Alternative to Yield in Python

Question

Is there an alternative (even if longer) method of writing the yield part of this code? I'm not very familiar with the function and would rather keep it simple.

for i in range(0, len(lstInput) - intCount + 1):
    if intCount == 1:
        yield [lstInput[i]]
    else:
        current = lstInput[i]
        remainder = lstInput[i+1:]
        for rest in yieldLotto(remainder, intCount - 1):
            yield [current] + rest

[How about instead you learn what `yield` does?](http://stackoverflow.com/questions/231767/the-python-yield-keyword-explained) — jwodder, May 27 '14 at 04:47
Seriously, though, `yield` is not complex. Just take the 20 minutes to learn it completely and you'll never worry about it again. @jwodder's SO link is perfect and even has some links to slideshows if I remember correctly. (It's how I learned yield). — woot, May 27 '14 at 04:52

score 3 · Accepted Answer · answered May 27 '14 at 04:47

The alternative is to either embed the loops into your calling program, or to change it to return a list. This means everything will be going into memory, though.

def foo():
    return_list = []
    for i in range(0, len(lstInput) - intCount + 1):
        if intCount == 1:
            return_list.append([lstInput[i]])
        else:
            current = lstInput[i]
            remainder = lstInput[i+1:]
            for rest in yieldLotto(remainder, intCount - 1):
                return_list.append([current] + rest)
    return return_list

Honestly though, I think yield is better and it's an important feature of Python. I suggest learning it.

Alternative to Yield in Python

1 Answers1