按对象特征创建列表的字典 [英] Create dictionary of lists by object feature

问题描述

可以说我有一个类：

class Cat(object):
     color = "White"     # "Black", "Red" or whatever
     name = "..."
     ....

如果我有猫的名单，我想通过颜色将它们分开，放在列表的字典：

If I have a list of cats, I want to separate them by color and put in dictionary of lists:

cats = [cat1, cat2, cat3, ...]
cats_by_color = {"White": [cat1, cat3, ...], "Black": [...], ...}

现在我这样做：

cats_by_color ={}
for cat in cats:
    if cat.color not in cats_by_color.keys():    # add new key if needed
        cats_by_color[cat.color] = []
    cats_by_colors[cat.color].append(cat)        # add cat to right list

和我有一个强烈的感觉，有做这样的事情更pythonish方式。如何通过一些单行办呢？

And I have a strong feeling that there is a "more pythonish way" of doing such things. How to do it by some one-liner?

推荐答案

要做到在一个班轮，你需要首先对列表进行排序，然后使用的 itertools.groupby（） ：

To do it in a one-liner, you'd need to sort the list first, then use itertools.groupby():

from operator import attrgetter
from itertools import groupby

color = attrgetter('color')
cats_by_color = {k: list(g) for k, g in groupby(sorted(cats, key=color), color)}

不过，排序不仅仅是直循环更昂贵;这不再是一个班轮，但使用 collections.defaultdict（） 至少使得紧凑：

However, sorting is more costly than just a straight loop; this is no longer a one-liner, but using collections.defaultdict() at least makes it compact:

from collections import defaultdict

cats_by_color = defaultdict(list)

for cat in cats:
    cats_by_colors[cat.color].append(cat)

下面 defaultdict 将创建一个新的列表对象键是没有在字典中，仅仅通过努力访问密钥。

Here defaultdict will create a new list object for keys that are not yet in the dictionary, merely by trying to access the key.

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