在Python中删除字典值中的重复项

Question

Sorry the topic's title is vague, I find it hard to explain.

I have a dictionary in which each value is a list of items. I wish to remove the duplicated items, so that each item will appear minimum times (preferable once) in the lists.

Consider the dictionary:

example_dictionary = {"weapon1":[1,2,3],"weapon2":[2,3],"weapon3":[2,3]}

'weapon2' and 'weapon3' have the same values, so it should result in:

result_dictionary = {"weapon1":[1],"weapon2":[3],"weapon3":[2]}

since I don't mind the order, it can also result in:

result_dictionary = {"weapon1":[1],"weapon2":[2],"weapon3":[3]}

But when "there's no choice" it should leave the value. Consider this new dictionary:

example_dictionary = {"weapon1":[1,2,3],"weapon2":[2,3],"weapon3":[2,3],"weapon4":[3]}

now, since it cannot assign either '2' or '3' only once without leaving a key empty, a possible output would be:

result_dictionary = {"weapon1":[1],"weapon2":[3],"weapon3":[2],"weapon4":[3]}

I can relax the problem to only the first part and manage, though I prefer a solution to the two parts together

Answer 1

#!/usr/bin/env python3

example_dictionary = {"weapon1":[1,2,3],"weapon2":[2,3],"weapon3":[2,3]}

result = {}
used_values = []

def extract_semi_unique_value(my_list):
    for val in my_list:
        if val not in used_values:
            used_values.append(val)
            return val
    return my_list[0]

for key, value in example_dictionary.items():
    semi_unique_value = extract_semi_unique_value(value)
    result[key] = [semi_unique_value]

print(result)

Answer 2

This is probably not the most efficient solution possible. Because it involves iteration over all possible combinations, then it'll run quite slow for large targets.

It makes use of itertools.product() to get all possible combinations. Then in it, tries to find the combination with the most unique numbers (by testing the length of a set).

from itertools import product
def dedup(weapons):
    # get the keys and values ordered so we can join them back
    #  up again at the end
    keys, vals = zip(*weapons.items())

    # because sets remove all duplicates, whichever combo has
    #  the longest set is the most unique
    best = max(product(*vals), key=lambda combo: len(set(combo)))

    # combine the keys and whatever we found was the best combo
    return {k: [v] for k, v in zip(keys, best)}

From the examples:

dedup({"weapon1":[1,2,3],"weapon2":[2,3],"weapon3":[2,3]})
#: {'weapon1': 1, 'weapon2': 2, 'weapon3': 3}
dedup({"weapon1":[1,2,3],"weapon2":[2,3],"weapon3":[2,3],"weapon4":[3]})
#: {'weapon1': 1, 'weapon2': 2, 'weapon3': 2, 'weapon4': 3}

Answer 3

this could help

import itertools
res = {'weapon1': [1, 2, 3], 'weapon2': [2, 3], 'weapon3': [2, 3]}
r = [[x] for x in list(set(list(itertools.chain.from_iterable(res.values()))))]
r2 = [x for x in res.keys()]
r3 = list(itertools.product(r2,r))
r4 = dict([r3[x] for x in range(0,len(r3)) if not x%4])