Re: remove-duplicates performances

[Thierry Volpiatto]

David Kastrup <address@hidden> writes:

> Thierry Volpiatto <address@hidden> writes:
>
>> David Kastrup <address@hidden> writes:
>>
>>> I've found the following in some file of mine:
>>>
>>> (defun uniquify (list predicate)
>>>   (let* ((p list) lst (x1 (make-symbol "x1"))
>>>      (x2 (make-symbol "x2")))
>>>     (while p
>>>       (push p lst)
>>>       (setq p (cdr p)))
>>> ;;;    (princ lst)(princ "\n")
>>>     (setq lst
>>>       (sort lst `(lambda(,x1 ,x2)
>>>                    (funcall ',predicate (car ,x1) (car ,x2)))))
>>> ;;; lst now contains all sorted sublists, with equal cars being
>>> ;;; sorted in order of increasing length (from end of list to start).
>>> ;;
>>>
>>>     (while (cdr lst)
>>>       (unless (funcall predicate (car (car lst)) (car (cadr lst)))
>>>     (setcar (car lst) x1))
>>>       (setq lst (cdr lst)))
>>>     (delq x1 list)))
>>>
>>> (uniquify '(2 1 2 1 2) '<)
>>> (uniquify '(4 7 3 26 4 2 6 24 4 5 2 3 2 4 6) '<)
>>
>> This is nice and very instructive (at least for me) thanks.
>> It is not as performant as the version with hash-table,
>
> Well, the sorting function is a mess due to not being compiled and
> fearing dynamic binding.  If you byte-compile something like
>
> (defun uniquify (list predicate)
>   (let* ((p list) lst (sentinel (list nil)))
>     (while p
>       (push p lst)
>       (setq p (cdr p)))
>     (setq lst
>       (sort lst (lambda(x1 x2)
>                      (funcall predicate (car x1) (car x2)))))
> ;;; lst now contains all sorted sublists, with equal cars being
> ;;; sorted in order of increasing length (from end of list to start).
> ;;
>     (while (cdr lst)
>       (unless (funcall predicate (car (car lst)) (car (cadr lst)))
>       (setcar (car lst) sentinel))
>       (setq lst (cdr lst)))
>     (delq sentinel list)))
>
> the behavior is likely better.
Yes, that's not really a problem, the results are very acceptable
compared to remove-duplicates (68s!).

>> but very usable: 0.3 <=> 0.13 with same test on list with 20000
>> elements.  However, isn't it a problem when we want to remove
>> duplicate in a list type alist e.g ((a . 1) (b . 2) (a . 1) (c . 3) (b
>> . 2)...)
>
> Why?  You need a predicate < both for sorting and for telling
> inequality.  As long as you define a suitable predicate for that
> purpose, what should go wrong?  Any elements for which
> (or (predicate a b) (predicate b a)) is nil will be considered
> duplicate.
Yes, i understand that, what i mean is you have to write a predicate
each time, which could be inconvenient, instead of using :test 'equal.

-- 
A+ Thierry
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