toy list processing problem: collect similar terms

namekuseijin namekuseijin at gmail.com
Thu Sep 30 14:35:20 CEST 2010


On 29 set, 11:04, w_a_x_man <w_a_x_... at yahoo.com> wrote:
> On Sep 26, 9:24 am, p... at informatimago.com (Pascal J. Bourguignon)
> wrote:
>
>
>
> > Xah Lee <xah... at gmail.com> writes:
> > > here's a interesting toy list processing problem.
>
> > > I have a list of lists, where each sublist is labelled by
> > > a number. I need to collect together the contents of all sublists
> > > sharing
> > > the same label. So if I have the list
>
> > > ((0 a b) (1 c d) (2 e f) (3 g h) (1 i j) (2 k l) (4 m n) (2 o p) (4 q
> > > r) (5 s t))
>
> > > where the first element of each sublist is the label, I need to
> > > produce:
>
> > > output:
> > > ((a b) (c d i j) (e f k l o p) (g h) (m n q r) (s t))
>
> > > a Mathematica solution is here:
> > >http://xahlee.org/UnixResource_dir/writ/notations_mma.html
>
> > > R5RS Scheme lisp solution:
> > >http://xahlee.org/UnixResource_dir/writ/Sourav_Mukherjee_sourav.work_...
> > > by Sourav Mukherjee
>
> > > also, a Common Lisp solution can be found here:
> > >http://groups.google.com/group/comp.lang.lisp/browse_frm/thread/5d1de...
>
> > It's too complex. Just write:
>
> > (let ((list '((0 a b) (1 c d) (2 e f) (3 g h) (1 i j) (2 k l) (4 m n)
> >               (2 o p) (4 q r) (5 s t))))
>
> >   (mapcar (lambda (class) (reduce (function append) class :key (function rest)))
> >            (com.informatimago.common-lisp.list:equivalence-classes list :key (function first)))
>
> >    )
>
> > --> ((S T) (Q R M N) (G H) (O P K L E F) (I J C D) (A B))
>
> > --
> > __Pascal Bourguignon__                    http://www.informatimago.com/
>
> Ruby:
>
> [[0, 'a', 'b'], [1, 'c', 'd'], [2, 'e', 'f'], [3, 'g', 'h'], [1,
> 'i', 'j'], [2, 'k', 'l'], [4, 'm', 'n'], [2, 'o', 'p'], [4, 'q', 'r'],
> [5, 's', 't']].
> group_by{|x| x.first}.values.map{|x| x.map{|y| y[1..-1]}.flatten}
>
>     ==>[["s", "t"], ["a", "b"], ["c", "d", "i", "j"],
>  ["e", "f", "k", "l", "o", "p"],
>  ["g", "h"], ["m", "n", "q", "r"]]

cool, it comes with order all fucked up.  This is something I was
criticized for before, though not all that important to most
functional processing.  Not the case here, though.

here's a scheme version that is hopefully better than the given one:

(define (dig in)

  (if (null? in) '()

    (let* ((n         (first-of-first in))

           (all-n     (filter in (lambda (x)      (eq? n (first x)))))

           (all-but-n (filter in (lambda (x) (not (eq? n (first
x)))))))

       (pair

          (fold all-n
             (lambda (i o) (pair (second i) (pair (third i) o))))

          (dig all-but-n)))))


; given these aliases to lisp n00bs

(define pair cons)

(define first car)

(define rest cdr)

(define first-of-first caar)

(define second cadr)

(define third caddr)


; and these well-known functions
 (non-tail-recursive for benefit of n00bs)
(define (fold ls f) ; AKA reduce

  (if (null? ls) '()

      (f (first ls) (fold (rest ls) f))))


(define (filter ls f)

  (fold ls (lambda (i o) (if (f i) (pair i o) o))))



; testing
(let ((in '((0 a b) (1 c d) (2 e f) (3 g h) (1 i j) (2 k l) (4 m n)
            (2 o p) (4 q r) (5 s t))))
  (display (dig in))
  (newline))




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