list comprehension to merge various lists in python

Question

I need to plot a lot of data samples, each stored in a list of integers. I want to create a list from a lot of concatenated lists, in order to plot it with enumerate(big_list) in order to get a fixed-offset x coordinate. My current code is:

biglist = []
for n in xrange(number_of_lists):
    biglist.extend(recordings[n][chosen_channel])
for x,y in enumerate(biglist):
    print x,y

Notes: number_of_lists and chosen_channel are integer parameters defined elsewhere, and print x,y is for example (actually there are other statements to plot the points.

My question is: is there a better way, for example, list comprehensions or other operation, to achieve the same result (merged list) without the loop and the pre-declared empty list?

Thanks

@ Falmarri: the code I made seems non-pythonic, so I am looking for a way to avoid the typical empty-list declaration followed by a loop to fill the list one element at a time. — heltonbiker
– heltonbiker, Commented Feb 25, 2011 at 5:30

John La Rooy · Accepted Answer · 2011-02-25 05:18:27Z

4

import itertools
for x,y in enumerate(itertools.chain(*(recordings[n][chosen_channel] for n in xrange(number_of_lists))):
    print x,y

You can think of itertools.chain() as managing an iterator over the individual lists. It remembers which list and where in the list you are. This saves you all memory you would need to create the big list.

answered Feb 25, 2011 at 5:18

John La Rooy

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2 Comments

heltonbiker Over a year ago

The answer worked, and was what I asked for, indeed. In the other hand, it seem so verbose... But I could not figure out anything better (except using numpy.ndarray for my recordings, so I could just slice things right through...). Many thanks! (my current code incorporated your one-liner ;o)

Andrew Clark Over a year ago

If you don't like the one line version you can split it up, big_iter = itertools.chain(*(recordings[n][chosen_channel] for n in xrange(number_of_lists)), for x,y in enumerate(big_iter).

user2665694 · Accepted Answer · 2011-02-25 05:08:47Z

3

>>> import itertools
>>> l1 = [2,3,4,5]
>>> l2=[9,8,7]
>>> itertools.chain(l1,l2)
<itertools.chain object at 0x100429f90>
>>> list(itertools.chain(l1,l2))
[2, 3, 4, 5, 9, 8, 7]

answered Feb 25, 2011 at 5:08

user2665694

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