python - 条件和 itertools.groupby 问题

Question

我正在使用 groupby 来解析单词列表并按长度将它们组织成列表。例如:

from itertools import groupby

words = ['this', 'that', 'them', 'who', 'what', 'where', 'whyfore']

for key, group in groupby(sorted(words, key = len), len):
    print key, list(group)

3 ['who']
4 ['this', 'that', 'them', 'what']
5 ['where']
7 ['whyfore']

获取列表的长度也有效:

for key, group in groupby(sorted(words, key = len), len):
    print len(list(group))

1
4
1
1

问题是，如果我像这样预先在前面加上条件，这就是结果:

for key, group in groupby(sorted(words, key = len), len):
    if len(list(group)) > 1:
        print list(group)

输出:

[]

这是为什么呢？

Answer 1

每个group都是一个可迭代对象，将其转换为列表会耗尽它。您不能将可迭代对象转换为列表两次。

将列表存储为新变量:

for key, group in groupby(sorted(words, key = len), len):
    grouplist = list(group)
    if len(grouplist) > 1:
        print grouplist

现在您只需使用一次可迭代对象:

>>> for key, group in groupby(sorted(words, key = len), len):
...     grouplist = list(group)
...     if len(grouplist) > 1:
...         print grouplist
... 
['this', 'that', 'them', 'what']