List of tuples to dictionary
Question:
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:
Just call dict()
on the list of tuples directly
>>> my_list = [('a', 1), ('b', 2)]
>>> dict(my_list)
{'a': 1, 'b': 2}
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]])
With dict
comprehension:
h = {k:v for k,v in l}
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]}
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]}
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.
Just call dict()
on the list of tuples directly
>>> my_list = [('a', 1), ('b', 2)]
>>> dict(my_list)
{'a': 1, 'b': 2}
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]])
With dict
comprehension:
h = {k:v for k,v in l}
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]}
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]}