Get the first element of each tuple in a list in Python

Use a list comprehension:

res_list = [x[0] for x in rows]

Below is a demonstration:

>>> rows = [(1, 2), (3, 4), (5, 6)]
>>> [x[0] for x in rows]
[1, 3, 5]
>>>

Alternately, you could use unpacking instead of x[0]:

res_list = [x for x,_ in rows]

Below is a demonstration:

>>> lst = [(1, 2), (3, 4), (5, 6)]
>>> [x for x,_ in lst]
[1, 3, 5]
>>>

Both methods practically do the same thing, so you can choose whichever you like.

If you don't want to use list comprehension by some reasons, you can use map and operator.itemgetter:

>>> from operator import itemgetter
>>> rows = [(1, 2), (3, 4), (5, 6)]
>>> map(itemgetter(1), rows)
[2, 4, 6]
>>>

You can use list comprehension:

res_list = [i[0] for i in rows]

This should make the trick

res_list = [x[0] for x in rows]

c.f. http://docs.python.org/3/tutorial/datastructures.html#list-comprehensions

For a discussion on why to prefer comprehensions over higher-order functions such as map, go to http://www.artima.com/weblogs/viewpost.jsp?thread=98196.