Python:如何按特征或属性对对象列表进行分组？

Question

我想将对象列表分成子列表,其中具有相同属性/特征的对象保留在同一子列表中.

假设我们有一个字符串列表:

["This", "is", "a", "sentence", "of", "seven", "words"]

我们希望根据字符串的长度分隔字符串,如下所示:

[['sentence'], ['a'], ['is', 'of'], ['This'], ['seven', 'words']]

我目前提出的计划是这样的

sentence = ["This", "is", "a", "sentence", "of", "seven", "words"]
word_len_dict = {}
for word in sentence:
    if len(word) not in word_len_dict.keys():
        word_len_dict[len(word)] = [word]
    else:
        word_len_dict[len(word)].append(word)


print word_len_dict.values()

我想知道是否有更好的方法来实现这一目标？

Answer 1

使用defaultdict(list)，您可以省略密钥存在检查：

from collections import defaultdict

word_len_dict = defaultdict(list)

for word in sentence:
    word_len_dict[len(word)].append(word)

Answer 2

看看itertools.groupby().请注意,您的列表必须先排序(比您的方法OP贵).

>>> from itertools import groupby
>>> l = ["This", "is", "a", "sentence", "of", "seven", "words"]
>>> print [list(g[1]) for g in groupby(sorted(l, key=len), len)]
[['a'], ['is', 'of'], ['This'], ['seven', 'words'], ['sentence']]

或者如果你想要一本字典 - >

>>> {k:list(g) for k, g in groupby(sorted(l, key=len), len)}
{8: ['sentence'], 1: ['a'], 2: ['is', 'of'], 4: ['This'], 5: ['seven', 'words']}