(->> "aaaabbbcca"
     (partition-by identity)
     (map (juxt (comp str first) count)))

(for [p (partition-by identity "aaaabbbcca" )]
  [(-> p first str) (count p)])

The submitter also gave another solution, ``a bit chattier but also more legible:''

It is worth noting that comp is the functional composition, and juxt, with the rough type (a->b * a->c) -> a -> (b * c), is the combinator that takes a tuple of functions and a value, applies each function to it and returns the results as a tuple.

There were three solutions in Ruby -- all one-liners, and rather similar. The first is probably the easiest to understand.

"aaaabbbcca".split("").chunk_while {|a,b| a==b}.map {|a| [a.first, a.size]}

"aaaabbbcca".chars.slice_when {|c,n| c != n }.map {|s| [s[0], s.size]}

'aaaabbbcca'.chars.chunk(&:itself).map{|x,y| [x, y.size]}

Two APL solutions:

(((1∘+ +\) ⊢ ≠ ¯1∘⌽) (((⊃,≢) ⊢) ⌸) ⊢) 'aaaabbbcca'

f←(⊃,≢)¨⊢⊆⍨1+(+\0,2≠/⊢)
⊢x←f'aaaabbbcca'

``I know this is a Clojure post, but out of curiosity I wrote it in Go (what I've been working in lately) just to see if I could solve it quickly. Took about 10 minutes, subtract 3-4 for fighting with runes. It's not particularly elegant, but it works.''

type result struct {
        letter string
        count  int
        }

func main() {
   const input = "aaaabbbcca"
   var ret []result

   currentLetter := string(input[0])
   countCurrentLetter := 1

   for _, elem := range input[1:] {
           elemAsString := string(elem)
           if currentLetter == elemAsString {
                   countCurrentLetter++
           } else {
                   ret = append(ret, result{currentLetter, countCurrentLetter})
                   currentLetter = elemAsString
                   countCurrentLetter = 1
           }
   }
   ret = append(ret, result{currentLetter, countCurrentLetter})
   fmt.Printf("%+v", ret)
}

The salient feature of this and similar solutions is explicit iteration: the for loop.

(for/fold ([p #f]
           [cnt 0]
           [counts null]
           #:result (cdr (reverse (cons `(,p ,cnt) counts))))
          ([c (in-string "aaaabbbcca")])
  (if (equal? p c)
      (values c (add1 cnt) counts)
      (values c 1 (cons `(,p ,cnt) counts))))

The solution is pure functional -- yet closely resembling the Go solution above.

"aaaabbbcca".subst( /(.) $0*/ , {"({$0},{$/.chars})" }, :g )

Raku is the artist formerly known as Perl 6. Perl has always been known for its (ir)regular expressions, whose power is on display.

L = "aaaabbbcca",
lists:reverse(lists:foldl(
    fun
        (Elt, [ {[Elt], N} | R ]) -> [ {[Elt], N + 1} | R];
        (Elt, Acc) -> [{[Elt], 1} | Acc]
    end,
    [],
    L
)).

It is rather similar to the Racket solution. We see the characteristic for Erlang non-linear pattern matching: the variable Elt appears twice in the first pattern.

[(k, len(list(g))) for k, g in itertools.groupby("aaaabbbcca")]

fn run_length_encode(ins: &str) -> Vec<(char, usize)> {
    let mut out = vec![];
    let mut i = ins.chars();
    if let Some(mut c) = i.next() {
        let mut count = 1;
        for new_c in i {
            if new_c == c {
                count += 1;
            } else {
                out.push((c, count));
                count = 1;
                c = new_c;
            }
        }
        out.push((c, count));
    }
    out}

It is quite like the solution in Go -- but cleaner and with a better handling of edge cases. The author did grasp the nature of the problem: run-length encoding.

function rle(itr)
   acc = Tuple{eltype(itr), Int}[]
   x = iterate(itr)
   isnothing(x) && return acc # iterable is empty
   last, state = x
   n = 1
   for v in Iterators.rest(itr, state)
           if last == v
                   n += 1
           else
                   push!(acc, (last, n))
                   last = v
                   n = 1
           end
   end
   push!(acc, (last, n))
   return acc
end

It mirrors the Rust solution above.

main = interact soln

soln = show . reverse . go []

go counts [] = counts
go counts (x:xs) = go (inc x counts) xs

inc c ((x,ct):xs) | c == x = (x,ct+1):xs
inc c xs = (c, 1):xs

Although this solution relies on the explicit recursion rather than fold, it is rather similar to the Erlang solution. (A different person later mentioned another solution, using group).