Efficient way to find key by value in Python dict where dict values are iterables-CodePudding

I have an iterable of unique numbers:

lst = [14, 11, 8, 55]

where every value is somewhere among numbers of dict's iterable-values, say lists:

dict_itms.items() = dict_items([(1, [0, 1, 2, 3]), (2, [11, 14, 12]), (3, [30, 8, 42]), (4, [55, 6])])

I have to find each lst element in a dict such a way that, finally, I would have a list of keys pairwise against each element in lst.

This method:

keys_ = []
for a in lst:
    for k, v in dict_itms.items():
        if a in v:
            keys_  = [k]
            break
        else:
            continue

gives: [2, 2, 3, 4]

Is there more efficient way to find every key pairwise against each number to find?

CodePudding user response：

You can use any in a list comprehension:

print([k for k,v in dict_itms.items() if any(x in lst for x in v)])

Output:

[2, 3, 4]

Update

According to this answer not set(v).isdisjoint(lst) is the fastest:

print([k for k,v in dict_itms.items() if not set(v).isdisjoint(lst)])

CodePudding user response：

A simple and Pythonic implementation:

d = dict([(1, [0, 1, 2, 3]), (2, [11, 14, 12]), (3, [30, 8, 42]), (4, [55, 6])])

xs = [14, 11, 8, 55]

keys = [k for k, v in d.items() if set(v).intersection(xs)]
print(keys)

However, this doesn't duplicate the 2 key, which your example does - not sure if that's behaviour you need?