如何按值对计数器进行排序？- Python
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/20950650/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to sort Counter by value? - python
提问by alvas
Other than doing list comprehensions of reversed list comprehension, is there a pythonic way to sort Counter by value? If so, it is faster than this:
除了对反向列表理解进行列表理解之外，是否有一种 Pythonic 方法可以按值对 Counter 进行排序？如果是这样，它比这更快：
>>> from collections import Counter
>>> x = Counter({'a':5, 'b':3, 'c':7})
>>> sorted(x)
['a', 'b', 'c']
>>> sorted(x.items())
[('a', 5), ('b', 3), ('c', 7)]
>>> [(l,k) for k,l in sorted([(j,i) for i,j in x.items()])]
[('b', 3), ('a', 5), ('c', 7)]
>>> [(l,k) for k,l in sorted([(j,i) for i,j in x.items()], reverse=True)]
[('c', 7), ('a', 5), ('b', 3)
采纳答案by Martijn Pieters
Use the Counter.most_common()method, it'll sort the items for you:
使用该Counter.most_common()方法，它会为您排序项目：
>>> from collections import Counter
>>> x = Counter({'a':5, 'b':3, 'c':7})
>>> x.most_common()
[('c', 7), ('a', 5), ('b', 3)]
It'll do so in the most efficient manner possible; if you ask for a Top N instead of all values, a heapqis used instead of a straight sort:
它将以最有效的方式进行；如果您要求 Top N 而不是所有值，heapq则使用 a 而不是直接排序：
>>> x.most_common(1)
[('c', 7)]
Outside of counters, sorting can always be adjusted based on a keyfunction; .sort()and sorted()both take callable that lets you specify a value on which to sort the input sequence; sorted(x, key=x.get, reverse=True)would give you the same sorting as x.most_common(), but only return the keys, for example:
在计数器之外，排序总是可以根据key函数进行调整；.sort()并且sorted()都采用 callable ，它允许您指定一个值来对输入序列进行排序；sorted(x, key=x.get, reverse=True)会给你与 相同的排序x.most_common()，但只返回键，例如：
>>> sorted(x, key=x.get, reverse=True)
['c', 'a', 'b']
or you can sort on only the value given (key, value)pairs:
或者您可以仅对给定的值(key, value)对进行排序：
>>> sorted(x.items(), key=lambda pair: pair[1], reverse=True)
[('c', 7), ('a', 5), ('b', 3)]
See the Python sorting howtofor more information.
有关更多信息，请参阅Python 排序方法。
回答by Inbar Rose
Yes:
是的：
>>> from collections import Counter
>>> x = Counter({'a':5, 'b':3, 'c':7})
Using the sorted keyword key and a lambda function:
使用 sorted 关键字 key 和 lambda 函数：
>>> sorted(x.items(), key=lambda i: i[1])
[('b', 3), ('a', 5), ('c', 7)]
>>> sorted(x.items(), key=lambda i: i[1], reverse=True)
[('c', 7), ('a', 5), ('b', 3)]
This works for all dictionaries. However Counterhas a special function which already gives you the sorted items (from most frequent, to least frequent). It's called most_common():
这适用于所有词典。但是Counter有一个特殊的功能，它已经给你排序的项目（从最频繁到最不频繁）。它被称为most_common()：
>>> x.most_common()
[('c', 7), ('a', 5), ('b', 3)]
>>> list(reversed(x.most_common())) # in order of least to most
[('b', 3), ('a', 5), ('c', 7)]
You can also specify how many items you want to see:
您还可以指定要查看的项目数量：
>>> x.most_common(2) # specify number you want
[('c', 7), ('a', 5)]
回答by Hooked
A rather nice addition to @MartijnPietersanswer is to get back a dictionarysorted by occurrence since Collections.most_commononly returns a tuple. I often couple this with a json output for handy log files:
@MartijnPieters答案的一个相当不错的补充是取回按出现次数排序的字典，因为Collections.most_common只返回一个元组。我经常将它与方便的日志文件的 json 输出结合起来：
from collections import Counter, OrderedDict
x = Counter({'a':5, 'b':3, 'c':7})
y = OrderedDict(x.most_common())
With the output:
随着输出：
OrderedDict([('c', 7), ('a', 5), ('b', 3)])
{
"c": 7,
"a": 5,
"b": 3
}
回答by Alex Seam
More general sorted, where the keykeyword defines the sorting method, minus before numerical type indicates descending:
更一般的sorted，其中key关键字定义了排序方式，数字类型前减号表示降序：
>>> x = Counter({'a':5, 'b':3, 'c':7})
>>> sorted(x.items(), key=lambda k: -k[1]) # Ascending
[('c', 7), ('a', 5), ('b', 3)]
