More elegant way to avoid this hacky implementation of single line reduce for grouping a collection?

[MRAB]

On 2019-01-25 22:58, Travis Griggs wrote:
> Yesterday, I was pondering how to implement groupby, more in the vein of how Kotlin, Swift, Objc, Smalltalk do it, where order doesn’t matter. For example:
> 
>      def groupby(iterable, groupfunc):
>          result = defaultdict(list)
> 	for each in iterable:
>              result[groupfunc(each)].append(each)
>          return result
> 
>      original = [1, 2, 3, 4, 5, 1, 2, 4, 2]
>      groupby(original, lambda x: str(x)) ==> {‘1’: [1, 1], ‘2’: [2, 2, 2], ‘3’: [3], ‘4’: [4, 4], ‘5’: [5]}
> 
> Easy enough, but I found myself obsessing about doing it with a reduce. At one point, I lost sight of whether that was even a better idea or not (the above is pretty simple); I just wanted to know if I could do it. My naive attempt didn’t work so well:
> 
>      grouped = reduce(
>          lambda grouper, each: grouper[str(each)].append(each),
>          allValues,
>          defaultdict(list))
> 
> Since the result of the append() function is None, the second reduction fails, because the accumulator ceases to be a dictionary.
> 
> I persisted and came up with the following piece of evil, using a tuple to move the dict reference from reduction to reduction, but also force the (ignored) side effect of updating the same dict:
> 
>      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)))