Python 3 changed the behavior of dict.keys such that it now returns a dict_keys object, which is iterable but not indexable (it's like the old dict.iterkeys, which is gone now). You can get the Python 2 result back with an explicit call to list:

>>> b = { 'video':0, 'music':23 }>>> k = list(b.keys())>>> k['music', 'video']

or just

>>> list(b)['music', 'video']

If you assigned k like so:

k = list(b.keys())

your code will work.

As the error says, the dict_keys type does not support indexing.

This is one of the breaking changes between Python 2 and 3.

In Python 2:

>>> help(dict.keys)keys(...)    D.keys() -> list of D's keys

In Python 3:

>>> help(dict.keys)keys(...)    D.keys() -> a set-like object providing a view on D's keys

This change in behavior makes a lot of sense since a dict is semantically unordered and its keys are unique - just like a set.

This change means that you don't have to create a new list of keys every time you want to do some kind of set comparison with a dict's keys.

Getting the same behavior in 2 and 3

To help transition to Python 3, Python 2.7 has another dict method, viewkeys. The viewkeys method is most similar to Python 3's dict.keys method:

>>> d{'a': None, 'c': None, 'b': None, 'd': None}>>> for k in d.viewkeys(): print k... acbd>>> d.viewkeys() & set('abc')set(['a', 'c', 'b'])

In Python 3, the closest analog to the old behavior is to pass dict.keys() to list:

>>> d{'d': None, 'a': None, 'c': None, 'b': None}>>> list(d.keys())['d', 'a', 'c', 'b']

Or just pass the dict to list, since a dict will iterate over its keys anyways:

>>> list(d)['d', 'a', 'c', 'b']

You could create a utility functions to abstract the behavior over 2 and 3:

if hasattr(dict, 'viewkeys'): # Python 2.7    def keys(d):        return d.viewkeys()else:  # Python 3    def keys(d):        return d.keys()

And pass a dict to list to get the list form, and in both 2 and 3, you'll get the same output:

>>> d{'b': None, 'a': None, 'c': None, 'd': None}>>> keys(d)dict_keys(['b', 'a', 'c', 'd'])>>> list(d)['b', 'a', 'c', 'd']