How to get flat list in list comprehension from lists of arbitrary number of elements?

Question

I have something like this:

[e for e in ([n for n in xrange(random.randrange(1, 5))] for x in xrange(10))]

It produces:

[[0, 1, 2, 3], [0, 1, 2], [0], [0], [0, 1], [0], [0, 1], [0, 1, 2, 3], [0, 1, 2], [0, 1, 2]]

And I need the same but in flat structure.

For now I use something like:

l = []
[l.extend(e) for e in ([n for n in xrange(random.randrange(1, 5))] for x in xrange(10))]

But is there something less obsucre to achieve this 'unpacking' of arbitrary length list inside comprehension?

• Stephane Rolland: dupplicate of stackoverflow.com/questions/716477/join-list-of-lists-in-python

Answer 1

Use this list comprehension:

In [8]: [y for x in xrange(10) for y in xrange(random.randrange(1, 5))]
Out[8]: [0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 0, 1, 2, 3, 0, 0, 1, 0]

The above list comprehension is equivalent to this(but LC are much faster):

In [9]: lis=[]
In [10]: for x in xrange(10):
 ....: for y in xrange(random.randrange(1, 5)):
 ....: lis.append(y)
 ....: 

• Gareth Latty: +1 - The best option here is to make it the way you want, not flatten it after.
• Stephane Rolland: +1 I didn't know whe could nest for loops into list comprehension
• Gill Bates: xrange and random is just for illustration! My case requires that elems that pops from data srouce is arbitrary length lists
• Ashwini Chaudhary: @GillBates for y in xrange(random.randrange(1, 5)) iterates over that arbitrary lenght list to return a flattened list.
• Gill Bates: Iterates and generates it, in my case of course Im not generating data inside list comprehension

Answer 2

The best way to flatten any iterable in a generic situation is itertools.chain.from_iterable():

>>> import random
>>> from itertools import chain
>>> x = [e for e in ([n for n in xrange(random.randrange(1, 5))] 
... for x in xrange(10))]
>>> list(chain.from_iterable(x))
[0, 0, 0, 1, 2, 3, 0, 1, 2, 3, 0, 0, 1, 2, 3, 0, 1, 0, 1, 0, 0, 1, 2]

This said, it's preferable to avoid the extra work in this case by just making it flat to begin with.

Answer 3

You can use numpy flatten():

import numpy as np
l = [e for e in ([n for n in xrange(random.randrange(1, 5))] for x in xrange(10))]
a = np.asarray(l)
l = list(a.flatten(l))
print l

Answer 4

import itertools
l = [e for e in ([n for n in xrange(random.randrange(1, 5))] for x in xrange(10))]
result = list(itertools.chain(*l))

and then print result gives:

[0,1,2,3,0,1,2,0...]

the use of the * in chain(*l) is inspired from this question join list of lists in python .

• Gareth Latty: This is slower than itertools.chain.from_iterable() that was designed to do this more efficiently.