Question

我有一个如下所示的列表：

[
  [1, "a", 319],
  [1, "b", 991],
  [2, "Cd", 19],
  [3, "88x", 2891],
  [3, "foo", 11]
]

最多＆＃34; pythonic＆＃34;将此转化为：

[
  [1, [["a", 319], ["b", 991]]],
  [2, [["Cd", 19]]],
  [3, [["88x", 2891], ["foo", 11]]]
]

我知道如何使用for循环这样做，但是我采用了更清洁的方法。

Answer 1

这是一个选项，假设列表按照第一个元素排序：

from itertools import groupby    
[[k, [x[1:] for x in g]] for k, g in groupby(lst, key = lambda x: x[0])]

#[[1, [['a', 319], ['b', 991]]],
# [2, [['Cd', 19]]],
# [3, [['88x', 2891], ['foo', 11]]]]

Answer 2

如果您不介意类似字典的对象，也可以使用collections.defaultdict：

>>> l = [
...   [1, "a", 319],
...   [1, "b", 991],
...   [2, "Cd", 19],
...   [3, "88x", 2891],
...   [3, "foo", 11]
... ]
>>> import collections
>>> d = collections.defaultdict(list)
>>> for k, *v in l:
...     d[k].append(v)
...
>>> import pprint
>>> pprint.pprint(d)
{1: [['a', 319], ['b', 991]],
 2: [['Cd', 19]],
 3: [['88x', 2891], ['foo', 11]]}

Pythonic按键分组列表的方法

2 个答案: