Get the key corresponding to the minimum value within a dictionary Get the key corresponding to the minimum value within a dictionary python python

Get the key corresponding to the minimum value within a dictionary


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))


matomo