如何按频率对字典值进行排序

Question

Because I made my dict file with append list. I have the following dict file:

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

How do I sort the values by frequency within the list so I get output like:

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

The order doesn't matter for strings of the same frequency

I tried

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

and got

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

Something is wrong with the "2" in the first list.

Thanks.

Answer 1

One way using collections.Counter with dict comprehension:

from collections import Counter

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

Or without Counter , using list.count :

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

Then do sort:

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

Output:

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

Note:

key for sorted returns tuple of (count, itself) so that same items remain grouped.

Answer 2

Here is a way to do it without collections module in python.

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.

Output:

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

You can read more about sorted() from this link