How can I make a dictionary (dict) from separate lists of keys and values?


I want to combine these:

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

Into a single dictionary:

{'name': 'Monty', 'age': 42, 'food': 'spam'}
Asked By: Guido



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.

Answered By: Dan Lenski
keys = ('name', 'age', 'food')
values = ('Monty', 42, 'spam')
out = dict(zip(keys, values))


{'food': 'spam', 'age': 42, 'name': 'Monty'}
Answered By: iny

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.

Answered By: Mike Davis

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.

Answered By: jfs

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'}
Answered By: Brendan Berg

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))}
Answered By: exploitprotocol

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)}


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'}
Answered By: kiriloff

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))}
Answered By: Polla A. Fattah

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'}


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))

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

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)}))
>>> min(timeit.repeat(lambda: dict(zip(l1, l2))))

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_

print (d1)

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

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.

Answered By: Akash Nayak
  • 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')
Answered By: AbstProcDo

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))

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]}
Answered By: Cyd

I had this doubt while I was trying to solve a graph-related problem. The issue I had was I needed to define an empty adjacency list and wanted to initialize all the nodes with an empty list, that’s when I thought how about I check if it is fast enough, I mean if it will be worth doing a zip operation rather than simple assignment key-value pair. After all most of the times, the time factor is an important ice breaker. So I performed timeit operation for both approaches.

import timeit
def dictionary_creation(n_nodes):
    dummy_dict = dict()
    for node in range(n_nodes):
        dummy_dict[node] = []
    return dummy_dict

def dictionary_creation_1(n_nodes):
    keys = list(range(n_nodes))
    values = [[] for i in range(n_nodes)]
    graph = dict(zip(keys, values))
    return graph

def wrapper(func, *args, **kwargs):
    def wrapped():
        return func(*args, **kwargs)
    return wrapped

iteration = wrapper(dictionary_creation, n_nodes)
shorthand = wrapper(dictionary_creation_1, n_nodes)

for trail in range(1, 8):
    print(f'Itertion: {timeit.timeit(iteration, number=trails)}nShorthand: {timeit.timeit(shorthand, number=trails)}')

For n_nodes = 10,000,000
I get,

Iteration: 2.825081646999024
Shorthand: 3.535717916001886

Iteration: 5.051560923002398
Shorthand: 6.255070794999483

Iteration: 6.52859034499852
Shorthand: 8.221581164998497

Iteration: 8.683652416999394
Shorthand: 12.599181543999293

Iteration: 11.587241565001023
Shorthand: 15.27298851100204

Iteration: 14.816342867001367
Shorthand: 17.162912737003353

Iteration: 16.645022411001264
Shorthand: 19.976680120998935

You can clearly see after a certain point, iteration approach at n_th step overtakes the time taken by shorthand approach at n-1_th step.

Answered By: Mayank Prakash

Solution as dictionary comprehension with enumerate:

dict = {item : values[index] for index, item in enumerate(keys)}

Solution as for loop with enumerate:

dict = {}
for index, item in enumerate(keys):
    dict[item] = values[index]
Answered By: jay123

Although there are multiple ways of doing this but i think most fundamental way of approaching it; creating a loop and dictionary and store values into that dictionary. In the recursive approach the idea is still same it but instead of using a loop, the function called itself until it reaches to the end. Of course there are other approaches like using dict(zip(key, value)) and etc. These aren’t the most effective solutions.

y = [1,2,3,4]
x = ["a","b","c","d"]

# This below is a brute force method
obj = {}
for i in range(len(y)):
    obj[y[i]] = x[i]

# Recursive approach 
obj = {}
def map_two_lists(a,b,j=0):
    if j < len(a):
        obj[b[j]] = a[j]
        j +=1
        map_two_lists(a, b, j)
        return obj

res = map_two_lists(x,y)

Both the results should print

{1: 'a', 2: 'b', 3: 'c', 4: 'd'}  
Answered By: Franco

If you are working with more than 1 set of values and wish to have a list of dicts you can use this:

def as_dict_list(data: list, columns: list):
    return [dict((zip(columns, row))) for row in data]

Real-life example would be a list of tuples from a db query paired to a tuple of columns from the same query. Other answers only provided for 1 to 1.

Answered By: DonkeyKong
keys = ['name', 'age', 'food']
values = ['Monty', 42, 'spam']
dic = {}
c = 0
for i in keys:
    dic[i] = values[c]
    c += 1

{'name': 'Monty', 'age': 42, 'food': 'spam'}
Answered By: Zeinab Mardi

It can be done by the following way.

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

dict = {}

for i in range(len(keys)):
    dict[keys[i]] = values[i]

{'name': 'Monty', 'age': 42, 'food': 'spam'}
Answered By: J.Jai

All answers sum up:

l = [1, 5, 8, 9]
ll = [3, 7, 10, 11]


dict(zip(l,ll)) # {1: 3, 5: 7, 8: 10, 9: 11}

#if you want to play with key or value @recommended

{k:v*10 for k, v in zip(l, ll)} #{1: 30, 5: 70, 8: 100, 9: 110}


d = {}
for k in l:
    d[k] = ll[c] #setting up keys from the second list values
    c += 1
{1: 3, 5: 7, 8: 10, 9: 11}


d = {}
for i,k in enumerate(l):
    d[k] = ll[i]
{1: 3, 5: 7, 8: 10, 9: 11}
Answered By: guest
    import pprint
    def makeDictUsingAlternateLists1(**rest):
        print("*rest.keys() : ",*rest.keys())
        print("rest.keys() : ",rest.keys())
        print("*rest.values() : ",*rest.values())
        print("**rest.keys() : ",rest.keys())
        print("**rest.values() : ",rest.values())
        [print(a) for a in zip(*rest.values())]
        [ print(dict(zip(rest.keys(),a))) for a in zip(*rest.values())]
        finalRes= [ dict( zip( rest.keys(),a))  for a in zip(*rest.values())] 
        return finalRes
    l = makeDictUsingAlternateLists1(p=p,q=q,r=r,s=s)
*rest.keys() :  p q r s
rest.keys() :  dict_keys(['p', 'q', 'r', 's'])
*rest.values() :  ['A', 'B', 'C'] [5, 2, 7] ['M', 'F', 'M'] ['Sovabazaar', 'Shyambazaar', 'Bagbazaar', 'Hatkhola']
**rest.keys() :  dict_keys(['p', 'q', 'r', 's'])
**rest.values() :  dict_values([['A', 'B', 'C'], [5, 2, 7], ['M', 'F', 'M'], ['Sovabazaar', 'Shyambazaar', 'Bagbazaar', 'Hatkhola']])
('A', 5, 'M', 'Sovabazaar')
('B', 2, 'F', 'Shyambazaar')
('C', 7, 'M', 'Bagbazaar')
{'p': 'A', 'q': 5, 'r': 'M', 's': 'Sovabazaar'}
{'p': 'B', 'q': 2, 'r': 'F', 's': 'Shyambazaar'}
{'p': 'C', 'q': 7, 'r': 'M', 's': 'Bagbazaar'}
[{'p': 'A', 'q': 5, 'r': 'M', 's': 'Sovabazaar'},
 {'p': 'B', 'q': 2, 'r': 'F', 's': 'Shyambazaar'},
 {'p': 'C', 'q': 7, 'r': 'M', 's': 'Bagbazaar'}]
Answered By: Soudipta Dutta
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