Stoping generator in first answer, use return instead

Question

I am using too much to my taste the pattern (after every possible solution branch of the search). This is the code to find boggle words in given square. It had a bug if the words are not preselected to include only those whose letter pairs are neighbours, which I fixed now by changing comparrision not pos to pos is None.

def word_path(word,used=[],pos=None):
    if not word:
        yield False
        return
    else:
        correct_neighbour = [neigh for p,neigh in neighbour_set
                  if (not pos or pos==p) and (neigh not in used) and boggle[neigh]==word[0] ]
        for i in correct_neighbour:
            used_copy=used[:]+[i]
            if boggle[i]==word:
                yield used_copy
                return
            else:
                for solution in  word_path(word[1:],used_copy,pos=i) or (False,):
                    if solution:
                        yield solution
                    return

Is there any better alternative to make generator which stops after any answer was found?

Solution based on why not use return

Finally it got me and returned sequence is iterator no matter that it was returned not yielded value. so I changed my word_path code to use return and cleaned up the expressions generally. Instead of giving None or False the function returns (False,). Then I have not problem with None not accepted for for statement.

def word_path(word,used=[],pos=None):
if word:
    correct_neighbour = [neigh
                         for p,neigh in neighbour_set
                         if ((pos is None or pos==p) and
                             (neigh not in used) and
                             boggle[neigh]==word[0]
                             )
                         ]
    for i in correct_neighbour:
        used_copy=used[:]+[i]
        if len(word)==1:
            if boggle[i]==word:
                return (used_copy,)
        else:
            for solution in  word_path(word[1:],used_copy,pos=i):
                if solution:
                    return (solution,)
return (False,)

possible duplicate of [Yielding until all needed values are yielded](http://stackoverflow.com/questions/3324947/yielding-until-all-needed-values-are-yielded) — S.Lott, Jul 24 '10 at 12:21
@S.Lott: It isn't a duplicate. This question wants to stop producing at some point, whereas the alleged duplicate wants the caller to stop the generator when it wants to stop consuming. — Marcelo Cantos, Jul 24 '10 at 12:28
I'm not sure I understand your problem. While not exactly common, it is perfectly normal and OK to return immediately after a yield to signal that there are no more values forthcoming. — Marcelo Cantos, Jul 24 '10 at 12:30

score 3 · Accepted Answer · answered Jul 24 '10 at 13:56

3

Why do you make it a generator in the first place when you only want one answer? Just search for answers and return the first one instead of yielding it.

answered Jul 24 '10 at 13:56

score 1 · Answer 2 · answered Jul 24 '10 at 13:20

1

return iter([anwser])

answered Jul 24 '10 at 13:20

Tomasz Wysocki

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Stoping generator in first answer, use return instead

2 Answers2