Python: get key with the least value from a dictionary

Question

If I have a Python dictionary, how do I get the key to the entry which contains the minimum value?

I was thinking about something to do with the min() function...

Given the input:

{320:1, 321:0, 322:3}

It would return 321.

Any ideas?
Thanks!

Answer 1

Edit: this is an answer to the OP's original question about the minimal key, not the minimal answer.

---

You can get the keys of the dict using the keys function, and you're right about using min to find the minimum of that list.

Answer 2

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.

Answer 3

Is this what you are looking for?

d = dict()
d[15.0]='fifteen'
d[14.0]='fourteen'
d[14.5]='fourteenandhalf'

print d[min(d.keys())]

Prints 'fourteen'

Answer 4

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

Answer 5

# python 
d={320:1, 321:0, 322:3}
reduce(lambda x,y: x if d[x]<=d[y] else y, d.iterkeys())
  321

Answer 6

>>> d = {320:1, 321:0, 322:3}
>>> min(d, key=lambda k: d[k]) 
321

Answer 7

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

Answer 8

If you are not sure that you have not multiple minimum values, I would suggest:

d = {320:1, 321:0, 322:3, 323:0}
print ', '.join(str(key) for min_value in (min(d.values()),) for key in d if d[key]==min_value)

"""Output:
321, 323
"""

Answer 9

Another approach to addressing the issue of multiple keys with the same min value:

>>> dd = {320:1, 321:0, 322:3, 323:0}
>>>
>>> from itertools import groupby
>>> from operator import itemgetter
>>>
>>> print [v for k,v in groupby(sorted((v,k) for k,v in dd.iteritems()), key=itemgetter(0)).next()[1]]
[321, 323]