suppose we have:
a = [[1, 2, 3], [4, 5, 6]]
What is the fastest way to access the array such that we get the first element in each list, other than looping.
i would like the result to be giving me... 1,4
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4 Answers
using zip(*a)
a = [[1, 2, 3], [2, 3, 4]]
result = zip(*a)[0]
print result
2 Comments
mgilson
As a side note, this doesn't work in python3 as
zip returns an iterable object... next(iter(zip(*a))) should work just about anywhere though.
ToolmakerSteve
list(zip(*a))[0] works in python2.7, and IMHO will also work in python3.0, because list() accepts an iterable and produces a list.A quick and easy way is to just extract a[0][0] and a[1][0], but depending on what you are using it for, this might not work all the time.
Comments
Without looping, you need to unroll the loop as ethg242 does. This has the disadvantage of only working for a fixed length of a
Here is a list comprehension
[i[0] for i in a]
It's also possible to use map(), but this also has an implicit loop
from operator import itemgetter
map(itemgetter(0), a)
Comments
You might want to consider numpy:
>>> import numpy as np
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = np.array(a)
>>> b[:,0]
array([1, 4])
