python - How to get the groups generated by "groupby()" as lists? -

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i testing itertools.groupby() , try groups lists can't figure out how make work.

using examples here, in how use python's itertools.groupby()?

from itertools import groupby  things = [("animal", "bear"), ("animal", "duck"), ("plant", "cactus"),          ("vehicle", "speed boat"), ("vehicle", "school bus")]

i tried (python 3.5):

g = groupby(things, lambda x: x[0]) ll = list(g) list(tuple(ll[0])[1])

i thought should first group ("animal") list ['bear', 'duck']. empty list on repl.

what doing wrong?

how should extract 3 groups lists?

if want groups, without keys, need realize group generators go, per docs:

because source shared, when groupby() object advanced, previous group no longer visible. so, if data needed later, should stored list.

this means when try list-ify groupby generator first using ll = list(g), before converting individual group generators, last group generator invalid/empty.

(note list 1 option; tuple or other container works too).

so properly, you'd make sure listify each group generator before moving on next:

from operator import itemgetter  # nicer ad-hoc lambdas  # make key, group generator gen = groupby(things, key=itemgetter(0))  # strip keys; care group generators # in python 2, you'd use future_builtins.map, because non-generator map break groups = map(itemgetter(1), gen)  # convert them list 1 one before next group pulled groups = map(list, groups)  # , listify result (to run out generator , # results, assuming need them list groups = list(groups)

as one-liner:

groups = list(map(list, map(itemgetter(1), groupby(things, key=itemgetter(0)))))

or because many maps gets rather ugly/non-pythonic, , list comprehensions let nifty stuff unpacking named values, can simplify to:

groups = [list(g) k, g in groupby(things, key=itemgetter(0))]

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