Python list of tuples merge 2nd element with unique first element

问题

Given a list of tuples like so:

a = [ ( "x", 1, ), ( "x", 2, ), ( "y", 1, ), ( "y", 3, ), ( "y", 4, ) ]

What would be the easiest way to filter for unique first element and merge the second element. An output like so would be desired.

b = [ ( "x", 1, 2 ), ( "y", 1, 3, 4 ) ]

Thanks,

回答1:

You can use a defaultdict:

>>> from collections import defaultdict
>>> d = defaultdict(tuple)
>>> a = [('x', 1), ('x', 2), ('y', 1), ('y', 3), ('y', 4)]
>>> for tup in a:
...     d[tup[0]] += (tup[1],)
...
>>> [tuple(x for y in i for x in y) for i in d.items()]
[('y', 1, 3, 4), ('x', 1, 2)]

回答2:

>>> a = [("x", 1,), ("x", 2,), ("y", 1,), ("y", 3,), ("y", 4,)]
>>> d = {}
>>> for k, v in a:
...     d.setdefault(k, [k]).append(v)
>>> b = map(tuple, d.values())
>>> b
[('y', 1, 3, 4), ('x', 1, 2)]

回答3:

This is what I came up with:

[tuple(list(el) + [q[1] for q in a if q[0]==el]) for el in set([q[0] for q in a])]

回答4:

In addition to previous answers, another one-liner:

>>> a = [ ( "x", 1, ), ( "x", 2, ), ( "y", 1, ), ( "y", 3, ), ( "y", 4, ) ]
>>> from itertools import groupby
>>> [(key,) + tuple(elem for _, elem in group) for key, group in groupby(a, lambda pair: pair[0])]
[('x', 1, 2), ('y', 1, 3, 4)]

回答5:

One way is by using list comprehension expression with itertools.groupby , itertools.chain and operator.itemgetter as:

>>> from itertools import groupby, chain
>>> from operator import itemgetter

>>> my_list = [ ( "x", 1, ), ( "x", 2, ), ( "y", 1, ), ( "y", 3, ), ( "y", 4, ) ]

>>> [set(chain(*i)) for _, i in groupby(sorted(my_list), key=itemgetter(0))]
[set(['x', 2, 1]), set(['y', 1, 3, 4])]

Note: set are unordered in nature, so they won't preserve the position of elements. Do not use set if the position matters.

来源：https://stackoverflow.com/questions/18909280/python-list-of-tuples-merge-2nd-element-with-unique-first-element