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

Question

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?

Are the dictionary keys what you actually want? Does it matter if they are random or an arithmetic progression? — Ben
– Ben, Commented Jun 1, 2012 at 14:42
Loads of answers guessing your keys are stored in a variable or similar and you are looping. Are your keys static like that? Or are they generated? — Martijn Pieters
– Martijn Pieters, Commented Jun 1, 2012 at 14:47
@MartijnPieters All (most?) of the answers utilizing keys stored in an iterable can be made static by just replacing the variable name with a list or tuple literal. Using a variable just makes the representation more elegant. — JAB
– JAB, Commented Jun 1, 2012 at 16:48
@JAB: Have you seen my answer, below? :-) It actually uses static list in a variable in one example. Contrast this with generating the keys (keys = ['key%02i' for i in range(10)]). — Martijn Pieters
– Martijn Pieters, Commented Jun 1, 2012 at 16:52

Jakob Bowyer · Accepted Answer · 2012-06-01 15:47:16Z

8

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)

edited Jun 1, 2012 at 15:47

answered Jun 1, 2012 at 14:42

Jakob Bowyer

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Fred Foo · Accepted Answer · 2012-06-01 14:43:51Z

7

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", ...]}

answered Jun 1, 2012 at 14:43

Fred Foo

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Martijn Pieters · Accepted Answer · 2012-06-01 15:10:52Z

4

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}

edited Jun 1, 2012 at 15:10

answered Jun 1, 2012 at 14:42

Martijn Pieters

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1 Comment

georg Over a year ago

This is definitely most efficient, as the poster asked.

yasar · Accepted Answer · 2012-06-01 14:42:51Z

2

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

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

answered Jun 1, 2012 at 14:42

yasar

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JAB · Accepted Answer · 2012-06-01 14:44:11Z

1

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

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

answered Jun 1, 2012 at 14:44

JAB

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JAB Over a year ago

@MartijnPieters Thanks, simplified my answer now that I have the confirmation. [] could also work in place of list(), as well, as used in someone else's answer.

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