def groupby(key, seq):

""" Group a collection by a key function

>>> names = ['Alice', 'Bob', 'Charlie', 'Dan', 'Edith', 'Frank']

>>> groupby(len, names) # doctest: +SKIP

{3: ['Bob', 'Dan'], 5: ['Alice', 'Edith', 'Frank'], 7: ['Charlie']}

>>> iseven = lambda x: x % 2 == 0

>>> groupby(iseven, [1, 2, 3, 4, 5, 6, 7, 8]) # doctest: +SKIP

{False: [1, 3, 5, 7], True: [2, 4, 6, 8]}

Non-callable keys imply grouping on a member.

>>> groupby('gender', [{'name': 'Alice', 'gender': 'F'},

... {'name': 'Bob', 'gender': 'M'},

... {'name': 'Charlie', 'gender': 'M'}]) # doctest:+SKIP

{'F': [{'gender': 'F', 'name': 'Alice'}],

'M': [{'gender': 'M', 'name': 'Bob'},

{'gender': 'M', 'name': 'Charlie'}]}

See Also:

countby

"""

if not callable(key):

key = getter(key)

d = collections.defaultdict(lambda: [].append)

for item in seq:

d[key(item)](item)

rv = {}

for k, v in iteritems(d):

rv[k] = v.__self__

return rv

On Thursday, June 28, 2018, Chris Barker via Python-ideas <python-ideas@python.org> wrote:

On Thu, Jun 28, 2018 at 8:25 AM, Nicolas Rolin <nicolas.rolin@tiime.fr> wrote:
I use list and dict comprehension a lot, and a problem I often have is to do the equivalent of a group_by operation (to use sql terminology).

I don't know from SQL, so "group by" doesn't mean anything to me, but this:

For example if I have a list of tuples (student, school) and I want to have the list of students by school the only option I'm left with is to write

    student_by_school = defaultdict(list)
    for student, school in student_school_list:
        student_by_school[school].append(student)

seems to me that the issue here is that there is not way to have a "defaultdict comprehension"

I can't think of syntactically clean way to make that possible, though.

Could itertools.groupby help here? It seems to work, but boy! it's ugly:

In [45]: student_school_list

Out[45]:

[('Fred', 'SchoolA'),

('Bob', 'SchoolB'),

('Mary', 'SchoolA'),

('Jane', 'SchoolB'),

('Nancy', 'SchoolC')]

In [46]: {a:[t[0] for t in b] for a,b in groupby(sorted(student_school_list, key=lambda t: t[1]), key=lambda t: t[

...: 1])}

...:

...:

...:

...:

...:

...:

...:

Out[46]: {'SchoolA': ['Fred', 'Mary'], 'SchoolB': ['Bob', 'Jane'], 'SchoolC': ['Nancy']}

-CHB

--

Christopher Barker, Ph.D.
Oceanographer

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