Python 元組列表將第二個元素與唯一的第一個元素合並

Question

給定一個像這樣的元組列表：

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

過濾唯一的第一個元素並合並第二個元素的最簡單方法是什么。 需要這樣的輸出。

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

謝謝，

Answer 1

>>> 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)]

Answer 2

您可以使用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)]

Answer 3

這就是我想出的：

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

Answer 4

除了前面的答案，另一個單行：

>>> 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)]

Answer 5

一種方法是將列表理解表達式與itertools.groupby 、 itertools.chain和operator.itemgetter一起使用，如下所示：

>>> 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])]

注意： set本質上是無序的，因此它們不會保留元素的位置。 如果位置很重要，請不要使用set 。