How to create a value: index dict from a python list?

Question

I have a list:

a = ['a', 'b', 'c']

I want to get the dict:

b = {'a':0, 'b':1, 'c':2}

What is the most pythonic way to get that?

What have you tried? We can then tell you whether it is Pythonic. — Andy G, Jan 04 '18 at 11:53

score 5 · Accepted Answer · answered Jan 04 '18 at 11:53

5

Does the following help you?

b = dict(map(lambda t: (t[1], t[0]), enumerate(a)))

Example:

>>> dict(map(lambda t: (t[1], t[0]), enumerate(['a', 'b', 'c'])))
{'b': 1, 'c': 2, 'a': 0}

answered Jan 04 '18 at 11:53

tgwtdt

is your answer faster or the one from Dark? – yukashima huksay Jan 04 '18 at 12:00

score 2 · Answer 2 · answered Jan 04 '18 at 11:57

2

The simplest (and imo most pythonic) way of doing this would be:

b = dict(enumerate(a))

answered Jan 04 '18 at 11:57

L3viathan

1

You have to swap key and values. – yukashima huksay Jan 04 '18 at 11:59
2

In your question the target dictionary has the numbers as keys. – L3viathan Jan 04 '18 at 12:00
Oh you are right. The title is correct but I wrote it wrong in the question body. sorry. – yukashima huksay Jan 04 '18 at 12:26

score 2 · Answer 3 · answered Jan 04 '18 at 11:59

2

a = ['a','b','c']
b = dict(enumerate(a))
print(b)

It will produce:

{0: 'a', 1: 'b', 2: 'c'}

answered Jan 04 '18 at 11:59

Nabeel Ahmad Khan

score 0 · Answer 4 · answered Jan 04 '18 at 11:59

0

use index:

a = ['a', 'b', 'c']
dict_a = {}
for x in a:
    dict_a[a.index(x)] = x
print(dict_a)

answered Jan 04 '18 at 11:59

keramat

4 Answers4