Tim Chase wrote: > On 2013-06-07 23:46, Jason Swails wrote: >> On Fri, Jun 7, 2013 at 10:32 PM, Tim Chase >> > def calculate(params): >> > a = b = 0 >> > if some_calculation(params): >> > a += 1 >> > if other_calculation(params): >> > b += 1 >> > return (a, b) >> > >> > alpha = beta = 0 >> > temp_a, temp_b = calculate(...) >> > alpha += temp_a >> > beta += temp_b >> > >> > Is there a better way to do this without holding each temporary >> > result before using it to increment? >> >> alpha = beta = 0 >> alpha, beta = (sum(x) for x in zip( (alpha, beta), >> calculate(...) ) ) > > Yeah, I came up with something similar, but it was so opaque I fell > back to the more readable version I had above. With only the two > variables to increment in this case, the overhead of duplicating code > doesn't seem as worth it as it might be if there were umpteen > counters being returned as a tuple and umpteen corresponding > variables being updated (at which point, it might make more sense to > switch to another data structure like a dict). Ah well. Glad to see > at least I'm not the only one stymied by trying to make it more > pythonic while at least keeping it readable. > > -tkc
You can hide the complexity in a custom class: >>> class T(tuple): ... def __add__(self, other): ... return T((a+b) for a, b in zip(self, other)) ... >>> t = T((0, 0)) >>> for pair in [(1, 10), (2, 20), (3, 30)]: ... t += pair ... >>> t (6, 60) (If you are already using numpy you can do the above with a numpy.array instead of writing your own T.) -- http://mail.python.org/mailman/listinfo/python-list
