Use a dict comprehension (Python 2.7 and later):
{key: value for key, value in zip(keys, values)}
Alternatively, use the dict constructor:
pairs = [('a', 1), ('b', 2)]
dict(pairs) # โ {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs) # โ {'a': 11, 'b': 12}
Given separate lists of keys and values, use the dict constructor with zip:
keys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values)) # โ {'a': 1, 'b': 2}
Answer from fortran on Stack OverflowPython Dictionary Comprehension - Stack Overflow
Dictionary Comprehension not iterating over all of list
dictionary comprehension -- order of keys created?
List comprehension unpacking JSON: handling missing values (KeyError)
Use a dict comprehension (Python 2.7 and later):
{key: value for key, value in zip(keys, values)}
Alternatively, use the dict constructor:
pairs = [('a', 1), ('b', 2)]
dict(pairs) # โ {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs) # โ {'a': 11, 'b': 12}
Given separate lists of keys and values, use the dict constructor with zip:
keys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values)) # โ {'a': 1, 'b': 2}
In Python 3 and Python 2.7+, dictionary comprehensions look like the below:
d = {k:v for k, v in iterable}
For Python 2.6 or earlier, see fortran's answer.
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(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(lambda: 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)
{}