Taking elements two by two from lists

Question

I have lists like:

['a', '2', 'b', '1', 'c', '4']
['d', '5', 'e', '7', 'f', '4', 'g', '6']

And I want to make a dictionary consist of keys as letters and values as numbers. I mean:

{'a': 2, 'b': 1, 'c': 4, 'd':5, 'e':7, 'f':4, 'g':6}

Answer 1

You can try:

>>> l = ['a', '2', 'b', '1', 'c', '4']
>>> it = iter(l)
>>> dict(zip(it, it))
{'a': '2', 'c': '4', 'b': '1'}

First you create an iterator out of the list. Then with zip of the iterator with itself you take pair of values from the list. Finally, with dict you transform these tuples to your wanted dictionary.

If you also want to do the string to number conversion, then use:

{x: int(y) for x, y in zip(it, it)}

EDIT

If you don't want to use zip then:

{x: int(next(it)) for x in it}

Answer 2

l = ['a', '2', 'b', '1', 'c', '4']
d = {k:v for k,v in zip(l[::2], l[1::2])}

Or if you want the numbers to be actual numbers:

l = ['a', '2', 'b', '1', 'c', '4']
d = {k:int(v) for k,v in zip(l[::2], l[1::2])}

Use float(v) instead of int(v) if the numbers have the potential to be floating-point values instead of whole numbers.

Without using any built-in functions:

l = ['a', '2', 'b', '1', 'c', '4']
d = {}
l1 = l[::2]
l2 = l[1::2]
idx = 0
while 1:
    try:
        d[l1[idx]] = l2[idx]
        idx += 1
    except IndexError:
        break