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python - 按最频繁的字段对 namedtupled 列表进行排序

问题描述

有哪些优雅而快速的简单方法可以按列表中namedtuple最常见的元素对列表进行排序？

例如，我们有这个列表

character_list = [
    Element(id=1, character='A'),
    Element(id=2, character='B'),
    Element(id=3, character='B'),
    Element(id=4, character='C'),
    Element(id=5, character='D'),
    Element(id=6, character='E'),
    Element(id=7, character='F'),
    Element(id=8, character='H'),
    Element(id=9, character='I'),
    Element(id=10, character='J'),
    Element(id=11, character='K'),
    Element(id=12, character='L'),
    Element(id=13, character='M'),
    Element(id=14, character='J'),
    Element(id=15, character='N'),
    Element(id=16, character='J')]

并这样排序？

character_list = [
    Element(id=10, character='J'),
    Element(id=14, character='J'),
    Element(id=16, character='J'),
    Element(id=2, character='B'),
    Element(id=3, character='B'),
    Element(id=1, character='A'),
    Element(id=4, character='C'),
    Element(id=5, character='D'),
    Element(id=6, character='E'),
    Element(id=7, character='F'),
    Element(id=8, character='H'),
    Element(id=9, character='I'),
    Element(id=11, character='K'),
    Element(id=12, character='L'),
    Element(id=13, character='M'),
    Element(id=14, character='J'),
    Element(id=15, character='N')]

试试这个，但似乎没有我要找的结果

sorted(character_list, key=lambda x: character_list.count(x.character))

标签： pythonpython-3.xsortingnamedtuple

解决方案

x.character永远不在您的列表中。无论如何，list.count像这样使用是非常低效的。排序是 O(N*log N)，但是，如果你的 key-function 使用list.count，它将使一切恶化到 O(N**2)。

相反，构建一个计数字典并使用该字典，这将保持您的 O(N*log N) 性能。所以给出：

>>> from pprint import pprint
>>> pprint(character_list)
[Element(id=1, character='A'),
 Element(id=2, character='B'),
 Element(id=3, character='B'),
 Element(id=4, character='C'),
 Element(id=5, character='D'),
 Element(id=6, character='E'),
 Element(id=7, character='F'),
 Element(id=8, character='H'),
 Element(id=9, character='I'),
 Element(id=10, character='J'),
 Element(id=11, character='K'),
 Element(id=12, character='L'),
 Element(id=13, character='M'),
 Element(id=14, character='J'),
 Element(id=15, character='N'),
 Element(id=16, character='J')]

然后

>>> from collections import Counter
>>> counts = Counter(e.character for e in character_list)
>>> counts
Counter({'J': 3, 'B': 2, 'A': 1, 'C': 1, 'D': 1, 'E': 1, 'F': 1, 'H': 1, 'I': 1, 'K': 1, 'L': 1, 'M': 1, 'N': 1})

最后，

>>> def keyfunc(e):
...     return counts[e.character]
...
>>> sorted_character_list = sorted(character_list, key=keyfunc, reverse=True)
>>> pprint(sorted_character_list)
[Element(id=10, character='J'),
 Element(id=14, character='J'),
 Element(id=16, character='J'),
 Element(id=2, character='B'),
 Element(id=3, character='B'),
 Element(id=1, character='A'),
 Element(id=4, character='C'),
 Element(id=5, character='D'),
 Element(id=6, character='E'),
 Element(id=7, character='F'),
 Element(id=8, character='H'),
 Element(id=9, character='I'),
 Element(id=11, character='K'),
 Element(id=12, character='L'),
 Element(id=13, character='M'),
 Element(id=15, character='N')]

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