python - One liner: creating a dictionary from list with indices as keys

97
votes

I want to create a dictionary out of a given list, in just one line. The keys of the dictionary will be indices, and values will be the elements of the list. Something like this:

a = [51,27,13,56]         #given list

d = one-line-statement    #one line statement to create dictionary

print(d)

Output:

{0:51, 1:27, 2:13, 3:56}

I don't have any specific requirements as to why I want one line. I'm just exploring python, and wondering if that is possible.

5
pythonlistdictionarypython-3.x
What advantage do you think you'll gain by having this dictionary? Index lookups won't be any faster with a dictionary. - Martijn Pieters♦
@MartijnPieters: Well, I might need, e.g, this: {[x[0]:51, x[1]:27, x[2]:13, x[3]:56}. - Nawaz
then just use zip(): dict(zip(x, a)). - Martijn Pieters♦

5 Answers

162
votes
a = [51,27,13,56]
b = dict(enumerate(a))
print(b)

will produce

{0: 51, 1: 27, 2: 13, 3: 56}

enumerate(sequence, start=0)

Return an enumerate object. sequence must be a sequence, an iterator, or some other object which supports iteration. The next() method of the iterator returned by enumerate() returns a tuple containing a count (from start which defaults to 0) and the values obtained from iterating over sequence:

54
votes

With another constructor, you have

a = [51,27,13,56]         #given list
d={i:x for i,x in enumerate(a)}
print(d)
15
votes

Try enumerate: it will return a list (or iterator) of tuples (i, a[i]), from which you can build a dict:

a = [51,27,13,56]  
b = dict(enumerate(a))
print b
14
votes
{x:a[x] for x in range(len(a))}
3
votes

Simply use list comprehension.

a = [51,27,13,56]  
b = dict( [ (i,a[i]) for i in range(len(a)) ] )
print b