12.477856   8.522470   7.389083  13.549351  24.626431  1.748501   100   c  
    joris(xl)   7.922274   5.449662   4.418423   5.964554   9.855588  1.491417   100  b   

Answer 3

You can gain a lot of speed by using %% and %/%:

x <- matrix(rep(c(F,T,T),10), nrow=10)

z <- which(t(x))-1
((z%%ncol(x))+1)[match(1:nrow(x), (z%/%ncol(x))+1)]

This can be adapted as needed: if you want to do this for columns, you don't have to transpose the matrix.

Tried out on a 1,000,000 X 5 matrix :

x <- matrix(sample(c(F,T),5000000,replace=T), ncol=5)

system.time(apply(x,1,function(y) which(y)[1]))

#>   user  system elapsed 
#>  12.61    0.07   12.70 

system.time({
 z <- which(t(x))-1
 (z%%ncol(x)+1)[match(1:nrow(x), (z%/%ncol(x))+1)]}
)

#>   user  system elapsed 
#>   1.11    0.00    1.11 

You could gain quite a lot this way.