Sometimes, when rewriting recursive functions as generators, I miss the brevity of return.
"""
Returns a list of all length n strings that can be made out of a's and/or b's.
"""
def ab_star(n):
if n == 0:
return [""]
results = []
for s in ab_star(n - 1):
results.append("a" + s)
results.append("b" + s)
return results
turns into
"""
Generator for all length n strings that can be made out of a's and/or b's.
"""
def ab_star(n):
if n == 0:
yield ""
else:
for s in ab_star(n - 1):
yield "a" + s
yield "b" + s
It's that else that bugs me. I wish there was a way to say "yield, and this is it, so exit the function". Is there a way?
Don't miss return, use it.
You can return right after you yield.
def ab_star(n):
if n == 0:
yield ""
return
for s in ab_star(n - 1):
yield "a" + s
yield "b" + s
An alternative is to use return in both cases, where the first case returns a sequence of length 1, and the second returns a generator-expression:
def ab_star(n):
if n == 0:
return ( "", )
return ( c+s for s in ab_star(n - 1) for c in 'ab' )
This avoidance of yield avoids the limitation that you cannot use both return <value> and yield in the same function.
(This works in your case because your function doesn't have to be a generator. Since you only iterate over the results, it can also return a tuple.)
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