CODEWITHSUNDEEP
schemeracket
Fanhanes
Fanhanes

Reputation: 45

count number of element in multiple sublists scheme

i would like to ask you for help with my task i am trying to solve. Function should work like this. Input is sublist ((1 10 250) (1 10 250) (250 10 250) (1 10 255))) and output should be ((1 . 3) (10 . 4) (250 . 4) (255 . 1)) so it output is actually histogram in text format.

I am using this code with implementation of flatten function which is making from sublists one list. But it is counting number of sublists and not every element in sublist.

 (define (run-length-encode lst )
    (define (rle val-lst cur-val cur-cnt acc)
        (if (pair? val-lst)
          (let ((new-val (car val-lst)))
           (if (eq? new-val cur-val)
              (rle (cdr val-lst) cur-val (+ cur-cnt 1) acc)
              (rle (cdr val-lst) new-val 1 (cons (cons cur-cnt cur-val) acc))))
        (cons (cons cur-cnt cur-val) acc)))
          (if (pair? lst)
           (reverse (rle (cdr lst) (car lst) 1 '()))
           '()))

Flatten function: 

(define (flatten lst)
  (if (not (list? lst))
      (list lst)
      (apply append (map flatten lst))))

Output:

> (run-length-encode '((1 10 250) (1 10 250) (250 10 250) (1 10 255)))
(250 10 1 1 250 10 1 1 250 10 250 1 255 10 1 1)

Thanks for help on this. Jan

Upvotes: 1

Views: 242

Answers (1)

Óscar López
Óscar López

Reputation: 235994

In Racket there's a built-in flatten procedure, there's no need to rewrite it. Coupled with good old bagify, we can solve the problem using simple procedure composition - you should avoid trying to do everything in a single procedure, it'll be confusing:

#lang racket

(define (bagify lst)
  (foldl (lambda (key ht)
           (hash-update ht key add1 0))
         #hash() lst))

(define (run-length-encode lst)
  (hash->list
   (bagify (flatten lst))))

It works as expected:

(run-length-encode '((1 10 250) (1 10 250) (250 10 250) (1 10 255)))
=> '((1 . 3) (250 . 4) (10 . 4) (255 . 1))

Upvotes: 1

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