Check whether two strings are anagrams

Question

What is the most straightforward way to check whether two strings are anagrams? i.e. they share the same letters as well as number of occurrences of each these letters (and possibly other characters).

Something like this:

s1 = "elbow"
s2 = "below"

is_anagram(s1, s2)
# [1] TRUE

Answer 1

One way to do it is:

s1 = "elbow"
s2 = "below"

is_anagram <- function(s1, s2){
  s1_sorted <- sort(strsplit(s1, "")[[1]])
  s2_sorted <- sort(strsplit(s2, "")[[1]])
  identical(s1_sorted, s2_sorted)
}

#> is_anagram(s1, s2)
#> [1] TRUE

Answer 2

You can try the code below

> do.call(identical,Map(function(x) sort(utf8ToInt(x)), list(s1, s2)))
[1] TRUE

---

If you want to generalize the case to more than two strings, e.g.,

s1 <- "elbow"
s2 <- "below"
s3 <- "owlbe"

then we can try

> lst <- list(s1, s2, s3)

> all(apply(table(stack(Map(utf8ToInt, setNames(lst, seq_along(lst))))), 1, var) == 0)
[1] TRUE

or

> lst <- list(s1, s2, s3)

> m <- as.data.frame.matrix(table(stack(Map(utf8ToInt, setNames(lst, seq_along(lst))))))

> identical(do.call(pmin, m), do.call(pmax, m))
[1] TRUE

Answer 3

A simple way to achieve that:

library(tidyverse)

s1 = "elbow"
s2 = "below"

is_anagram <- function(s1, s2){
  identical(str_split(s1, "") %>% table, str_split(s2, "") %>% table)
}

is_anagram(s1, s2)

#> [1] TRUE