如果未设置最小键，则返回与字典的最小值关联的键

Question

I have a dictionary and a set. I want to return the key associated with the minimum value if that key is not already in the set without creating a second dictionary used to delete items that are already in the set.

Example: I have dictionary

d = {1:10, 2:20, 3:30, 4:40}

and a set

s = set([1,2])

It should return 3

Answer 1

Use the set operations to subtract the set s from the set of keys of d , then use min with key=d.get to get the remaining key with the smallest value:

d = {1:10, 2:20, 3:30, 4:40}
s = set([1,2])

print min(set(d) - s, key=d.get)

prints 3

If the result must be None when no key was found use:

q = set(d) - s
print min(q, key=d.get) if q else None

Rather than the ValueError from min .

Answer 2

You could use min with generator expression that returns keys not in s and key parameter that gets the value from d :

>>> d = {1:10, 2:20, 3:30, 4:40}
>>> s = set([1,2])
>>> min((k for k in d if k not in s), key=d.get)
3

Above approach wouldn't create any intermediate containers. In case there's a possibility that s matches keys from d default value can be used:

>>> d = {1:10, 2:20, 3:30, 4:40}
>>> s = set([1,2,3,4])
>>> min((k for k in d if k not in s), key=d.get, default=-1)
-1

Answer 3

Set operations work nicely here.

d = {1:10, 2:20, 3:30, 4:40}
s = set([1,2])

def find_min(d, s):
    keys = set(d)
    difference = keys-s
    return min(difference) if difference else None

>>> print find_min(d, s)
3

alternatively, you could have difference = filter(lambda e: e not in s, d)