在Python 3中排序字典列表

问题描述 投票:2回答:2

我正在尝试排序字典列表。我的目标是以相同的方式对具有多个(可能是相同的)键的字典进行排序,即使字典的顺序不同,或者键在字典中的顺序不同。

在Python 2中,我使用了以下内容:

a = [{1: 2, 7: 8}, {7: 8, 3: 4}, {5: 6}]
b = [{3: 4, 7: 8}, {7: 8, 1: 2}, {5: 6}]
a.sort()
b.sort()
a
Out[20]: [{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]
b
Out[21]: [{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]

这成功实现了我创建两个看起来完全相同的排序字典的目标。

我试图在Python 3中做同样的事情,其中​​.sort()不适用于字典列表。

我尝试过不同的方法。

1.

sorted(a, key=lambda d: max(d.keys()))

这不起作用:

a = [{1: 2, 7: 8}, {3: 4, 7: 8}, {5: 6}]
b = [{3: 4, 7: 8}, {1: 2, 7: 8}, {5: 6}]
a2 = sorted(a, key=lambda d: max(d.keys()))
b2 = sorted(b, key=lambda d: max(d.keys()))
a2
Out[1]: [{5: 6}, {1: 2, 7: 8}, {7: 8, 3: 4}]
b2
Out[2]: [{5: 6}, {3: 4, 7: 8}, {7: 8, 1: 2}]

2.

a2 = sorted([list(zip(x.keys(),x.values())) for x in a])
a3 = [{k: v for (k,v) in x} for x in a2]

这不起作用:

a = [{1: 2, 7: 8}, {7: 8, 3: 4}, {5: 6}]
b = [{3: 4, 7: 8}, {7: 8, 1: 2}, {5: 6}]
a2 = sorted([list(zip(x.keys(),x.values())) for x in a])
a3 = [{k: v for (k,v) in x} for x in a2]
b2 = sorted([list(zip(x.keys(),x.values())) for x in b])
b3 = [{k: v for (k,v) in x} for x in b2]
a3
Out[1]: [{1: 2, 7: 8}, {5: 6}, {7: 8, 3: 4}]
b3
Out[2]: [{3: 4, 7: 8}, {5: 6}, {7: 8, 1: 2}]

有谁知道如何在Python 3中获得Python 2结果?

python python-3.x list sorting dictionary
2个回答
1
投票

您需要在排序时使用每个dict的所有项目,并且d.items()以任意顺序迭代项目,因此需要对它们进行排序。这是你的排序键:

a = [{1: 2, 7: 8}, {7: 8, 3: 4}, {5: 6}]
b = [{3: 4, 7: 8}, {7: 8, 1: 2}, {5: 6}]
sorted(a, key=lambda d: sorted(d.items()))
Out: [{1: 2, 7: 8}, {3: 4, 7: 8}, {5: 6}]

sorted(b, key=lambda d: sorted(d.items()))
Out: [{1: 2, 7: 8}, {3: 4, 7: 8}, {5: 6}]

x = [{1: 2},{1: 3}]
y = [{1: 3},{1: 2}]
sorted(x, key=lambda d: sorted(d.items()))
Out: [{1: 2}, {1: 3}]

sorted(y, key=lambda d: sorted(d.items()))
Out: [{1: 2}, {1: 3}]

请注意,sorted(a, key=...)创建了一个新列表,而a.sort(key=...)则进行了排序。


4
投票

对字典中所有keys进行排序可以通过以下方式完成:

a.sort(key=lambda d: d.keys())
b.sort(key=lambda d: d.keys())

要获得您想要的结果,我们需要按升序对键进行排序,如下所示:

a.sort(key=lambda d: sorted(list(d.keys()), reverse=True))
b.sort(key=lambda d: sorted(list(d.keys()), reverse=True))

这给出了以下结果:

>>> a
[{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]
>>> b
[{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]

编辑:为了根据值进行排序(在评论中询问),以下内容可能有效:

a.sort(key=lambda d: sorted(list(d.keys()) + sorted(list(d.values())), reverse=True))
b.sort(key=lambda d: sorted(list(d.keys()) + sorted(list(d.values())), reverse=True))
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