Re: Accumulating in hash-table

On 2024-07-22, B. Pym <Nobody447095@here-nor-there.org> wrote:

(defun distribution1 (items values test)
  (let ((table (make-hash-table :test test)))
    (loop for item in items
          for value in values
          do (incf (gethash item table 0) value))
    (let ((items-list nil))
      (maphash (lambda (item sum-value)
                 (push (cons item sum-value) items-list))
               table)
      (sort items-list #'> :key #'cdr))))
 
An example call:
 
CL-USER 58 > (distribution1 '("a" "b" "c" "b" "a" "f" "e" "g"
  "h" "k" "z" "k" "r" "u" "f")
                            '(1 5 8 7 14 8 3 7 9 4 3 21 5 7 9)
                            #'equal)
(("k" . 25) ("f" . 17) ("a" . 15) ("b" . 12) ("h" . 9) ("c" . 8)
 ("g" . 7) ("u" . 7) ("r" . 5) ("e" . 3) ("z" . 3))

>
Gauche Scheme
>
(define (distribution1 items values test)
  (let1 table (make-hash-table test)
    (for-each
      (^(item value)
        (hash-table-update! table item (cut  + value <>) 0))
      items
      values)
    (sort (hash-table->alist table) > cdr)))
>
(distribution1 '("a" "b" "c" "b" "a" "f" "e" "g"
  "h" "k" "z" "k" "r" "u" "f")
  '(1 5 8 7 14 8 3 7 9 4 3 21 5 7 9)
  'equal?)
>
 ===>
(("k" . 25) ("f" . 17) ("a" . 15) ("b" . 12) ("h" . 9) ("c" . 8) ("g" . 7)
 ("u" . 7) ("r" . 5) ("z" . 3) ("e" . 3))

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1> (defun distrib (items values)
     (let ((h (hash)))
       (each ((i items) (v values))
         (inc [h i 0] v))
       [sort (hash-alist h) : car]))
distrib
2> (distrib '("a" "b" "c" "b" "a" "f" "e" "g"
              "h" "k" "z" "k" "r" "u" "f")
            '(1 5 8 7 14 8 3 7 9 4 3 21 5 7 9))
(("a" . 15) ("b" . 12) ("c" . 8) ("e" . 3) ("f" . 17) ("g" . 7)
 ("h" . 9) ("k" . 25) ("r" . 5) ("u" . 7) ("z" . 3))

Look how much the better code looks when you don't have silly
things like a for-each that takes a lambda, and having to
call a function with a functional argument to update a hash
cell.

Also, when you make equal hash tables default, most of the time
it's the right default. You can skip the test arguments and whatnot.

Names like "hash-table->alist" make my eyes bleed.

Oops, I sorted on the wrong thing.

1> (defun distrib (items values)
     (let ((h (hash)))
       (each ((i items) (v values))
         (inc [h i 0] v))
       [sort (hash-alist h) > cdr]))
distrib
2> (distrib '("a" "b" "c" "b" "a" "f" "e" "g"
              "h" "k" "z" "k" "r" "u" "f")
            '(1 5 8 7 14 8 3 7 9 4 3 21 5 7 9))
(("k" . 25) ("f" . 17) ("a" . 15) ("b" . 12) ("h" . 9) ("c" . 8)
 ("u" . 7) ("g" . 7) ("r" . 5) ("z" . 3) ("e" . 3))

Using group-reduce:

1> (defun distrib (items values)
     (flow [group-reduce (hash) car [mapf + use cdr]
                         [mapcar cons items values] 0]
           hash-alist
           (sort @1 > cdr)))
distrib
2> (distrib '("a" "b" "c" "b" "a" "f" "e" "g"
              "h" "k" "z" "k" "r" "u" "f")
            '(1 5 8 7 14 8 3 7 9 4 3 21 5 7 9))
(("k" . 25) ("f" . 17) ("a" . 15) ("b" . 12) ("h" . 9) ("c" . 8)
 ("u" . 7) ("g" . 7) ("r" . 5) ("z" . 3) ("e" . 3))

Using group-reduce on the keys, using pop to get the values,
so we don't cons up list of pairs up-front:

3> (defun distrib (items values)
     (flow [group-reduce (hash) identity [mapf + use (ret (pop values))]
                         items 0]
           hash-alist
           (sort @1 > cdr)))
distrib
4> (distrib '("a" "b" "c" "b" "a" "f" "e" "g"
              "h" "k" "z" "k" "r" "u" "f")
            '(1 5 8 7 14 8 3 7 9 4 3 21 5 7 9))
(("k" . 25) ("f" . 17) ("a" . 15) ("b" . 12) ("h" . 9) ("c" . 8)
 ("u" . 7) ("g" . 7) ("r" . 5) ("z" . 3) ("e" . 3))

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