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.
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)
{}
What is the syntax for dictionary comprehension in Python?
The syntax for dictionary comprehension in Python is as follows: {key:value for (key,value) in iterable}
What are the benefits of using dictionary comprehension in Python?
The benefits of using dictionary comprehension in Python are that it is concise, readable, and reduces the amount of code needed to create dictionaries. It also allows for filtering and transformation of dictionary elements in a single line of code.
How do I create a dictionary from a list using dictionary comprehension in Python?
You can create a dictionary from a list using dictionary comprehension in Python by iterating over the list and setting the list elements as keys and values in the new dictionary.
Use a dict comprehension (Python 2.7 and later):
Use with AI{key: value for key, value in zip(keys, values)}
Alternatively, use the dict constructor:
Use with AIpairs = [('a', 1), ('b', 2)]
dict(pairs) # → {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs) # → {'a': 11, 'b': 12}
Given separate lists of keys and values, use the dict constructor with zip:
Use with AIkeys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values)) # → {'a': 1, 'b': 2}
In Python 3 and Python 2.7+, dictionary comprehensions look like the below:
Use with AId = {k:v for k, v in iterable}
For Python 2.6 or earlier, see fortran's answer.
