How to 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?

Asked By: user2015601

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Answers:

This will convert the dict_keys object to a list:

list(newdict.keys())

On the other hand, you should ask yourself whether or not it matters. It is Pythonic to assume duck typingif 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.

Answered By: Chris

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']]
Answered By: Chris Barker

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

Answered By: Seb

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).

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]

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

Answered By: Cas

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.


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).


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!

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]

Operation no. Of characters
k = list(d) 9 characters (excluding whitespace)
k = [*d] 6 characters
*k, = d 5 characters
Answered By: cs95

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']
Answered By: Rahul

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 🙂

Answered By: Giuppox

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()]
Answered By: Neeraj Kumar

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

use inbuild list() function

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