如何访问列表列表中每个列表的第一个元素

Question

Example:

numbers = ['1','2','3']
letters = ['a','b','c']

I want to get [1,a] as a results. Yeah I can loop through it, but I'm wondering if there is a fast one line way of doing this.

EDIT EDIT !!!!

I made a horrible mistake in describing the problem.

I have access to the combined list (the list of lists of the question):

list_of_lists = [ numbers, letters]

which is equal to:

[ ['1','2','3'],['a','b','c']]

Sorry for the confusion. The end result is still the same, this would be ['1','a'].

Answer 1

Try a list comprehension :

# (numbers, letters) can be replaced with `list_of_lists` 
>>> [ x[0] for x in (numbers, letters) ] 
['1', 'a']

Answer 2

import operator
map(operator.itemgetter(0), [numbers, letters])

Answer 3

list_of_lists = [['1', '2', '3'], ['a', 'b', 'c']]
list_of_firsts = [l[0] for l in list_of_lists]

Answer 4

您可能正在寻找拉链

Answer 5

我会尝试：

zip(*list_of_lists)[0]