Getting key with maximum value in dictionary?

Question

I have a dictionary: keys are strings, values are integers. Example: stats = {'a':1000, 'b':3000, 'c': 100}.

I'd like to get 'b' as an answer, since it's the key with a higher value.

I did the following, using an intermediate list with reversed key-value tuples:

    inverse = [(value, key) for key, value in stats.items()]
    print max(inverse)[1]

Is that one the better (or even more elegant) approach?

Answer 1

You can use operator.itemgetter for that:

import operator
stats = {'a':1000, 'b':3000, 'c': 100}
max(stats.iteritems(), key=operator.itemgetter(1))[0]

And instead of building a new list in memory use stats.iteritems(). The key parameter to the max() function is a function that computes a key that is used to determine how to rank items.

Answer 2

Thanks, very elegant, I didn't remember that max allows a "key" parameter.

BTW, to get the right answer ('b') it has to be:

import operator
stats = {'a':1000, 'b':3000, 'c': 100}
max(stats.iteritems(), key=operator.itemgetter(1))[0]

Answer 3

Here is another one:

stats = {'a':1000, 'b':3000, 'c': 100}
max(stats.iterkeys(), key=lambda k: stats[k])

The key-function simply returns the value that should be used for ranking and max() returns the demanded element right away.

Answer 4

key,value = max(stats.iteritems(), key=lambda x:x[1])

If you don't care about value (I'd be surprised, but) you can do:

key,ignored = max(stats.iteritems(), key=lambda x:x[1])

I like the tuple unpacking better than a [0] subscript at the end of the expression. I never like the readability of lambda expressions very much, but find this one better than the operator.itemgetter(1) IMHO.

Answer 5

max(stats, key=stats.get)