List of tuples to dictionary

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.

Answers:

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Method 1

Just call dict() on the list of tuples directly

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

Method 2

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]}

Method 3

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]])

Method 4

With dict comprehension:

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

Method 5

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]}

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0