Python Dictionary Comprehension

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 .

---

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.

---

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

>>> dict.fromkeys(range(1, 11), True)
{1: True, 2: True, 3: True, 4: True, 5: True, 6: True, 7: True, 8: True, 9: True, 10: True}

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

Be careful using this with mutable objects though:

d = dict.fromkeys(range(10), [])
d[1].append(2)
print(d[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(lambda: True)

---

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

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

Or for python2.6:

dict((k, k) for k in range(10))

You could also create a clever 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)
{}

Neat.

---

>>> {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}

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