Re: How to mapcar or across a list?

["Pascal J. Bourguignon" <pjb@informatimago.com>]

Marcin Borkowski <mbork@mbork.pl> writes:

> Hi there,
>
> so here's my problem: I have a list of Boolean values, and I want to
> `mapcar' an `or' across it (IOW, I want to test whether at least one of
> them is true).  Of course, (apply #'or my-list) does not work.  Of
> course, I can (cl-reduce (lambda (x y) (or x y)) my-list) -- but is
> there a better method?
>
> BTW, my-list doesn't really exist: it is a result of `mapcar'ing
> a function taking some value and yielding a Boolean value, so bonus
> points if the method does not process the whole list.

It's too late, lisp is not a lazy language.

> (Note: I just noticed that my-list is in fact sorted so that all true
> values (if any) will occur at the beginning anyway, so I can just test
> the first one.  This means that my problem is purely academic, though
> still interesting, I guess.)

First you want to reduce a list of booleans to a single value, so what
makes you thing that mapcar is the right too for that?  There's the word
"map" in "mapcar", didn't you notice?

       map           reduce

    x1 --> r1       x1 -\
    x2 --> r2       x2 --+-> r
    x3 --> r3       x3 -/

Also, mapcar allocates a new list of same length as the original list
for the result, so it is costly.  At least, you could consider mapc.



With mapc (or mapcar) you could perform your task with some cl magic:

    (require 'cl)
    (setf lexical-binding t)

    (defun* or-all (list)
       (mapc (lambda (element)
                (when element
                  (return-from or-all t)))
             list)
       nil)

    (defun* and-all (list)
       (mapc (lambda (element)
                (unless element
                  (return-from and-all nil)))
             list)
       t)

    (or-all '(nil nil nil)) --> nil
    (or-all '(nil t   nil)) --> t 

    (and-all '(t   t   t)) --> t
    (and-all '(t   nil t)) --> nil 


But I fail to see how it's better than:

    (defun* or-all (list)
       (reduce (lambda (a b) (or a b)) list))

    (defun* and-all (list)
       (reduce (lambda (a b) (and a b)) list))


or than just:

   (some     'identity '(nil nil nil)) --> nil
   (some     'identity '(nil t   nil)) --> t
   (some     'identity '(t   t   t  )) --> t

   (every    'identity '(nil nil nil)) --> nil
   (every    'identity '(t   nil t  )) --> nil 
   (every    'identity '(t   t   t  )) --> t

There are also the negations:

   (notevery 'identity '(nil nil nil)) --> t
   (notevery 'identity '(nil t   nil)) --> t
   (notevery 'identity '(t   t   t  )) --> nil

   (notany   'identity '(nil nil nil)) --> t
   (notany   'identity '(t   nil t  )) --> nil 
   (notany   'identity '(t   t   t  )) --> nil


And since your booleans are computed by a function with mapcar, you
could avoid computing this function for all the elements of the original
list, and you could avoid allocating the temporary list by having some
call this function:


So instead of:

    (reduce (lambda (a b) (or a b))
            (mapcar 'complex-and-lengthy-predicate list))

use:

    (some 'complex-and-lengthy-predicate list)



For example:
    (some (lambda (x) (print x) (evenp x)) '(1 2 3 5 7 9 11))
prints:
    1

    2
--> t


Learn more about Common Lisp:
http://www.lispworks.com/documentation/HyperSpec/Front/
http://cliki.net/

-- 
__Pascal Bourguignon__                 http://www.informatimago.com/
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