CODEWITHSUNDEEP
python
idealistikz
idealistikz

Reputation: 1267

Most efficient way to create a dictionary of empty lists in Python?

I am initializing a dictionary of empty lists. Is there a more efficient method of initializing the dictionary than the following?

dictLists = {}
dictLists['xcg'] = []
dictLists['bsd'] = []
dictLists['ghf'] = []
dictLists['cda'] = []
...

Is there a way I do not have to write dictLists each time, or is this unavoidable?

Upvotes: 0

Views: 8683

Answers (5)

Jakob Bowyer
Jakob Bowyer

Reputation: 34718

You can use collections.defaultdict it allows you to set a factory method that returns specific values on missing keys.

a = collections.defaultdict(list)

Edit:

Here are my keys

b = ['a', 'b','c','d','e']

Now here is me using the "predefined keys"

for key in b:
    a[key].append(33)

Upvotes: 8

Martijn Pieters
Martijn Pieters

Reputation: 1124110

If you adding static keys, why not just directly put them in the dict constructor?

dictLists = dict(xcg=[], bsd=[], ghf=[])

or, if your keys are not always also valid python identifiers:

dictLists = {'xcg': [], 'bsd': [], 'ghf': []}

If, on the other hand, your keys are not static or easily stored in a sequence, looping over the keys could be more efficient if there are a lot of them. The following example still types out the keys variable making this no more efficient than the above methods, but if keys were generated in some way this could be more efficient:

keys = ['xcg', 'bsd', 'ghf', …]
dictLists = {key: [] for key in keys}

Upvotes: 4

JAB
JAB

Reputation: 21089

In the versions of Python that support dictionary comprehensions (Python 2.7+):

dictLists = {listKey: list() for listKey in listKeys}

Upvotes: 1

Fred Foo
Fred Foo

Reputation: 363787

If the keys are known in advance, you can do

dictLists = dict((key, []) for key in ["xcg", "bsd", ...])

Or, in Python >=2.7:

dictLists = {key: [] for key in ["xcg", "bsd", ...]}

Upvotes: 7

yasar
yasar

Reputation: 13758

You can also use a for loop for the keys to add:

for i in ("xcg", "bsd", "ghf", "cda"):
    dictList[i] = []

Upvotes: 2

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