How to filter out dictionaries with the same value for a certain key from a list?

Question

I have the following list of objects:

l = [{'name': 'Mike', 'age': 31},
     {'name': 'Peter', 'age': 29},
     {'name': 'Mike', 'age': 44}]

I want to filter it based on the name, and because "Mike" is a duplicate in this case, I want to remove all entries with name=Mike (regardless of the age).

(i.e. to get the same list but without the entries that have name=Mike)

What’s the best approach to do that?

Asked By: Mikael

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Source

Answer 1

You can use a Counter to get the counts of all names. Then filter the names that appear more than once:

from collections import Counter

l = [{'name': 'Mike', 'age': 31},
     {'name': 'Peter', 'age': 29},
     {'name': 'Mike', 'age': 44}]

names_count = Counter(d['name'] for d in l)
new_l = list(filter(lambda d: names_count[d['name']] == 1, l))
print(new_l)

Will give:

[{'name': 'Peter', 'age': 29}]

Answered By: Tomerikoo

Answer 2

Collecting the dicts in a dict where keys are names and values are either the only dict with that name or None if the name is duplicated.

l = [{'name': 'Mike', 'age': 31},
     {'name': 'Peter', 'age': 29},
     {'name': 'Mike', 'age': 44}]

D = {}
for d in l:
    n = d['name']
    D[n] = None if n in D else d
l[:] = filter(None, D.values())

print(l)

Answered By: Kelly Bundy

Answer 3

And if you want to implement it without using any function.

l = [{'name': 'Mike', 'age': 31},
     {'name': 'Peter', 'age': 29},
     {'name': 'Mike', 'age': 44}]

name_dict = {}
for i in l:
    if i['name'] not in name_dict:
        name_dict[i['name']] = 1
    else:
        name_dict[i['name']] += 1

new_l = [d for d in l if name_dict[d['name']] == 1]

print(new_l)

outputs:

[{'name': 'Peter', 'age': 29}]

Answered By: Ajeet Verma

How to filter out dictionaries with the same value for a certain key from a list?

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