On Mar 11, 2006, at 11:21 AM, Peter Otten wrote: > Am Freitag, 10. März 2006 19:38 schrieben Sie: > >>>> item = mylist.pop(random.randint(0,len(mylist))) >>> >>> This is broken because randint(a, b) may return b. >>> I prefer randrange(len(mylist)) over randint(0, len(mylist)-1) as >>> a fix. >> >> This brings up an interesting proposal. >> random.choice(seq) brings back a random element from a list, why not >> add an optional second argument which is a flag to pop the element >> instead of choosing? >> >> ie. >> >>>>> import random >>>>> def choice(seq, pop=False): >> >> ... if not pop: >> ... return seq[random.randrange(len(seq))] >> ... else: >> ... return seq.pop(random.randrange(len(seq))) >> ... >> >>>>> x = [1, 2, 3] >>>>> choice(x) >> >> 1 >> >>>>> x >> >> [1, 2, 3] >> >>>>> choice(x, True) >> >> 1 >> >>>>> x >> >> [2, 3] > > [The main reason I am answering your mail is because you may have > intended to > post on c.l.py] > > Regarding your enhancement, I don't see any use cases that aren't > handled by > random.sample() already. > > Regards, > Peter
I can see a use case. Think of a bag datastructure. You push things into some container and pop them out randomly. If random.choice was capable of 'pop' it would be implemented implicitly. random.sample, select elements from a list, but the original list remains intact. This would not be the desired 'bag' behavior. --- Andrew Gwozdziewycz [EMAIL PROTECTED] http://ihadagreatview.org http://and.rovir.us -- http://mail.python.org/mailman/listinfo/python-list
