How to return dictionary keys as a list in Python?

Question

In Python 2.7, I could get dictionary keys, values, or items as a list:

>>> newdict = {1:0, 2:0, 3:0}
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Answer 1

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!

Answer 2

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

Answer 3

I can think of 2 ways in which we can extract the keys from the dictionary.

Method 1: - To get the keys using .keys() method and then convert it to list.

some_dict = {1: 'one', 2: 'two', 3: 'three'}
list_of_keys = list(some_dict.keys())
print(list_of_keys)
-->[1,2,3]

Method 2: - To create an empty list and then append keys to the list via a loop. You can get the values with this loop as well (use .keys() for just keys and .items() for both keys and values extraction)

list_of_keys = []
list_of_values = []
for key,val in some_dict.items():
    list_of_keys.append(key)
    list_of_values.append(val)

print(list_of_keys)
-->[1,2,3]

print(list_of_values)
-->['one','two','three']

Answer 4

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).

Answer 5

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().

Answer 6

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.