One liner: creating a dictionary from list with indices as keys
PythonListDictionaryPython 3.xPython Problem Overview
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.
Python Solutions
Solution 1 - Python
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:
Solution 2 - Python
With another constructor, you have
a = [51,27,13,56] #given list
d={i:x for i,x in enumerate(a)}
print(d)
Solution 3 - Python
{x:a[x] for x in range(len(a))}
Solution 4 - Python
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
Solution 5 - Python
Simply use list comprehension.
a = [51,27,13,56]
b = dict( [ (i,a[i]) for i in range(len(a)) ] )
print b