Count occurrences of objects in a list in python

Question

Say I have the following list:

result = [{"name": "a", "number": 1},
{"name": "a", "number": 2},
{"name": "b", "number": 1},
{"name": "a", "number": 1}]

Can I turn it into something like:

result = [{"name": "a", "number": 1, "count": 2},
{"name": "a", "number": 2},
{"name": "b", "number": 1}]

I tried using the Count class but I couldn't make it work with dictionaries.

Answer 1

You can use collections.Counter with list comprehension:

from collections import Counter
[dict(tuple(t) + (('count', c),)) for t, c in Counter(frozenset(d.items()) for d in result).items()]

This returns:

[{'number': 1, 'name': 'a', 'count': 2}, {'number': 2, 'name': 'a', 'count': 1}, {'number': 1, 'name': 'b', 'count': 1}]

Answer 2

THis will create a empty dictionary with keys from unique elements in result and initialise all default values to 0.

_dict = dict.fromkeys(set([e["name"] for e in result]), 0)
output: {'b': 0, 'a': 0}

This will count the number of element in result for the keys present in _dict.

[_dict.update({element["name"]: _dict[element["name"]]+1}) for element in result]

Since your "number" is same for whole list,

[{"name": key, "number": 1, "count": _dict[key]}  for key in _dict.keys()]
output:[{'name': 'b', 'number': 1, 'count': 1},
  {'name': 'a', 'number': 1, 'count': 2}]

if number is not same for all the keys, remove all the duplicate dicts from the result list.

no_dups = [i for n, i in enumerate(result) if i not in result[n + 1:]]
output: [{'name': 'b', 'number': 1}, {'name': 'a', 'number': 1}]

create another dict with values as their number :

new = {} 
[new.update({i["name"]: i["number"]}) for i in b]

Repeat the last step, like this

[{"name": key, "number": new[key], "count": _dict[key]}  for key in _dict.keys()]

Answer 3

One way is to use collections.Counter to count dictionaries by ('name', 'number') , then add counts conditional on the count being greater than 1. This can be achieved using a list comprehension:

from collections import Counter
from operator import itemgetter

keys = ('name', 'number')
c = Counter(map(itemgetter(*keys), L))

res = [{**dict(zip(keys, k)), **({'count': v} if v > 1 else {})} \
       for k, v in c.items()]

Result:

[{'count': 2, 'name': 'a', 'number': 1},
 {'name': 'a', 'number': 2},
 {'name': 'b', 'number': 1}]

Answer 4

The Counter class relies on your objects being hashable to count them. Thus, a work-around could be to cast your dictionaries to their immutable equivalents.

The immutable equivalent of a dict is a frozenset (to account for unorderness of dict ) of tuple .

You can then count then and form back a list of dictionaries from the counter.

from collections import Counter

result = [{"name": "a", "number": 1},
    {"name": "a", "number": 2},
    {"name": "b", "number": 1},
    {"name": "a", "number": 1}]

frozen_result = map(lambda d: frozenset(d.items()), result)

count = Counter(frozen_result)

new_result = [dict(k, count=v) for k, v in count.items()]

print(new_result)

Output

[{'number': 1, 'name': 'a', 'count': 2},
 {'number': 2, 'name': 'a', 'count': 1},
 {'name': 'b', 'number': 1, 'count': 1}]