How do I access the 1st element of each list in a lists of lists

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'].

http://stackoverflow.com/questions/2054416/getting-the-first-elements-per-row-in-an-array-in-python. Someone else asked for the answer to the same homework problem before you. — S.Lott, Jan 27 '11 at 20:19
Is your question related to some bigger context, if so please elaborate us. Just commented because you are refering to looping, but clearly the solution you are looking for doesen't need any looping, thanks — eat, Jan 27 '11 at 20:48
OK took me forever to find the add comment. Well I've looked at the homework and that's exactly how I should do it. I'm using the solution to grab the description names from a MySQL description query. You know, cause they decided that the attributes should be part of it too. — Peter, Jan 27 '11 at 22:05

miku · Answer 1 · 2011-01-27T20:34:13.430

7

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

edited Jan 27 '11 at 20:34

answered Jan 27 '11 at 20:18

miku

Why upvoted? Clearly OP's question (as stated now) has a constant time solution. Or is your answer some kind of joke? thanks – eat Jan 27 '11 at 20:57
1

@eat: No, no joke, just the an answer, that worked with the previous version of the question (and still works). Besides, it's in the same complexity class as your answer. – miku Jan 27 '11 at 21:03
@The MYYN OK, then perhaps I'll misunderstod something. Do you consider my answer giving wrong solution? Why list comprehension seems to be the right answer? Clearly no iterations needed. I'm just concernd because I felt that OP was looking for something not stated in the question yet ;-). – eat Jan 27 '11 at 21:21
@eat: No, your answer is fine. I would have upvoted it, if it just said `[ list_of_list[0][0], list_of_list[1][0] ]`, but the rest is a bit noisy (at the moment). To me, list comprehension looks a bit cleaner (and it is actually shorter). – miku Jan 27 '11 at 21:27

score 2 · Answer 2 · answered Jan 27 '11 at 20:20

2

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

answered Jan 27 '11 at 20:20

ephemient

score 0 · Answer 3 · answered Jan 27 '11 at 20:19

0

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

answered Jan 27 '11 at 20:19

Jay Conrod

Confusion · Answer 4 · 2011-01-27T20:21:47.813

-1

You may be looking for zip

edited Jan 27 '11 at 20:21

answered Jan 27 '11 at 20:18

Confusion

score -1 · Answer 5 · answered Jan 27 '11 at 21:04

-1

I would try:

zip(*list_of_lists)[0]

answered Jan 27 '11 at 21:04

rubik

1

At least go for `itertools.izip(*list_of_lists).next()`. Your solution loops through the entire lists. – Rosh Oxymoron Jan 27 '11 at 21:13

5 Answers5