128
votes

Here's how I'm currently converting a list of tuples to dictionary in Python:

l = [('a',1),('b',2)]
h = {}
[h.update({k:v}) for k,v in l]
> [None, None]
h
> {'a': 1, 'b': 2}

Is there a better way? It seems like there should be a one-liner to do this.

5

5 Answers

221
votes

Just call dict() on the list of tuples directly

>>> my_list = [('a', 1), ('b', 2)]
>>> dict(my_list)
{'a': 1, 'b': 2}
24
votes

The dict constructor accepts input exactly as you have it (key/value tuples).

>>> l = [('a',1),('b',2)]
>>> d = dict(l)
>>> d
{'a': 1, 'b': 2}

From the documentation:

For example, these all return a dictionary equal to {"one": 1, "two": 2}:

dict(one=1, two=2)
dict({'one': 1, 'two': 2})
dict(zip(('one', 'two'), (1, 2)))
dict([['two', 2], ['one', 1]])
19
votes

With dict comprehension:

h = {k:v for k,v in l}
17
votes

It seems everyone here assumes the list of tuples have one to one mapping between key and values (e.g. it does not have duplicated keys for the dictionary). As this is the first question coming up searching on this topic, I post an answer for a more general case where we have to deal with duplicates:

mylist = [(a,1),(a,2),(b,3)]    
result = {}
for i in mylist:  
   result.setdefault(i[0],[]).append(i[1])
print(result)
>>> result = {a:[1,2], b:[3]}
0
votes

Functional decision for @pegah answer:

from itertools import groupby

mylist = [('a', 1), ('b', 3), ('a', 2), ('b', 4)]
#mylist = iter([('a', 1), ('b', 3), ('a', 2), ('b', 4)])

result = { k : [*map(lambda v: v[1], values)]
    for k, values in groupby(sorted(mylist, key=lambda x: x[0]), lambda x: x[0])
    }

print(result)
# {'a': [1, 2], 'b': [3, 4]}