Get the nth element from the inner list of a list of lists in Python

Question

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First items in inner list efficiently as possible

Lets say I have:

a = [ [1,2], [2,9], [3,7] ]

I want to retrieve the first element of each of the inner lists:

b = [1,2,3]

Without having to do this (my current hack):

for inner in a:
    b.append(inner[0])

I'm sure there's a one liner for it but I don't really know what i'm looking for.

score 72 · Accepted Answer · answered Nov 02 '12 at 02:31

72

Simply change your list comp to be:

b = [el[0] for el in a]

Or:

from operator import itemgetter
b = map(itemgetter(0), a)

Or, if you're dealing with "proper arrays":

import numpy as np
a = [ [1,2], [2,9], [3,7] ]
na = np.array(a)
print na[:,0]
# array([1, 2, 3])

And zip:

print zip(*a)[0]

answered Nov 02 '12 at 02:31

Jon Clements

1

I guess there's no other alternative left to achieve this. :) – Ashwini Chaudhary Nov 02 '12 at 02:46
@AshwiniChaudhary I'm struggling as to what else I can think of... gimme a bit ;) - do you reckon `map(next, map(iter, a))` is pushing it? – Jon Clements Nov 02 '12 at 02:47
it's nice, but slow compared to the others. – Ashwini Chaudhary Nov 02 '12 at 08:34
I like that `zip(*a)[0]` suggestion! – Dannid Apr 12 '18 at 00:01
1

In Python 3, zip() returns an iterator, not a list, so you need list(zip(*a))[0], which ironically returns a tuple, so you need another list() wrapped around it if you need a list returned. – kevin_theinfinityfund Jul 02 '20 at 07:13

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