CODEWITHSUNDEEP
pythonpython-2.7
ierdna
ierdna

Reputation: 6293

Shorthand to create a dictionary of lists?

I have a list of keys and want to create a dictionary that looks as follows:

d = {"a":[] ,"b":[] ,"c":[] ,"d":[]}

If i try something like this:

keys = ['a','b','c','d']
d = dict(zip(keys,[[]]*len(keys)))

it creates the dictionary just fine, except all lists are references to the same object, so if I try d['c'].append('one') then it gets appended to all four lists.

I solved it using dict comprehension (for those interested: d = {k : [] for k in keys}) but wondering if there's a more efficient way?

Edit: yes I KNOW it's a dict comprehension. That was a typo obviously.

Upvotes: 3

Views: 4353

Answers (3)

Moinuddin Quadri
Moinuddin Quadri

Reputation: 48077

Dict comprehensions are pretty efficient (and it is not the list comprehension as mentioned by you). I would suggest to go with the dict comprehension. Surely there are few alternatives to achieve this, but they won't be as efficient as the one you mentioned.

Below is the timeit comparison of your alternative solution using zip (even though I know that it doesn't works the way you expect) and the one using dict comparison on Python 2.7:

# Using dict comprehension --> 0.415 usec    
mquadri$ python -m "timeit" -s "keys=keys = ['a','b','c','d']" "{key: [] for key in keys}"
1000000 loops, best of 3: 0.415 usec per loop

# Using `dict`-`zip` conversion --> 1.48 usec    
mquadri$ python -m "timeit" -s "keys=keys = ['a','b','c','d']" "dict(zip(keys,[[]]*len(keys)))"
1000000 loops, best of 3: 1.48 usec per loop

As you can see, dict comprehension approach is about three times faster and I doubt that there is any other more efficient way to achieve this.

Upvotes: 3

Vaishali
Vaishali

Reputation: 38415

keys = ['a','b','c','d']
d = dict((k,[]) for k in keys)
d['c'].append('one')
print(d)

You get

{'c': ['one'], 'a': [], 'b': [], 'd': []}

Upvotes: 1

Rafael Barros
Rafael Barros

Reputation: 2881

How about defaultdict:

from collections import defaultdict

data = defaultdict(list)

data['key'].append('a')
data['key'] # prints ['a']

This way you directly skip the dict iteration in the beginning.

Upvotes: 4

Related Questions