Home How to create a dictionary with the unique items of a list as key and the count of unique items as value?

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How to create a dictionary with the unique items of a list as key and the count of unique items as value?

May 25, 2022

I know that if I have a dictionary filled with with x = {unique : count} then I can use prop_dict = dict((k, round(v/sum(x.values()),2)) for k,v in x.items()), but I do not know how to get there from an array/list.

Here is some sample data:

arr = [0., 137.,   3.,   1.,   1.,   5.,   2.,   2.,   8.,  31., 155.,
       3., 233.,  72., 302.,  66., 416.,   1., 148., 200., 237., 238.,
       354., 383., 422., 192.,  48.,  78., 136.,  15., 111.,   5.,  21.]

>Solution :

Python has the collections.Counter class that will solve that for you:


In [1]: from collections import Counter                                                                                     

In [2]: arr = [0., 137.,   3.,   1.,   1.,   5.,   2.,   2.,   8.,  31., 155., 
   ...:        3., 233.,  72., 302.,  66., 416.,   1., 148., 200., 237., 238., 
   ...:        354., 383., 422., 192.,  48.,  78., 136.,  15., 111.,   5.,  21.]                                            

In [3]: Counter(arr)                                                                                                        
Out[3]: 
Counter({0.0: 1,
         137.0: 1,
         3.0: 2,
         1.0: 3,
         5.0: 2,
         2.0: 2,
         8.0: 1,
         31.0: 1,
         155.0: 1,
         233.0: 1,
         72.0: 1,
         302.0: 1,
         66.0: 1,
         416.0: 1,
         148.0: 1,
         200.0: 1,
         237.0: 1,
         238.0: 1,
         354.0: 1,
         383.0: 1,
         422.0: 1,
         192.0: 1,
         48.0: 1,
         78.0: 1,
         136.0: 1,
         15.0: 1,
         111.0: 1,
         21.0: 1})