Create a dictionary with comprehension

Use a dict comprehension (Python 2.7 and later):

{key: value for key, value in zip(keys, values)}

Alternatively, use the dict constructor:

pairs = [('a', 1), ('b', 2)]
dict(pairs)                          # → {'a': 1, 'b': 2}
dict((k, v + 10) for k, v in pairs)  # → {'a': 11, 'b': 12}

Given separate lists of keys and values, use the dict constructor with zip:

keys = ['a', 'b']
values = [1, 2]
dict(zip(keys, values))              # → {'a': 1, 'b': 2}

In Python 3 and Python 2.7+, dictionary comprehensions look like the below:

d = {k:v for k, v in iterable}

For Python 2.6 or earlier, see fortran's answer.

In fact, you don't even need to iterate over the iterable if it already comprehends some kind of mapping, the dict constructor doing it graciously for you:

>>> ts = [(1, 2), (3, 4), (5, 6)]
>>> dict(ts)
{1: 2, 3: 4, 5: 6}
>>> gen = ((i, i+1) for i in range(1, 6, 2))
>>> gen
<generator object <genexpr> at 0xb7201c5c>
>>> dict(gen)
{1: 2, 3: 4, 5: 6}

In Python 2.7, it goes like:

>>> list1, list2 = ['a', 'b', 'c'], [1,2,3]
>>> dict( zip( list1, list2))
{'a': 1, 'c': 3, 'b': 2}

Zip them!