Convert two lists into a dictionary

Question

Imagine that you have:

keys = ['name', 'age', 'food']
values = ['Monty', 42, 'spam']

What is the simplest way to produce the following dictionary?

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

Answer 1

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: https://docs.python.org/3/library/functions.html#func-dict

Answer 2

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.

Answer 3

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 - Python 2.7 and 3, dict comprehension:

A possible improvement on 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)}

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:

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))

Python 3

In Python 3, zip becomes the same function that was in the itertools module, so that is simply:

new_dict = dict(zip(keys, values))

A dict comprehension would be more performant though (see performance review at the end of this answer).

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 3.4.3, on Ubuntu 14.04, ordered from fastest to slowest:

>>> min(timeit.repeat(lambda: {k: v for k, v in zip(keys, values)}))
0.7836067057214677
>>> min(timeit.repeat(lambda: dict(zip(keys, values))))
1.0321204089559615
>>> min(timeit.repeat(lambda: {keys[i]: values[i] for i in range(len(keys))}))
1.0714934510178864
>>> min(timeit.repeat(lambda: dict([(k, v) for k, v in zip(keys, values)])))
1.6110592018812895
>>> min(timeit.repeat(lambda: dict((k, v) for k, v in zip(keys, values))))
1.7361853648908436

Answer 4

>>> keys = ('name', 'age', 'food')
>>> values = ('Monty', 42, 'spam')
>>> dict(zip(keys, values))
{'food': 'spam', 'age': 42, 'name': 'Monty'}

Answer 5

You can also use dictionary comprehensions in Python ≥ 2.7:

>>> keys = ('name', 'age', 'food')
>>> values = ('Monty', 42, 'spam')
>>> {k: v for k, v in zip(keys, values)}
{'food': 'spam', 'age': 42, 'name': 'Monty'}

Answer 6

A more natural way is to use dictionary comprehension

keys = ('name', 'age', 'food')
values = ('Monty', 42, 'spam')    
dict = {keys[i]: values[i] for i in range(len(keys))}

Answer 7

If you need to transform keys or values before creating a dictionary then a generator expression could be used. Example:

>>> adict = dict((str(k), v) for k, v in zip(['a', 1, 'b'], [2, 'c', 3])) 

Take a look Code Like a Pythonista: Idiomatic Python.

Answer 8

with Python 3.x, goes for dict comprehensions

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

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

print(dic)

More on dict comprehensions here, an example is there:

>>> print {i : chr(65+i) for i in range(4)}
    {0 : 'A', 1 : 'B', 2 : 'C', 3 : 'D'}

Answer 9

For those who need simple code and aren’t familiar with zip:

List1 = ['This', 'is', 'a', 'list']
List2 = ['Put', 'this', 'into', 'dictionary']

This can be done by one line of code:

d = {List1[n]: List2[n] for n in range(len(List1))}

Answer 10

• 2018-04-18

The best solution is still:

In [92]: keys = ('name', 'age', 'food')
...: values = ('Monty', 42, 'spam')
...: 

In [93]: dt = dict(zip(keys, values))
In [94]: dt
Out[94]: {'age': 42, 'food': 'spam', 'name': 'Monty'}

Tranpose it:

    lst = [('name', 'Monty'), ('age', 42), ('food', 'spam')]
    keys, values = zip(*lst)
    In [101]: keys
    Out[101]: ('name', 'age', 'food')
    In [102]: values
    Out[102]: ('Monty', 42, 'spam')

Answer 11

you can use this below code:

dict(zip(['name', 'age', 'food'], ['Monty', 42, 'spam']))

But make sure that length of the lists will be same.if length is not same.then zip function turncate the longer one.

Answer 12

Here is also an example of adding a list value in you dictionary

list1 = ["Name", "Surname", "Age"]
list2 = [["Cyd", "JEDD", "JESS"], ["DEY", "AUDIJE", "PONGARON"], [21, 32, 47]]
dic = dict(zip(list1, list2))
print(dic)

always make sure the your "Key"(list1) is always in the first parameter.

{'Name': ['Cyd', 'JEDD', 'JESS'], 'Surname': ['DEY', 'AUDIJE', 'PONGARON'], 'Age': [21, 32, 47]}

Answer 13

method without zip function

l1 = [1,2,3,4,5]
l2 = ['a','b','c','d','e']
d1 = {}
for l1_ in l1:
    for l2_ in l2:
        d1[l1_] = l2_
        l2.remove(l2_)
        break  

print (d1)


{1: 'd', 2: 'b', 3: 'e', 4: 'a', 5: 'c'}