如何从列表字典的值中获取所有键？

Question

I have a dictionary in a format like this

d = {
    "Fruit_1" : ["mango", "apple"],
    "Fruit_2" : ["apple"],
    "Fruit_3" : ["mango", "banana", "apple", "kiwi", "orange"]
}

I'm passing a value as "mango" and I want to get all corresponding keys where only mango occurs. I am not able to get corresponding keys where ever value occurs.

Answer 1

Iterate in d.items and check mango existence in value.

In [21]: [key for key,value in d.items() if 'mango' in value]
Out[21]: ['Fruit_1', 'Fruit_3']

Answer 2

You can do this perhaps:

d = {
    "Fruit_1" : ["mango", "apple"],
    "Fruit_2" : ["apple"],
    "Fruit_3" : ["mango", "banana", "apple", "kiwi", "orange"]
}

# list comprehension
mango_keys = [fruit for fruit in d.keys() if "mango" in d[fruit]]
print(mango_keys)       


# ['Fruit_1', 'Fruit_3']         


# or more traditional for-loop (but non pythonic)

for fruit in d.keys():
    if "mango" in d[fruit]:
        print(fruit)

Answer 3

The naive approaches (looping through all items and looking for the fruit) work but have a high complexity, mostly if you have to perform a lot of requests. You could slightly improve it by replacing your list values by a set (for faster in lookup), but that would still be slow ( O(n**2) => O(n) but room for improvement).

If you want to be able to perform those queries a lot of times, it would be better to rebuild the dictionary so lookup is very fast once built, using collections.defaultdict

d = {
    "Fruit_1" : ["mango", "apple"],
    "Fruit_2" : ["apple"],
    "Fruit_3" : ["mango", "banana", "apple", "kiwi", "orange"]
}

import collections

newd = collections.defaultdict(list)

for k,vl in d.items():
    for v in vl:
        newd[v].append(k)

print(newd)
print(newd["mango"])

this is the rebuilt dict:

defaultdict(<class 'list'>, {'apple': ['Fruit_2', 'Fruit_3', 'Fruit_1'], 'orange': ['Fruit_3'], 'banana': ['Fruit_3'], 'kiwi': ['Fruit_3'], 'mango': ['Fruit_3', 'Fruit_1']})

this is the query for "mango":

['Fruit_3', 'Fruit_1']

Answer 4

Like this?

>>> d = {
...     "Fruit_1" : ["mango", "apple"],
...     "Fruit_2" : ["apple"],
...     "Fruit_3" : ["mango", "banana", "apple", "kiwi", "orange"]
... }
>>> 
>>> [key for key, value in d.items() if 'mango' in value]
['Fruit_1', 'Fruit_3']

The idea is to iterate over the (key, value) itempairs and check each value for the existence of 'mango' . If yes, keep the key.

Since you are new to Python here's the tradidtional for -loop logic:

>>> result = []
>>> for key, value in d.items():
...     if 'mango' in value:
...         result.append(key)
... 
>>> result
['Fruit_1', 'Fruit_3']

Answer 5

For a single query, you can use a list comprehension. This will has O( n ) time complexity each time you search a value:

res = [k for k, v in d.items() if 'mango' in v]

For multiple queries, you can use a defaultdict of set objects via a one-off O( n ) cost:

from collections import defaultdict

dd = defaultdict(set)

for k, v in d.items():
    for fruit in v:
        dd[fruit].add(k)

print(dd)

defaultdict({'mango': {'Fruit_1', 'Fruit_3'},
             'apple': {'Fruit_1', 'Fruit_2', 'Fruit_3'},
             'banana': {'Fruit_3'},
             'kiwi': {'Fruit_3'},
             'orange': {'Fruit_3'}})

You can then use dd['mango'] to extract relevant keys.