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?

dupplicate of http://stackoverflow.com/questions/716477/join-list-of-lists-in-python — Stephane Rolland, Apr 07 '13 at 13:55

Ashwini Chaudhary · Answer 1 · 2013-04-07T13:46:00.663

7

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)
   ....:

edited Apr 07 '13 at 13:46

answered Apr 07 '13 at 13:38

Ashwini Chaudhary

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+1 - The best option here is to make it the way you want, not flatten it after. – Gareth Latty Apr 07 '13 at 13:41
+1 I didn't know whe could nest for loops into list comprehension – Stephane Rolland Apr 07 '13 at 13:49
xrange and random is just for illustration! My case requires that elems that pops from data srouce is arbitrary length lists – Gill Bates Apr 07 '13 at 13:49
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@GillBates `for y in xrange(random.randrange(1, 5))` iterates over that arbitrary lenght list to return a flattened list. – Ashwini Chaudhary Apr 07 '13 at 13:58
Iterates and generates it, in my case of course Im not generating data inside list comprehension – Gill Bates Apr 07 '13 at 14:10

score 3 · Accepted Answer · edited May 23 '17 at 12:29

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.

score 3 · Answer 3 · answered Apr 07 '13 at 13:38

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

answered Apr 07 '13 at 13:38

ASGM

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score 0 · Answer 4 · edited May 23 '17 at 11:56

0

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 .

edited May 23 '17 at 11:56

Community

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answered Apr 07 '13 at 13:45

Stephane Rolland

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This is slower than `itertools.chain.from_iterable()` that was designed to do this more efficiently. – Gareth Latty Apr 07 '13 at 13:54

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

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