Zip on single element in nested sequence.

[rhubarbpieguy]

How do I create the list c=[2,3,4] from the list a=[[1,2,3],[2,3,4],[3,4,5]]? In other words, how do I create a list from element [1] of a list containing nested sequences?

I know b,c,d=zip(*a) produces b=[1,2,3], c=[2,3,4], and d=[3,4,5].  That works well, but I don't need b and d.  It seems I should be able to use zip(*a[1]) or something of the sort but I have no success.

This is a minor problem but I must be missing something obvious.

my code here

[buran]

a = [[1,2,3],[4,5,6],[7,8,9]]
b = zip(*a)[1]
print b
c =  [x[1] for x in a]
print c

Output:
(2, 5, 8)
[2, 5, 8]

by the way, no need for zip for the unpacking:

b, c, d = a

EDIT: Above applies when you want the i-th element of each element in a. In this case i=1

If you need the 1st element of a:

b = a[1]

Output:
[4, 5, 6]

[rhubarbpieguy]

The list comprehension works and is what I've used. However,zip(*a)[1] produces:

TypeError: 'zip' object is not subscriptable

[micseydel]

In Python 3, zip() doesn't return a list, it returns a specialized iterator. So you can't just ask for the nth item (1th item here). You'll have to operate on the iterator.

[nilamo]

In other words, list(zip(*a))[1].

[micseydel]

In other words, list(zip(*a))[1].

That would be equivalent to Python 2, yes. Would be ok for small values of a. Unfortunately Python doesn't appear to have a very clean way of getting the nth value of an iterator so this is probably the cleanest thing to do.

[nilamo]

next() I guess?

So...

zipped = zip(*a) #maybe iter(zip(*a)), I didn't actually try
# ignore the first
next(zipped)
second_elem = next(zipped)

[micseydel]

Yeah, that would work for n=2, but it doesn't seem super elegant to me. I wish there was something in itertools to get the nth item of an iterator.

[ichabod801]

compress?

compress(zip(*a), [0] * (n - 1) + [1])

[snippsat]

With Toolz a little nicer.
pip install toolz

>>> from toolz import nth
>>> a = [[1,2,3],[4,5,6],[7,8,9]]
>>> nth(1, zip(*a))
(2, 5, 8)