I'd like to get from this:
keys = [1,2,3]
to this:
{1: None, 2: None, 3: None}
Is there a pythonic way of doing it?
This is an ugly way to do it:
>>> keys = [1,2,3]
>>> dict[[[1,2]]]
{1: 2}
>>> dict[zip[keys, [None]*len[keys]]]
{1: None, 2: None, 3: None}
evtoh
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asked Feb 11, 2010 at 2:43
Juanjo ContiJuanjo Conti
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dict.fromkeys directly solves the problem:
>>> dict.fromkeys[[1, 2, 3, 4]]
{1: None, 2: None, 3: None, 4: None}
This is actually a classmethod, so it works for dict-subclasses [like collections.defaultdict] as well.
The optional second argument, which defaults to None, specifies the value to use for the keys. Note that the same object will be used for each key, which can cause problems with mutable values:
>>> x = dict.fromkeys[[1, 2, 3, 4], []]
>>> x[1].append['test']
>>> x
{1: ['test'], 2: ['test'], 3: ['test'], 4: ['test']}
answered Feb 11, 2010 at 2:45
Thomas WoutersThomas Wouters
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Use a dict comprehension:
>>> keys = [1,2,3,5,6,7]
>>> {key: None for key in keys}
{1: None, 2: None, 3: None, 5: None, 6: None, 7: None}
The value expression is evaluated each time, so this can be used to create a dict with separate lists [say] as values:
>>> x = {key: [] for key in [1, 2, 3, 4]}
>>> x[1] = 'test'
>>> x
{1: 'test', 2: [], 3: [], 4: []}
answered Feb 11, 2010 at 11:18
Adrien PlissonAdrien Plisson
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dict.fromkeys[keys, None]
answered Feb 11, 2010 at 2:48
Dominic CooneyDominic Cooney
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A list comprehension can be used to build a list of key-value pairs, which can then be passed to the dict constructor. Thus:
>>> keys = {"a", "b", "c", "d"}
>>> d = dict[[[key, []] for key in keys]]
>>> d
{'d': [], 'c': [], 'a': [], 'b': []}
The value expression is evaluated each time, creating separate lists in the above example:
>>> d['a'].append['test']
>>> d
{'d': [], 'c': [], 'a': ['test'], 'b': []}
answered Aug 24, 2015 at 12:31
3
Simply iterate and add the values to an empty dictionary:
d = {}
for i in keys:
d[i] = None
answered Feb 11, 2010 at 4:59
inspectorG4dgetinspectorG4dget
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0
In many workflows where you want to attach a default / initial value for arbitrary keys, you don't need to hash each key individually ahead of time. You can use collections.defaultdict. For example:
from collections import defaultdict
d = defaultdict[lambda: None]
print[d[1]] # None
print[d[2]] # None
print[d[3]] # None
This is more efficient, it saves having to hash all your keys at instantiation. Moreover, defaultdict is a subclass of dict, so there's usually no need to
convert back to a regular dictionary.
For workflows where you require controls on permissible keys, you can use dict.fromkeys as per the accepted answer:
d = dict.fromkeys[[1, 2, 3, 4]]
answered Dec 17, 2018 at 11:47
jppjpp
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Just because it's fun how the dict constructor works nicely with
zip, you can repeat the default value and zip it to the keys:
from itertools import repeat
keys = [1, 2, 3]
default_value = None
d = dict[zip[keys, repeat[default_value]]]
print[d]
Will give:
{1: None, 2: None, 3: None}
repeat creates an infinite iterator of the element passed to it but as zip stops on the shortest iterable all works well.
answered Mar 9 at 8:45
TomerikooTomerikoo
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