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

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.

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

Output:

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

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.

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.

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]}

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.

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

print(dic)
{'name': 'Monty', 'age': 42, 'food': 'spam'}

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]
    
print(dict)

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

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]

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.

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]
print(obj)

# 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)
print(res)

Both the results should print

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