How do I convert two lists into a dictionary?

Like this:

keys = ['a', 'b', 'c']
values = [1, 2, 3]
dictionary = dict(zip(keys, values))
print(dictionary) # {'a': 1, 'b': 2, 'c': 3}

Voila :-) The pairwise dict constructor and zip function are awesomely useful.

Try this:

>>> import itertools
>>> keys = ('name', 'age', 'food')
>>> values = ('Monty', 42, 'spam')
>>> adict = dict(itertools.izip(keys,values))
>>> adict
{'food': 'spam', 'age': 42, 'name': 'Monty'}

In Python 2, it's also more economical in memory consumption compared to zip.

Imagine that you have:

keys = ('name', 'age', 'food')
values = ('Monty', 42, 'spam')

What is the simplest way to produce the following dictionary ?

dict = {'name' : 'Monty', 'age' : 42, 'food' : 'spam'}

Most performant, dict constructor with zip

new_dict = dict(zip(keys, values))

In Python 3, zip now returns a lazy iterator, and this is now the most performant approach.

dict(zip(keys, values)) does require the one-time global lookup each for dict and zip, but it doesn't form any unnecessary intermediate data-structures or have to deal with local lookups in function application.

Runner-up, dict comprehension:

A close runner-up to using the dict constructor is to use the native syntax of a dict comprehension (not a list comprehension, as others have mistakenly put it):

new_dict = {k: v for k, v in zip(keys, values)}

Choose this when you need to map or filter based on the keys or value.

In Python 2, zip returns a list, to avoid creating an unnecessary list, use izip instead (aliased to zip can reduce code changes when you move to Python 3).

from itertools import izip as zip

So that is still (2.7):

new_dict = {k: v for k, v in zip(keys, values)}

Python 2, ideal for <= 2.6

izip from itertools becomes zip in Python 3. izip is better than zip for Python 2 (because it avoids the unnecessary list creation), and ideal for 2.6 or below:

from itertools import izip
new_dict = dict(izip(keys, values))

Result for all cases:

In all cases:

>>> new_dict
{'age': 42, 'name': 'Monty', 'food': 'spam'}

Explanation:

If we look at the help on dict we see that it takes a variety of forms of arguments:

>>> help(dict)

class dict(object)
 |  dict() -> new empty dictionary
 |  dict(mapping) -> new dictionary initialized from a mapping object's
 |      (key, value) pairs
 |  dict(iterable) -> new dictionary initialized as if via:
 |      d = {}
 |      for k, v in iterable:
 |          d[k] = v
 |  dict(**kwargs) -> new dictionary initialized with the name=value pairs
 |      in the keyword argument list.  For example:  dict(one=1, two=2)

The optimal approach is to use an iterable while avoiding creating unnecessary data structures. In Python 2, zip creates an unnecessary list:

>>> zip(keys, values)
[('name', 'Monty'), ('age', 42), ('food', 'spam')]

In Python 3, the equivalent would be:

>>> list(zip(keys, values))
[('name', 'Monty'), ('age', 42), ('food', 'spam')]

and Python 3's zip merely creates an iterable object:

>>> zip(keys, values)
<zip object at 0x7f0e2ad029c8>

Since we want to avoid creating unnecessary data structures, we usually want to avoid Python 2's zip (since it creates an unnecessary list).

Less performant alternatives:

This is a generator expression being passed to the dict constructor:

generator_expression = ((k, v) for k, v in zip(keys, values))
dict(generator_expression)

or equivalently:

dict((k, v) for k, v in zip(keys, values))

And this is a list comprehension being passed to the dict constructor:

dict([(k, v) for k, v in zip(keys, values)])

In the first two cases, an extra layer of non-operative (thus unnecessary) computation is placed over the zip iterable, and in the case of the list comprehension, an extra list is unnecessarily created. I would expect all of them to be less performant, and certainly not more-so.

Performance review:

In 64 bit Python 3.8.2 provided by Nix, on Ubuntu 16.04, ordered from fastest to slowest:

>>> min(timeit.repeat(lambda: dict(zip(keys, values))))
0.6695233230129816
>>> min(timeit.repeat(lambda: {k: v for k, v in zip(keys, values)}))
0.6941362579818815
>>> min(timeit.repeat(lambda: {keys[i]: values[i] for i in range(len(keys))}))
0.8782548159942962
>>> 
>>> min(timeit.repeat(lambda: dict([(k, v) for k, v in zip(keys, values)])))
1.077607496001292
>>> min(timeit.repeat(lambda: dict((k, v) for k, v in zip(keys, values))))
1.1840861019445583

dict(zip(keys, values)) wins even with small sets of keys and values, but for larger sets, the differences in performance will become greater.

A commenter said:

min seems like a bad way to compare performance. Surely mean and/or max would be much more useful indicators for real usage.

We use min because these algorithms are deterministic. We want to know the performance of the algorithms under the best conditions possible.

If the operating system hangs for any reason, it has nothing to do with what we're trying to compare, so we need to exclude those kinds of results from our analysis.

If we used mean, those kinds of events would skew our results greatly, and if we used max we will only get the most extreme result - the one most likely affected by such an event.

A commenter also says:

In python 3.6.8, using mean values, the dict comprehension is indeed still faster, by about 30% for these small lists. For larger lists (10k random numbers), the dict call is about 10% faster.

I presume we mean dict(zip(... with 10k random numbers. That does sound like a fairly unusual use case. It does makes sense that the most direct calls would dominate in large datasets, and I wouldn't be surprised if OS hangs are dominating given how long it would take to run that test, further skewing your numbers. And if you use mean or max I would consider your results meaningless.

Let's use a more realistic size on our top examples:

import numpy
import timeit
l1 = list(numpy.random.random(100))
l2 = list(numpy.random.random(100))

And we see here that dict(zip(... does indeed run faster for larger datasets by about 20%.

>>> min(timeit.repeat(lambda: {k: v for k, v in zip(l1, l2)}))
9.698965263989521
>>> min(timeit.repeat(lambda: dict(zip(l1, l2))))
7.9965161079890095