How do I return dictionary keys as a list in Python?

Question

With Python 2.7, I can get dictionary keys, values, or items as a list:

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

With Python >= 3.3, I get:

>>> newdict.keys()
dict_keys([1, 2, 3])

How do I get a plain list of keys with Python 3?

Answer 1

This will convert the dict_keys object to a list:

list(newdict.keys())

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On the other hand, you should ask yourself whether or not it matters. It is Pythonic to assume duck typing -- if it looks like a duck and it quacks like a duck, it is a duck. The dict_keys object can be iterated over just like a list. For instance:

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

Note that dict_keys doesn't support insertion newdict[k] = v, though you may not need it.

Answer 2

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 3

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

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

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']]

Answer 6

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

Answer 7

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 8

Yes, There is a better and simplest way to do this in python3.X

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use inbuild list() function

#Devil
newdict = {1:0, 2:0, 3:0}
key_list = list(newdict)
print(key_list) 
#[1, 2, 3] 

Answer 9

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 10

Beyond the classic (and probably more correct) way to do this (some_dict.keys()) there is also a more "cool" and surely more interesting way to do this:

some_dict = { "foo": "bar", "cool": "python!" }
print( [*some_dict] == ["foo", "cool"] )         # True  

Note: this solution shouldn't be used in a develop environment; I showed it here just because I thought it was quite interesting from the *-operator-over-dictionary side of view. Also, I'm not sure whether this is a documented feature or not, and its behaviour may change in later versions :)

Answer 11

You can you use simple method like below

keys = newdict.keys()
print(keys)

Answer 12

Get a list of keys with specific values

You can select a subset of the keys that satisfies a specific condition. For example, if you want to select the list of keys where the corresponding values are not None, then use

[k for k,v in newdict.items() if v is not None]

Slice the list of keys in a dictionary

Since Python 3.7, dicts preserve insertion order. So one use case of list(newdict) might be to select keys from a dictionary by its index or slice it (not how it's "supposed" to be used but certainly a possible question). Instead of converting to a list, use islice from the built-in itertools module, which is much more efficient since converting the keys into a list just to throw away most of it is very wasteful. For example, to select the second key in newdict:

from itertools import islice
next(islice(newdict, 1, 2))

or to slice the second to fifth key:

list(islice(newdict, 1, 6))

For large dicts, it's thousands of times faster than list(newdict)[1] etc.

Answer 13

This is the best way to get key List in one line of code

dict_variable = {1:"a",2:"b",3:"c"}  
[key_val for key_val in dict_variable.keys()]