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最初的数据是这样的：

dict = {<User: user2>: {'diff': 48, 'alike': 1}, <User: user3>: {'diff': 42, 'alike': 2}, <User: user4>: {'diff': 45, 'alike': 1}, <User: user5>: {'diff': 43, 'alike':
 2}, <User: user6>: {'diff': 46, 'alike': 1}, <User: user7>: {'diff': 46, 'alike': 1}, <User: user8>: {'diff': 49, 'alike': 1}, <User: user9>: {'diff': 50, 'ali
ke': 0}, <User: user10>: {'diff': 46, 'alike': 1}, <User: user11>: {'diff': 37, 'alike': 3}, <User: user12>: {'diff': 50, 'alike': 0}, <User: user13>: {'diff':
50, 'alike': 0}, <User: user14>: {'diff': 50, 'alike': 0}, <User: user15>: {'diff': 50, 'alike': 0}, <User: user16>: {'diff': 50, 'alike': 0}, <User: user17>: {
'diff': 50, 'alike': 0}, <User: user18>: {'diff': 50, 'alike': 0}, <User: user19>: {'diff': 50, 'alike': 0}, <User: user20>: {'diff': 50, 'alike': 0}}

然后我对其进行排序：

sorted(dict) == [{'diff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'d
    iff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'diff': 50, 'alike': 0}, {'diff': 45, 'alike': 1}, {'diff': 46, 'alike': 1}, {'diff'
    : 46, 'alike': 1}, {'diff': 46, 'alike': 1}, {'diff': 48, 'alike': 1}, {'diff': 49, 'alike': 1}, {'diff': 42, 'alike': 2}, {'diff': 43, 'alike': 2}, {'diff': 37
    , 'alike': 3}]

如何按“差异”对其进行排序？

4

2 回答 2

8

一方面，您没有对 adict进行排序 - 您正在对列表进行排序dicts- 这些是非常不同的事情，尤其是因为dictPython 中的 a 没有定义的顺序。

您可以使用sorted()和轻松做到这一点operator.itemgetter()：

import operator
sorted_dicts = sorted(dicts, key=operator.itemgetter("diff"))

sorted()内置函数接受关键字参数，这key是一个接受值的函数，并给出另一个值进行排序。在这里，我们使用 an从排序依据itemgetter()中获取所需的值。dict

编辑：

鉴于您的更改，有两个答案，因为您不清楚。如果您想要值列表，只需从原始值中提取它们dict：

sorted(users.values(), key=operator.itemgetter("diff"))

这就像服用一样简单dict.values()。自然，在 Python 2.x 下，您会想要使用viewitems()oriteritems()以获得良好的性能。

如果您想对dict自身进行排序，那就另当别论了，因为dicts （如我所说）本质上是无序的。

首先，我想指出这sorted(dict)不会产生您建议的输出 - dict 会遍历键，而不是默认的值：

users = {
    '<User: user10>': {'alike': 1, 'diff': 46},
    '<User: user11>': {'alike': 3, 'diff': 37},
    '<User: user12>': {'alike': 0, 'diff': 50},
    '<User: user13>': {'alike': 0, 'diff': 50},
    '<User: user14>': {'alike': 0, 'diff': 50},
    '<User: user15>': {'alike': 0, 'diff': 50},
    '<User: user16>': {'alike': 0, 'diff': 50},
    '<User: user17>': {'alike': 0, 'diff': 50},
    '<User: user18>': {'alike': 0, 'diff': 50},
    '<User: user19>': {'alike': 0, 'diff': 50},
    '<User: user20>': {'alike': 0, 'diff': 50},
    '<User: user2>': {'alike': 1, 'diff': 48},
    '<User: user3>': {'alike': 2, 'diff': 42},
    '<User: user4>': {'alike': 1, 'diff': 45},
    '<User: user5>': {'alike': 2, 'diff': 43},
    '<User: user6>': {'alike': 1, 'diff': 46},
    '<User: user7>': {'alike': 1, 'diff': 46},
    '<User: user8>': {'alike': 1, 'diff': 49},
    '<User: user9>': {'alike': 0, 'diff': 50}
}

print(sorted(users))

给我们：

['<User: user10>', '<User: user11>', '<User: user12>', '<User: user13>', '<User: user14>', '<User: user15>', '<User: user16>', '<User: user17>', '<User: user18>', '<User: user19>', '<User: user20>', '<User: user2>', '<User: user3>', '<User: user4>', '<User: user5>', '<User: user6>', '<User: user7>', '<User: user8>', '<User: user9>']

要产生一个 sorted dict，我们需要使用collections.OrderedDict()：

import collections

users = {
    '<User: user10>': {'alike': 1, 'diff': 46},
    '<User: user11>': {'alike': 3, 'diff': 37},
    '<User: user12>': {'alike': 0, 'diff': 50},
    '<User: user13>': {'alike': 0, 'diff': 50},
    '<User: user14>': {'alike': 0, 'diff': 50},
    '<User: user15>': {'alike': 0, 'diff': 50},
    '<User: user16>': {'alike': 0, 'diff': 50},
    '<User: user17>': {'alike': 0, 'diff': 50},
    '<User: user18>': {'alike': 0, 'diff': 50},
    '<User: user19>': {'alike': 0, 'diff': 50},
    '<User: user20>': {'alike': 0, 'diff': 50},
    '<User: user2>': {'alike': 1, 'diff': 48},
    '<User: user3>': {'alike': 2, 'diff': 42},
    '<User: user4>': {'alike': 1, 'diff': 45},
    '<User: user5>': {'alike': 2, 'diff': 43},
    '<User: user6>': {'alike': 1, 'diff': 46},
    '<User: user7>': {'alike': 1, 'diff': 46},
    '<User: user8>': {'alike': 1, 'diff': 49},
    '<User: user9>': {'alike': 0, 'diff': 50}
}

print(collections.OrderedDict(sorted(users.items(), key=lambda x: x[1]["diff"])))

这给了我们：

OrderedDict([('<User: user11>', {'diff': 37, 'alike': 3}), ('<User: user3>', {'diff': 42, 'alike': 2}), ('<User: user5>', {'diff': 43, 'alike': 2}), ('<User: user4>', {'diff': 45, 'alike': 1}), ('<User: user10>', {'diff': 46, 'alike': 1}), ('<User: user7>', {'diff': 46, 'alike': 1}), ('<User: user6>', {'diff': 46, 'alike': 1}), ('<User: user2>', {'diff': 48, 'alike': 1}), ('<User: user8>', {'diff': 49, 'alike': 1}), ('<User: user20>', {'diff': 50, 'alike': 0}), ('<User: user9>', {'diff': 50, 'alike': 0}), ('<User: user13>', {'diff': 50, 'alike': 0}), ('<User: user19>', {'diff': 50, 'alike': 0}), ('<User: user12>', {'diff': 50, 'alike': 0}), ('<User: user18>', {'diff': 50, 'alike': 0}), ('<User: user15>', {'diff': 50, 'alike': 0}), ('<User: user14>', {'diff': 50, 'alike': 0}), ('<User: user17>', {'diff': 50, 'alike': 0}), ('<User: user16>', {'diff': 50, 'alike': 0})])

请注意，这里我们必须对lambdakey 参数使用声明，因为itemgetter()不幸的是无法为我们获取多个级别的项目。

于 2012-05-12T02:53:56.213 回答
5 
sorted(your_list, key=lambda el: el["diff"])

关键字参数应该是一个函数，它接受您正在排序的列表中的key一个元素，并返回应该在排序比较中使用的值。 lambda是简写def- 也就是说，它定义了一个函数（有几个限制）。同一行可以这样写：

def get_key(el):
    return el["diff"]

sorted(your_list, key=get_key)
于 2012-05-12T02:52:33.330 回答