Question

有没有办法可以用list comprehension answered in this thread来创建字典？

listA = [
    "apple_v001",
    "apple_v002",
    "banana_v001",
    "orange_v001",
]
keywords = ["apple", "banana", "orange"]
[[item for item in listA if kw in item] for kw in keywords]
# Result: [['apple_v001', 'apple_v002'], ['banana_v001'], ['orange_v001']] #

我要做的是创建一个字典，使用关键字作为此结果的关键字。所以

dictA["apple"] = ['apple_v001', 'apple_v002']

等等。我试图做dict = {key，value for ...（iteration）}但总是得到语法错误。我真的不知道如何开始，任何帮助欣赏。

Answer 1

{kw: [item for item in listA if kw in item] for kw in keywords}

但这似乎不是创建这样一个字典的特别有效的方式

例如，这不需要事先提供关键字列表，而且效率相当高

>>> from itertools import groupby
>>> {k:list(g) for k,g in groupby(sorted(listA), key=lambda x:x.partition('_')[0])}
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

对于Python2.6，等价物是

dict((kw, [item for item in listA if kw in item]) for kw in keywords)

和

>>> from itertools import groupby
>>> dict((k,list(g)) for k,g in groupby(sorted(listA), key=lambda x:x.partition('_')[0]))
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

Answer 2

如果您不想使用一个班轮解决方案，请查看

In [58]: d
Out[58]: defaultdict(<type 'list'>, {})

In [59]: for elem in keywords:
   ....:     for item in listA:
   ....:         if item.startswith(elem):
   ....:             d[elem].append(item)
   ....:

In [60]: d
Out[60]: defaultdict(<type 'list'>, {'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']})

Answer 3

In the comments you mentioned that you use Python 2.6。在Python 2.6中没有字典理解，你可以使用dict()代替生成器表达式：

d = dict((kw, [item for item in listA if kw in item]) for kw in keywords)

这是一个可能更有效的版本：

import re
from collections import defaultdict

search_word = re.compile("(%s)" % "|".join(map(re.escape, keywords))).search

d = defaultdict(list)
for item in listA:
    m = search_word(item)
    if m:
       d[m.group(1)].append(item)

如果listA始终采用问题中给出的格式：

from collections import defaultdict

keywords = set(keywords)
d = defaultdict(list)
for item in listA:
    word = item.partition("_")[0]
    if word in keywords:
       d[word].append(item)

如果listA不包含不在keywords中的项目：

from collections import defaultdict

d = defaultdict(list)
for item in listA:
    d[item.partition('_')[0]].append(item)

Answer 4

您可以使用正则表达式：

>>> import re
>>> listA = [
...     "apple_v001",
...     "apple_v002",
...     "banana_v001",
...     "orange_v001",
... ]
>>> keywords = ["apple", "banana", "orange"]
>>> s=' '.join(listA)
>>> dict([(e,re.findall(r'{}_v\d+'.format(e),s)) for e in keywords])
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

或（发布Python 2.7）dict理解：

>>> {e:re.findall(r'{}_v\d+'.format(e),s) for e in keywords}
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

Python：使用多列表理解创建字典

4 个答案: