Convert a list to a dictionary in Python


Let's say I have a list a in Python whose entries conveniently map to a dictionary. Each even element represents the key to the dictionary, and the following odd element is the value

for example,

a = ['hello','world','1','2']

and I'd like to convert it to a dictionary b, where

b['hello'] = 'world'
b['1'] = '2'

What is the syntactically cleanest way to accomplish this?

4/27/2016 2:25:16 PM

Accepted Answer

b = dict(zip(a[::2], a[1::2]))

If a is large, you will probably want to do something like the following, which doesn't make any temporary lists like the above.

from itertools import izip
i = iter(a)
b = dict(izip(i, i))

In Python 3 you could also use a dict comprehension, but ironically I think the simplest way to do it will be with range() and len(), which would normally be a code smell.

b = {a[i]: a[i+1] for i in range(0, len(a), 2)}

So the iter()/izip() method is still probably the most Pythonic in Python 3, although as EOL notes in a comment, zip() is already lazy in Python 3 so you don't need izip().

i = iter(a)
b = dict(zip(i, i))

If you want it on one line, you'll have to cheat and use a semicolon. ;-)

1/7/2018 11:14:37 PM

Simple answer

Another option (courtesy of Alex Martelli

dict(x[i:i+2] for i in range(0, len(x), 2))

If you have this:

a = ['bi','double','duo','two']

and you want this (each element of the list keying a given value (2 in this case)):


you can use:

>>> dict((k,2) for k in a)
{'double': 2, 'bi': 2, 'two': 2, 'duo': 2}

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