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

Question

I am testing itertools.groupby() and try to get the groups as lists but can't figure out how to make it work.

using the examples here, in How do I 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 I should get the first group ("animal") as a list ['bear', 'duck']. But I just get an empty list on REPL.

What am I doing wrong?

How should I extract all three groups as lists?

Answer 1

If you just want the groups, without the keys, you need to realize the group generators as you go, per the docs:

Because the source is shared, when the groupby() object is advanced, the previous group is no longer visible. So, if that data is needed later, it should be stored as a list.

This means that when you try to list-ify the groupby generator first using ll = list(g), before converting the individual group generators, all but the last group generator will be invalid/empty.

(Note that list is just one option; a tuple or any other container works too).

So to do it properly, you'd make sure to listify each group generator before moving on to the next:

from operator import itemgetter  # Nicer than ad-hoc lambdas

# Make the key, group generator
gen = groupby(things, key=itemgetter(0))

# Strip the keys; you only care about the group generators
# In Python 2, you'd use future_builtins.map, because a non-generator map would break
groups = map(itemgetter(1), gen)

# Convert them to list one by one before the next group is pulled
groups = map(list, groups)

# And listify the result (to actually run out the generator and get all your
# results, assuming you need them as a list
groups = list(groups)

As a one-liner:

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

or because this many maps gets rather ugly/non-Pythonic, and list comprehensions let us do nifty stuff like unpacking to get named values, we can simplify to:

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

Answer 2

You could use a list comprehension as follows:

from itertools import groupby

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


g = groupby(things, lambda x: x[0])
answer = [list(group[1]) for group in g]
print(answer)

Output

[[('animal', 'bear'), ('animal', 'duck')],
 [('plant', 'cactus')],
 [('vehicle', 'speed boat'), ('vehicle', 'school bus')]]