How to return dictionary keys as a list in Python?

Question

Try list(newdict.keys()).

This will convert the dict_keys object to a list.

On the other hand, you should ask yourself whether or not it matters. The Pythonic way to code is to assume duck typing (if it looks like a duck and it quacks like a duck, it's a duck). The dict_keys object will act like a list for most purposes. For instance:

for key in newdict.keys():
  print(key)

Obviously, insertion operators may not work, but that doesn't make much sense for a list of dictionary keys anyway.

Python >= 3.5 alternative: unpack into a list literal [*newdict]

New unpacking generalizations (PEP 448) were introduced with Python 3.5 allowing you to now easily do:

>>> newdict = {1:0, 2:0, 3:0}
>>> [*newdict]
[1, 2, 3]

Unpacking with * works with any object that is iterable and, since dictionaries return their keys when iterated through, you can easily create a list by using it within a list literal.

Adding .keys() i.e [*newdict.keys()] might help in making your intent a bit more explicit though it will cost you a function look-up and invocation. (which, in all honesty, isn't something you should really be worried about).

_{The *iterable syntax is similar to doing list(iterable) and its behaviour was initially documented in the Calls section of the Python Reference manual. With PEP 448 the restriction on where *iterable could appear was loosened allowing it to also be placed in list, set and tuple literals, the reference manual on Expression lists was also updated to state this.}

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Though equivalent to list(newdict) with the difference that it's faster (at least for small dictionaries) because no function call is actually performed:

%timeit [*newdict]
1000000 loops, best of 3: 249 ns per loop

%timeit list(newdict)
1000000 loops, best of 3: 508 ns per loop

%timeit [k for k in newdict]
1000000 loops, best of 3: 574 ns per loop

with larger dictionaries the speed is pretty much the same (the overhead of iterating through a large collection trumps the small cost of a function call).

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In a similar fashion, you can create tuples and sets of dictionary keys:

>>> *newdict,
(1, 2, 3)
>>> {*newdict}
{1, 2, 3}

beware of the trailing comma in the tuple case!

list(newdict) works in both Python 2 and Python 3, providing a simple list of the keys in newdict. keys() isn't necessary.

You can also use a list comprehension:

>>> newdict = {1:0, 2:0, 3:0}
>>> [k  for  k in  newdict.keys()]
[1, 2, 3]

Or, shorter,

>>> [k  for  k in  newdict]
[1, 2, 3]

Note: Order is not guaranteed on versions under 3.7 (ordering is still only an implementation detail with CPython 3.6).

A bit off on the "duck typing" definition -- dict.keys() returns an iterable object, not a list-like object. It will work anywhere an iterable will work -- not any place a list will. a list is also an iterable, but an iterable is NOT a list (or sequence...)

In real use-cases, the most common thing to do with the keys in a dict is to iterate through them, so this makes sense. And if you do need them as a list you can call list().

Very similarly for zip() -- in the vast majority of cases, it is iterated through -- why create an entire new list of tuples just to iterate through it and then throw it away again?

This is part of a large trend in python to use more iterators (and generators), rather than copies of lists all over the place.

dict.keys() should work with comprehensions, though -- check carefully for typos or something... it works fine for me:

>>> d = dict(zip(['Sounder V Depth, F', 'Vessel Latitude, Degrees-Minutes'], [None, None]))
>>> [key.split(", ") for key in d.keys()]
[['Sounder V Depth', 'F'], ['Vessel Latitude', 'Degrees-Minutes']]

Converting to a list without using the keys method makes it more readable:

list(newdict)

and, when looping through dictionaries, there's no need for keys():

for key in newdict:
    print key

unless you are modifying it within the loop which would require a list of keys created beforehand:

for key in list(newdict):
    del newdict[key]

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On Python 2 there is a marginal performance gain using keys().

If you need to store the keys separately, here's a solution that requires less typing than every other solution presented thus far, using Extended Iterable Unpacking (Python3.x+):

newdict = {1: 0, 2: 0, 3: 0}
*k, = newdict

k
# [1, 2, 3]

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Operation	no. Of characters
k = list(d)	9 characters (excluding whitespace)
k = [*d]	6 characters
*k, = d	5 characters