Use enumerate:

In [3]: user_details = [{'name':x, 'rank':i} for i,x in enumerate(ranked_users)] 

In [4]: user_details
Out[4]: 
[{'name': 'jon', 'rank': 0},
 {'name': 'bob', 'rank': 1},
 {'name': 'jane', 'rank': 2},
 {'name': 'alice', 'rank': 3},
 {'name': 'chris', 'rank': 4}]

PS. My first answer was

user_details = map(lambda (i,x): {'name':x, 'rank':i}, enumerate(ranked_users))

I'd strongly recommend using a list comprehension or generator expression over map and lambda whenever possible. List comprehensions are more readable, and tend to be faster to boot.

You could do that in a one-liner using generators:

next(i for i,v in enumerate(l) if is_odd(v))

The nice thing about generators is that they only compute up to the requested amount. So requesting the first two indices is (almost) just as easy:

y = (i for i,v in enumerate(l) if is_odd(v))
x1 = next(y)
x2 = next(y)

Though, expect a StopIteration exception after the last index (that is how generators work). This is also convenient in your "take-first" approach, to know that no such value was found --- the list.index() function would raise ValueError here.

Python's closures are late binding. This means that the values of variables used in closures are looked up at the time the function is called.

To avoid the late binding effect you can use a lambda with a default arg:

index = 1
test = lambda t, index=index: t[index]+1  # binds index at definition time
index = 0
print(test([5, 0]))  # 1