How can we split an integer number into a vector of its constituent digits in R

Question

I think an example should make things clear enough.

I have

a_1 = 6547

and I want some function that transform a_1 into the following a_2

a_2 = c(6, 5, 4, 7)

Solution 1

Convert to character then split will do the trick

a <- 6547
as.numeric(strsplit(as.character(a), "")[[1]])
## [1] 6 5 4 7

Solution 2

(a %% c(1e4, 1e3, 1e2, 1e1)) %/% c(1e3, 1e2, 1e1, 1e0)

This is 3-4x as fast on my computer than doing a strsplit. But the strsplit is a lot more elegant and the difference decreases with longer vectors.

library(microbenchmark)
microbenchmark((a %% c(1e4, 1e3, 1e2, 1e1)) %/% c(1e3, 1e2, 1e1, 1e0))
# median of 1.56 seconds
microbenchmark(as.numeric(strsplit(as.character(a), "")[[1]]))
# median of 8.88 seconds

Edit: using Carl's insight, here is a more general version.

a <- 6547
dig <- ceiling(log10(a))
vec1 <- 10^(dig:1)
vec2 <- vec1/10
(a%%vec1)%/%vec2

Solution 3

This also works. It's slower than the other answers, but possibly easier to read...

library(stringr)
as.integer(unlist(str_split(a, "")))[-1]

Comments

[Carl Witthoft]

Found the link: stackoverflow.com/questions/18786432/…

[Carl Witthoft]

You should rewrite to handle the general case, by calculating log10(a) and using the (truncated or rounded) value to set the maximum power you wish to divide by. I wrote the code for that in some previous answer on SO but can't track it down just now.

[CHP]

Interesting approach. Would be interesting to see if this is faster for very long vectors?

[Arno Kalkman]

This code is actually incorrect and fails for numbers which are a power of 10, (100,1000,10000 etc.). This can easily be explained by the fact that ceiling(log10(1000)) == 3, and not 4. I think the code can be corrected by: dig <- floor(log10(a)+1). It also does not work for 0, but my change does not fix that.