For example, in dict1 the keys 1, 2, 3 all have the same value 'a', but the keys 3 and 5 have different values, 'b' and 'd'. What I want is:

If N keys have the same value and N >=3, then I want to remove all other elements from the dict and only keep those N key values, which means 'b' & 'd' have to be removed from the dict.

The following code works, but it seems very verbose. Is there a better way to do this?

from collections import defaultdict

dict1 = {1:'a', 2:'a', '3':'b', '4': 'a', '5':'d'}
l1 = [1, 2, 3, 4, 5]

dict2 = defaultdict(list)
for k, v in dict1.items():
    dict2[v].append(k)

to_be_removed = []
is_to_be_removed = False
for k, values in dict2.items():
    majority = len(values)
    if majority>=3:
        is_to_be_removed = True
    else:
        to_be_removed.extend(values)

if is_to_be_removed:
    for d in to_be_removed:
        del dict1[d]

print(f'New dict: {dict1}')

CodePudding user response：

You can use collections.Counter to get the frequency of every value, then use a dictionary comprehension to retain only the keys that have the desired corresponding value:

from collections import Counter

dict1 = {1:'a', 2:'a', '3':'b', '4': 'a', '5':'d'}
ctr = Counter(dict1.values())

result = {key: value for key, value in dict1.items() if ctr[value] >= 3}
print(result)

This outputs:

{1: 'a', 2: 'a', '4': 'a'}