flatten a list of variable length arrays

Question

how to convert this:

[2, 4, array([ 3.]), array([ 4.]), array([ 5.,  4.])]

to

[2,4,3,4,5,4]

I've tried searching but the solutions work for

[array([ 2.]), array([ 4.]), array([ 3.]), array([ 4.]), array([ 5.,  4.])]

and

[[2],[4],[3],[4],[5,4]]

What have you tried? >>> [2,4,array([3.])] Traceback (most recent call last): File "<stdin>", line 1, in <module> NameError: name 'array' is not defined — John Mee
– John Mee, Commented Dec 7, 2012 at 9:39
@iMom0 hmm, given the duplicate? you might be right. Still banging his drum but... you want to help him? — John Mee
– John Mee, Commented Dec 7, 2012 at 9:44
With variable lengths and nesting of numbers and lists I would try to make a recursive approach. — Alex
– Alex, Commented Dec 7, 2012 at 9:46

BenT · Accepted Answer · 2023-02-06 19:08:57Z

3

import itertools
import numpy as np

a = [2, 4, array([ 3.]), array([ 4.]), array([ 5.,  4.])]
list(itertools.chain.from_iterable(np.asarray(b).ravel() for b in a))

edited Feb 6, 2023 at 19:08

BenT

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answered Dec 7, 2012 at 9:56

Michael J. Barber

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Comments

Gurgeh · Accepted Answer · 2012-12-07 10:23:19Z

0

I like to use sum for flattening things. In your case, you want to convert the arrays to lists and the numbers to lists of numbers, first. Thus if the array is in a:

sum((list(x) if hasattr(x, '__iter__') else [x] for x in a), [])

answered Dec 7, 2012 at 10:23

Gurgeh

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2 Comments

Michael J. Barber Over a year ago

This will be very slow if you have many elements to deal with. With each step in the summation, a new list is allocated containing all elements seen so far. End result is worst-case O(n**2) time in the number of elements.

Gurgeh Over a year ago

Premature optimization. If performance on long lists were a requirement, he would have said so. My solution is both more general and easier to read, and thus maintain, than your solution. Though yours certainly is better if there is a chance that there are many elements.

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