如何按频率对字典值进行排序

Question

因为我用附加列表创建了我的dict 文件。我有以下 dict 文件：

dict_1 = {'a':["1","2","d","d","d","1","2","2"], 
          'b':["1","2","e","e","5","5","5","6"]}

如何在list 中按frequency 对值进行排序，以便得到如下输出：

dict_1 = {'a':["d","d","d","2","2","2","1","1"], 
          'b':["5","5","5","e","e","6","2","1"]}

相同频率的字符串顺序无关

我试过了

result=[]
for k,v in dict_1.items():
    result.append(sorted(v, key = v.count,
                                reverse = True))

得到了

[['2', 'd', 'd', 'd', '2', '2', '1', '1'],
 ['5', '5', '5', 'e', 'e', '1', '2', '6']]

第一个列表中的“2”有问题。

谢谢。

Answer 1

使用collections.Counter 和dict 理解的一种方式：

from collections import Counter

cnts = {k: Counter(v) for k, v in dict_1.items()}

或者没有Counter，使用list.count：

cnts = {k: {i: v.count(i) for i in set(v)} for k, v in dict_1.items()}

然后进行排序：

{k: sorted(v, key=lambda x: (cnts[k][x], x), reverse=True) for k, v in dict_1.items()}

输出：

{'a': ['d', 'd', 'd', '2', '2', '2', '1', '1'],
 'b': ['5', '5', '5', 'e', 'e', '6', '2', '1']}

注意：

key for sorted 返回 tuple of (count, itself)，以便相同的项目保持分组。

Answer 2

这是一种在 python 中不使用集合模块的方法。

dict_1 = {'a':["1","2","d","d","d","1","2","2"], 
          'b':["1","2","e","e","5","5","5","6"]}

for items in dict_1: # looping through the dictionary to get every key value pair in the dictionary

    l = dict_1.get(items) # getting the value of every key value pair in the dictionary. Here "l" is short for "list"
    print(sorted(l, key=l.count, reverse=True)) # sorting it according to frequency of the element in the list.

输出：

['2', 'd', 'd', 'd', '2', '2', '1', '1']
['5', '5', '5', 'e', 'e', '1', '2', '6']

你可以从this link阅读更多关于sorted()的信息