Cross 2 arrays

[dylan261999]

Hi,

With Matlab, I can run this easily :

a = [0 0 0 0 0]
b = [1 1 1 1]

a(1:1:end) = b

To obtain : a = [0 1 0 1 0 1 0 1 0]

How can I implement this with Python ?

Thanks

[deanhystad]

Something like this?

from itertools import zip_longest, chain


def interleave(a, b):
    return [x for x in chain(*zip_longest(a, b)) if x is not None]


a = [0, 0, 0, 0, 0]
b = [1, 1, 1, 1]


print(interleave(a, b))
print(interleave(b, a))

Output:
[0, 1, 0, 1, 0, 1, 0, 1, 0]
[1, 0, 1, 0, 1, 0, 1, 0, 0]

There are functions named interleave and interleave_longest in more_itertools. This is not a standard library. You can read about it here:

https://pypi.org/project/more-itertools/

[ndc85430]

Another, similar library is Toolz: https://pypi.org/project/toolz/.

[paul18fr]

It's not straigtforward (and it might be improved), but it does the job quit fastly even for huge vectors :-).

import time
import numpy as np

n,m=1_000_000, 4_000_000
a=np.ones((n), dtype=int)
b=2*np.ones((m), dtype=int)

t0=time.time()
La=len(a)
Lb=len(b)
if (Lb<La):
    c=np.concatenate((a[:Lb], b))
    c=np.reshape(c, (Lb, 2), order='f')
    c=np.reshape(c, (2*Lb), order='c')
    c=np.concatenate((c, a[Lb::]))
elif (Lb>La):
    c=np.concatenate((a, b[:La]))
    c=np.reshape(c, (La, 2), order='f')
    c=np.reshape(c, (2*La), order='c')
    c=np.concatenate((c, b[La::]))
else:
    c=np.concatenate((a, b))
    c=np.reshape(c, (La, 2), order='f')
    c=np.reshape(c, (2*La), order='c')
t1=time.time()
print(f"Duration={t1-t0}") 

[thensun]

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