Is it possible to create a dictionary comprehension in Python (for the keys)?
Without list comprehensions, you can use something like this:
l = []
fo
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)
{}