按字长对列表进行分组

Question

例如,我有一个清单,比方说

list = ['sight', 'first', 'love', 'was', 'at', 'It']

我想按字长对这个列表进行分组

newlist = [['sight', 'first'],['love'], ['was'], ['at', 'It']]

请帮帮我.升值!

Answer 1

用途itertools.groupby:

>>> from itertools import groupby
>>> lis = ['sight', 'first', 'love', 'was', 'at', 'It']
>>> [list(g) for k, g in groupby(lis, key=len)]
[['sight', 'first'], ['love'], ['was'], ['at', 'It']]

请注意,为了itertools.groupby正常工作,所有项目必须按长度排序,否则请先使用collections.defaultdict(O(N))或排序列表,然后使用itertools.groupby(O(NlogN)).:

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> lis = ['sight', 'first', 'foo', 'love', 'at', 'was', 'at', 'It']
>>> for x in lis:
...     d[len(x)].append(x)
...     
>>> d.values()
[['at', 'at', 'It'], ['foo', 'was'], ['love'], ['sight', 'first']]

如果您希望对最终输出列表进行排序,则可以按长度更好地对列表项进行排序并应用于该列表项itertools.groupby.

Answer 2

您可以使用临时字典然后按长度排序：

li=['sight', 'first', 'love', 'was', 'at', 'It']

d={}
for word in li:
    d.setdefault(len(word), []).append(word)

result=[d[n] for n in sorted(d, reverse=True)] 

print result  
# [['sight', 'first'], ['love'], ['was'], ['at', 'It']]

您可以使用 defaultdict：

from collections import defaultdict
d=defaultdict(list)
for word in li:
    d[len(word)].append(word)

result=[d[n] for n in sorted(d, reverse=True)] 
print result

或__missing__像这样使用：

class Dicto(dict):
    def __missing__(self, key):
        self[key]=[]
        return self[key]

d=Dicto()
for word in li:
    d[len(word)].append(word)

result=[d[n] for n in sorted(d, reverse=True)] 
print result