在字典中查找包含相等值的键列表

Question

I want to find a list (of lists) of all keys in a dictionary that contain values equal to other elements.

For example:

dict_with_dups = {
    "a": 1,
    "b": 1,
    "c": 1,
    "d": 2,
    "e": 3,
    "f": 3,
    "g": 4,
}

keys_with_same = locate_same_keys(dict_with_dups)

for key_list in keys_with_same:
    print(f"{key_list}")

The above should print this:

['a', 'b', 'c']
['e', 'f']

How do I most efficiently write the function locate_same_keys ?

Answer 1

You can find the duplicate values from the dictionary using a flipped dictionary .

You can create it by iterating over the original dictionary and adding each value to the flipped dictionary as a key, and it's key as it's value. Then, if the value appears again in the original dictionary, add it's key as another value in the flipped dictionary.

Then you can just go over each key in the flipped dictionary, check if it has more than 1 value and if so, print it:

dict_with_dups = {
    "a": 1,
    "b": 1,
    "c": 1,
    "d": 2,
    "e": 3,
    "f": 3,
    "g": 4,
}

# finding duplicate values from dictionary using flip 
flipped = {} 

# iterate over the original dictionary and check if each value is associated 
# with more than one key
for key, value in dict_with_dups.items(): 
    if value not in flipped: 
        flipped[value] = [key] 
    else: 
        flipped[value].append(key) 

# printing all values that are assosiated with more then one key
for key, value in flipped.items():
    if len(value)>1:
        print(value)

Output :

['a', 'b', 'c']
['e', 'f']

Regarding efficiency , creating the flipped dictionary requires going over all the key, value pairs in the original dictionary, so we get O(n) time complexity.

Answer 2

Iterate over dict items and for each value add key to proper list.

from collections import defaultdict

res = defaultdict(list)

for k, v in dict_with_dups.items():
    res[v].append(k)

for v in res.values():
    if len(v) > 1:
        print(v)

Answer 3

dict_with_dups = {
    "a": 1,
    "b": 1,
    "c": 1,
    "d": 2,
    "e": 3,
    "f": 3,
    "g": 4,
}
result = {}
for val in dict_with_dups:
    if dict_with_dups[val] in result:
        result[dict_with_dups[val]].append(val)
    else:
        result[dict_with_dups[val]] = [val]    

for key, value in result.items():
   if len(value)>1:
      print(value)

it will give u all list with the same keys like this

Answer 4

dict_with_dups = {
    "a": 1,
    "b": 1,
    "c": 1,
    "d": 2,
    "e": 3,
    "f": 3,
    "g": 4,
}

result = list(filter(lambda z: len(z)>1,
                     (list(map(lambda y: y[0],
                               filter(lambda x: x[1]==v, dict_with_dups.items())))
                      for v in set(dict_with_dups.values()))))

or

result = [z for z in ([x[0] for x in dict_with_dups.items() if x[1]==v]
                      for v in set(dict_with_dups.values()))
          if len(z)>1]

as suggested by @wjandrea.

For me, the first variant is more clear (but less concise).