replace element by element different arrays

Question

I have an array :

a = np.array([1,2,3,4,5,6,7,8])

The array may be reshaped to a = np.array([[1,2,3,4],[5,6,7,8]]), whatever is more convenient.

Now, I have an array :

b = np.array([[11,22,33,44], [55,66,77,88]])

I want to replace to each of these elements the corresponding elements from a.

The a array will always hold as many elements as b has.

So, array b will be :

[1,2,3,4], [5,6,7,8]

Note, that I must keep each b subarray dimension to (4,). I don't want to change it.So, the idx will take values from 0 to 3.I want to make a fit to every four values.

I am struggling with reshape, split,mask ..etc and I can't figure a way to do it.

import numpy as np

#a = np.array([[1,2,3,4],[5,6,7,8]])
a = np.array([1,2,3,4,5,6,7,8])


b = np.array([[11,22,33,44], [55,66,77,88]])

for arr in b:
    for idx, x in enumerate(arr):
        replace every arr[idx] with corresponding a value

@languitar:No, as I said, I will have a loop which goes down until it reaches shape (4,) and there I want to do the replacement. — George
– George, Commented Feb 15, 2017 at 10:03
"I want to replace to each of these arrays the corresponding elements from a." this is an unclear statement. How? What do you mean by "corresponding"? Is it always true len(b)==len(c), len(a)==2*len(b)? — Daniel F
– Daniel F, Commented Feb 15, 2017 at 10:04
@DanielForsman:That's correct.a will have twice the elements of b.And c will be always equal to b. — George
– George, Commented Feb 15, 2017 at 10:06

Daniel F · Accepted Answer · 2017-02-16 09:26:44Z

2

For your current case, what you want is probably:

b, c = list(a.reshape(2, -1))

This isn't the cleanest, but it is a one-liner. Turn your 1D array into a 2D array with with the first dimension as 2 with reshape(2, -1), then list splits it along the first dimension so you can directly assign them to b, c

You can also do it with the specialty function numpy.split

b, c = np.split(a, 2)

EDIT: Based on accepted solution, vectorized way to do this is

b = a.reshape(b.shape)

edited Feb 16, 2017 at 9:26

answered Feb 15, 2017 at 10:06

Daniel F

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4 Comments

George Over a year ago

Nice,thanks,but what can I do in this situation.(I edited my post)

George Over a year ago

Thanks for this solution.(upvoting)

Daniel F Over a year ago

Can't see your pastebin, but if you always want to split in half you'd want b=a.reshape(2,-1). if you always want 4 elements (and len(a) is always divisible by 4) you can do b=a.reshape(-1,4)

George Over a year ago

Thank you very much for the info!

Yoiro · Accepted Answer · 2017-02-15 12:32:22Z

1

The following worked for me:

i = 0
for arr in b:
    for idx, x in enumerate(arr):
        arr[idx] = a[i]
        print(arr[idx])
        i += 1

Output (arr[idx]): 1 2 3 4 5 6 7 8 If you type print(b) it'll output [[1 2 3 4] [5 6 7 8]]

answered Feb 15, 2017 at 12:32

Yoiro

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mwstc013 · Accepted Answer · 2017-02-15 10:06:40Z

0

b = a[:len(a)//2] c = a[len(a)//2:]

answered Feb 15, 2017 at 10:06

mwstc013

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Yoiro · Accepted Answer · 2017-02-15 10:14:06Z

0

Well, I'm quite new to Python but this worked for me:

    for i in range (0, len(a)//2):
        b[i] = a[i]
    for i in range (len(a)//2,len(a)):
        c[i-4] = a[i]

by printing the 3 arrays I have the following output:

[1 2 3 4 5 6 7 8]
[1 2 3 4]
[5 6 7 8]

But I would go for Daniel's solution (the split one): 1 liner, using numpy API, ...

answered Feb 15, 2017 at 10:14

Yoiro

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2 Comments

George Over a year ago

Hi, I edited my post.The thing is that I must replace at the same time only b array (edited in post)

Yoiro Over a year ago

I added another answer for more clarity ;)

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