Python3: Get count for same dictionaries within list-CodePudding

I have a list of dictionaries and want to get the unique dictionaries with count. i.e

[{'Basic': 100},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 200}]

Out put should be like

{'Basic': 100} -> 3
{'Basic': 100, 'Food Allowance': 1000} -> 3
{'Basic': 200} -> 1

OR

[index_no] -> count

CodePudding user response：

A quick and dirty solution would be to loop over the list of dictionaries and save each unique one and keep a counter. Something like this:

def count_and_set_dict(dict):
    set_dict = []
    count_dict = []

    for d in dict:
        if d not in set_dict:
            set_dict.append(d)
            count_dict.append(1)
        else:
            count_dict[set_dict.index(d)]  = 1

    for i in range(len(set_dict)):
        print(f"{set_dict[i]} --> {count_dict[i]}")

dicts = [{'Basic': 100},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 200}]

count_and_set_dict(dicts)

Output

{'Basic': 100} --> 3
{'Basic': 100, 'Food Allowance': 1000} --> 3
{'Basic': 200} --> 1

CodePudding user response：

My approach is getting unique dictionaries in the list and counting the elements in the original list by count()

dictlist = [{'Basic': 100},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 200}]

unique_dictlist = list(map(dict, set(tuple(sorted(sub.items())) for sub in dictlist))) #to get only unique values.
for d in unique_dictlist:
    print(d,"-->",dictlist.count(d))

result:

{'Basic': 200} -> 1
{'Basic': 100, 'Food Allowance': 1000} -> 3
{'Basic': 100} -> 3

CodePudding user response：

Using collections.Counter

Dictionary is not hashable so must convert each to a tuple

Code

from collections import Counter

lst = [{'Basic': 100},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 100},
{'Basic': 100, 'Food Allowance': 1000},
{'Basic': 200}]

# Output count of each dictionary
print(Counter([tuple(d.items()) for d in lst]))

Output

Counter({(('Basic', 100),): 3,
         (('Basic', 100), ('Food Allowance', 1000)): 3,
         (('Basic', 200),): 1})

CodePudding user response：

import ast
tmp = {}
for i in set([str(d) for d in df1]):
    tmp[i] = df1.count(ast.literal_eval(i))

Output:

{"{'Basic': 100, 'Food Allowance': 1000}": 3,
 "{'Basic': 100}": 3,
 "{'Basic': 200}": 1}

CodePudding user response：

You can use collections.Counter, however, it requires the list elements to be hashable, which dictionaries are clearly not. One solution might be converting your dictionaries to 'string dictionaries':

from collections import Counter
Counter([str(i) for i in your_list])

will return:

Counter({"{'Basic': 100}": 3,
     "{'Basic': 100, 'Food Allowance': 1000}": 3,
     "{'Basic': 200}": 1})