Concatenation of many lists in Python [duplicate]

Answer 1

[x for n in getNeighbors(vertex) for x in getNeighbors(n)]

or

sum(getNeighbors(n) for n in getNeighbors(vertex), [])

Answer 2

As usual, the itertools module contains a solution:

>>> l1=[1, 2, 3]

>>> l2=[4, 5, 6]

>>> l3=[7, 8, 9]

>>> import itertools

>>> list(itertools.chain(l1, l2, l3))
[1, 2, 3, 4, 5, 6, 7, 8, 9]

Answer 3

Using .extend() (update in place) combined with reduce instead of sum() (new object each time) should be more efficient however I'm too lazy to test that :)

mylist = [[1,2], [3,4], [5,6]] 
reduce(lambda acc_l, sl: acc_l.extend(sl) or acc_l, mylist)

Answer 4

If speed matters, it may be better to use this:

from operator import iadd
reduce(iadd, (getNeighbors(n) for n in getNeighbors(vertex)))

The point of this code is in concatenating whole lists by list.extend where list comprehension would add one item by one, as if calling list.append. That saves a bit of overhead, making the former (according to my measurements) about three times faster. (The iadd operator is normally written as += and does the same thing as list.extend.)

Using list comprehensions (the first solution by Ignacio) is still usually the right way, it is easier to read.

But definitely avoid using sum(..., []), because it runs in quadratic time. That is very impractical for many lists (more than a hundred or so).

Answer 5

I like itertools.chain approach because it runs in linear time (sum(...) runs in qudratic time) but @Jochen didn't show how to deal with lists of dynamic length. Here is solution for OP's question.

import itertools
list(itertools.chain(*[getNeighbors(n) for n in getNeighbors(vertex)]))

You can get rid of list(...) call if iterable is sufficient for you.

Answer 6

Appending lists can be done with + and sum():

>>> c = [[1, 2], [3, 4]]
>>> sum(c, [])
[1, 2, 3, 4]