按最常见字段排序的namedtupled列表

Question

按列表中最常用的元素对namedtuple列表进行排序的一些优雅，快捷的简便方法是什么？

例如，我们有此列表

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')]

并像这样排序？

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')]

尝试一下，但似乎没有我想要的结果

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

Answer 1

x.character是从不在列表中。无论如何，像这样使用list.count是非常无效。排序为O（N * log N），但是，如果键功能使用list.count，它将使所有内容恶化为O（N ** 2）。

而是建立一个计数字典，并使用该字典，这将保持您的O（N * log N）性能。因此给定：

>>> 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')]