Python Dictionary Comprehension

Question

Is it possible to create a dictionary comprehension in Python (for the keys)?

Without list comprehensions, you can use something like this:

l = []
for n in range(1, 11):
    l.append(n)

We can shorten this to a list comprehension: l = [n for n in range(1, 11)].

However, say I want to set a dictionary's keys to the same value. I can do:

d = {}
for n in range(1, 11):
     d[n] = True # same value for each

I've tried this:

d = {}
d[i for i in range(1, 11)] = True

However, I get a SyntaxError on the for.

In addition (I don't need this part, but just wondering), can you set a dictionary's keys to a bunch of different values, like this:

d = {}
for n in range(1, 11):
    d[n] = n

Is this possible with a dictionary comprehension?

d = {}
d[i for i in range(1, 11)] = [x for x in range(1, 11)]

This also raises a SyntaxError on the for.

Answer 1

There are dictionary comprehensions in Python 2.7+, but they don't work quite the way you're trying. Like a list comprehension, they create a new dictionary; you can't use them to add keys to an existing dictionary. Also, you have to specify the keys and values, although of course you can specify a dummy value if you like.

>>> d = {n: n**2 for n in range(5)}
>>> print d
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

If you want to set them all to True:

>>> d = {n: True for n in range(5)}
>>> print d
{0: True, 1: True, 2: True, 3: True, 4: True}

What you seem to be asking for is a way to set multiple keys at once on an existing dictionary. There's no direct shortcut for that. You can either loop like you already showed, or you could use a dictionary comprehension to create a new dict with the new values, and then do oldDict.update(newDict) to merge the new values into the old dict.

Answer 2

I really like the @mgilson comment, since if you have a two iterables, one that corresponds to the keys and the other the values, you can also do the following.

keys = ['a', 'b', 'c']
values = [1, 2, 3]
d = dict(zip(keys, values))

giving

d = {'a': 1, 'b': 2, 'c': 3}

Answer 3

you can't hash a list like that. try this instead, it uses tuples

d[tuple([i for i in range(1,11)])] = True

Answer 4

The main purpose of a list comprehension is to create a new list based on another one without changing or destroying the original list.

Instead of writing

    l = []
    for n in range(1, 11):
        l.append(n)

or

    l = [n for n in range(1, 11)]

you should write only

    l = range(1, 11)

In the two top code blocks you're creating a new list, iterating through it and just returning each element. It's just an expensive way of creating a list copy.

To get a new dictionary with all keys set to the same value based on another dict, do this:

    old_dict = {'a': 1, 'c': 3, 'b': 2}
    new_dict = { key:'your value here' for key in old_dict.keys()}

You're receiving a SyntaxError because when you write

    d = {}
    d[i for i in range(1, 11)] = True

you're basically saying: "Set my key 'i for i in range(1, 11)' to True" and "i for i in range(1, 11)" is not a valid key, it's just a syntax error. If dicts supported lists as keys, you would do something like

    d[[i for i in range(1, 11)]] = True

and not

    d[i for i in range(1, 11)] = True

but lists are not hashable, so you can't use them as dict keys.

Answer 5

>>> {i:i for i in range(1, 11)}
{1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9, 10: 10}

Answer 6

You can use the dict.fromkeys class method ...

>>> dict.fromkeys(range(5), True)
{0: True, 1: True, 2: True, 3: True, 4: True}

This is the fastest way to create a dictionary where all the keys map to the same value.

But do not use this with mutable objects:

d = dict.fromkeys(range(5), [])
# {0: [], 1: [], 2: [], 3: [], 4: []}
d[1].append(2)
# {0: [2], 1: [2], 2: [2], 3: [2], 4: [2]} !!!

If you don't actually need to initialize all the keys, a defaultdict might be useful as well:

from collections import defaultdict
d = defaultdict(True)

---

To answer the second part, a dict-comprehension is just what you need:

{k: k for k in range(10)}

You probably shouldn't do this but you could also create a subclass of dict which works somewhat like a defaultdict if you override __missing__:

>>> class KeyDict(dict):
...    def __missing__(self, key):
...       #self[key] = key  # Maybe add this also?
...       return key
... 
>>> d = KeyDict()
>>> d[1]
1
>>> d[2]
2
>>> d[3]
3
>>> print(d)
{}