Somehow my other response to the list got lost. I'm still learning Python, but
this seems much better than my first attempt:
def pr(x): print x
def cross(x,y): return [a+b for a in x for b in y]
x=map(pr, reduce(cross, [map(str,range(1,6))]*5))
-j
Stargaming wrote:
> Jason Nordwick schrieb:
>> Or without filter:
>>
>> from operator import add
>> def pr(x): print x
>> def cross(x,y): return reduce(add, [[a+b for b in y] for a in x])
>> x=map(pr, reduce(cross, [map(str,range(1,6))]*5))
>>
> [...]
>
> reduce(add, list) is the same as sum(list) and is only half as fast as sum:
> >>> from timeit import Timer
> >>> sum(Timer("sum(range(500))").repeat(200, 1000))/200
> 0.055693786500798058
> >>> sum(Timer("reduce(add, range(500))", "from operator import
> add").repeat(200, 1000))/200
> 0.10820861031220445
> >>>
>
> Also note that reduce will be removed in Python 3000.
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