Why does assigning a sliced numpy array to another variable keep slicing it?

Question

I've been experiencing some strange behavior with numpy arrays. Consider the code below:

import numpy as np

# 1
a = [np.arange(16)]
b = a
print(f'b = {b}')

# 2
b[0] = a[0][::2]
print(f'b = {b}')

# 3
b[0] = a[0][::2]
print(f'b = {b}')

b = [array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15])]
b = [array([ 0,  2,  4,  6,  8, 10, 12, 14])]
b = [array([ 0,  4,  8, 12])]

I'd expect b in section 2 to be the same as the b in section 3, but for some reason it seems to be different. This doesn't seem to happen when a is a 1D array—writing b = a[::2] twice gives the same value for b. Why is this happening when a is a 2D array?

a is not a 2D array. You made a list with an array inside. — user2357112
– user2357112, Commented Sep 13, 2020 at 4:23
It is working as designed. You are asking b[0] to be reassigned to every other value of a. Remember, you defined b = a so both b and a are same lists. Just referenced. — Joe Ferndz
– Joe Ferndz, Commented Sep 13, 2020 at 4:24
@JoeFerndz: b and a are the same list. They're not arrays. — user2357112
– user2357112, Commented Sep 13, 2020 at 4:24
I see, seems like an elementary mistake. But then why doesn't this happen when a = np.arange(16)? Is there something fundamentally different between a list and an array? — Oion Akif
– Oion Akif, Commented Sep 13, 2020 at 4:26

Joe Ferndz · Accepted Answer · 2020-09-13 04:35:05Z

1

In your code, you have given b = a So b is just making a reference of a. Every time you change the value of b, you are also making a change to value of a. For more details on assignment vs copy vs deep copy, please refer to this link

b[0] = a[0][::2]

The above code is asking b to have skipped values of b (a is same as b). So every time you do this, it is reducing it further.

If you want to make a copy, use

b = a.copy()

or

b = a[::]

edited Sep 13, 2020 at 4:35

answered Sep 13, 2020 at 4:28

Joe Ferndz

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2 Comments

Oion Akif Over a year ago

So why doesn't this happen with a = np.arange(16)? It seems as though b isn't made a reference of a if a is a numpy array?

Joe Ferndz Over a year ago

np.arange() Return evenly spaced values within a given interval. Here, you are just calling a numpy function to create an evenly spaced values. b=a is assignment of two variables. In this case you just make the reference of b and a as same.

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