An SQL query gives me a list of tuples, like this:
[(elt1, elt2), (elt1, elt2), (elt1, elt2), (elt1, elt2), (elt1, elt2), ...] I'd like to have all the first elements of each tuple. Right now I use this:
rows = cur.fetchall() res_list = [] for row in rows: res_list += [row[0]] But I think there might be a better syntax to do it. Do you know a better way?
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
The functional way of achieving this is to unzip the list using:
sample = [(2, 9), (2, 9), (8, 9), (10, 9), (23, 26), (1, 9), (43, 44)] first,snd = zip(*sample) print first,snd (2, 2, 8, 10, 23, 1, 43) (9, 9, 9, 9, 26, 9, 44) 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.