Get the key corresponding to the minimum value within a dictionary

Question

Best: min(d, key=d.get) -- no reason to interpose a useless lambda indirection layer or extract items or keys!

>>> d = {320: 1, 321: 0, 322: 3}
>>> min(d, key=d.get)
321

Here's an answer that actually gives the solution the OP asked for:

>>> d = {320:1, 321:0, 322:3}
>>> d.items()
[(320, 1), (321, 0), (322, 3)]
>>> # find the minimum by comparing the second element of each tuple
>>> min(d.items(), key=lambda x: x[1]) 
(321, 0)

Using d.iteritems() will be more efficient for larger dictionaries, however.

For multiple keys which have equal lowest value, you can use a list comprehension:

d = {320:1, 321:0, 322:3, 323:0}

minval = min(d.values())
res = [k for k, v in d.items() if v==minval]

[321, 323]

An equivalent functional version:

res = list(filter(lambda x: d[x]==minval, d))

min(d.items(), key=lambda x: x[1])[0]