MRAB wrote:
> On 2019-01-25 22:58, Travis Griggs wrote:
>>
>> grouped = reduce(
>> lambda accum, each: (accum[0],
>> accum[0][str(each)].append(each)), allValues,
>> (defaultdict(list), None))[0]
>>
>> My question, only for the sake of learning python3 fu/enlightenment, is
>> there a simpler way to do this with a reduce? I get there’s lots of way
>> to do a groupby. The pursuit here is what’s the
>> simplest/cleverest/sneakiest way to do it with reduce, especially if the
>> quality that gorupfunc (str() in this example) is only called once per
>> item is persevered.
>>
> How about this:
>
> grouped = lambda iterable, groupfunc: dict(reduce(
> lambda accum, each: accum[groupfunc(each)].append(each) or accum,
> iterable,
> defaultdict(list)))
The
false-expr or ret-val
expression could also be spelt
(false-expr, ret-val)[1]
I suppose the peephole optimiser could be taught to avoid building the
tuple, i. e.
def thelambda(accum, each):
return (accum[groupfunc(each)].append(each), accum)[1]
would be compiled to
def thelambda(accum, each):
accum[groupfunc(each)].append(each)
return accum
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