Why the original numpy array get changed while changing another array created from it?

Question

i have a numpy array r when i used to create another array r2 out of it and turning that new array r2 to zero it also changed the original array r

I have searched around the similar questions but did not turned around the any satisfying answer for this, so please consider suggesting an appropriate answer.

Original Array:

>>> r
array([[ 0,  1,  2,  3,  4,  5],
       [ 6,  7,  8,  9, 10, 11],
       [12, 13, 14, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35]])

another numpy array from original array r2 as follows:

>>> r2 = r[:3, :3]
>>> r2
array([[ 0,  1,  2],
       [ 6,  7,  8],
       [12, 13, 14]])

So, When i do set new array to r2 to zero

>>> r2[:]  = 0
>>> r2
array([[0, 0, 0],
       [0, 0, 0],
       [0, 0, 0]])

So, when i see original array then it also been looked change:

Array Changed after chanin the new array:

>>> r
array([[ 0,  0,  0,  3,  4,  5],
       [ 0,  0,  0,  9, 10, 11],
       [ 0,  0,  0, 15, 16, 17],
       [18, 19, 20, 21, 22, 23],
       [24, 25, 26, 27, 28, 29],
       [30, 30, 30, 30, 30, 30]])

Happy New Years in advanced, Guys!

Does this answer your question? List changes unexpectedly after assignment. Why is this and how can I prevent it? — Thomas Weller
– Thomas Weller, Commented Mar 4, 2022 at 12:02

javan.rajpopat · Accepted Answer · 2018-12-27 03:33:32Z

2

Explanation

r2 = r[:3, :3]

Doesn't create a new array, but renames the current array. What you need to do is known as 'deep copy'. Use, numpy.copy() to do what you need.

x = np.array([1, 2, 3])
y = x
z = np.copy(x)

x[0] = 10
x[0] == y[0]
True
x[0] == z[0]
False

1 Comment

hpaulj Over a year ago

r2 is a new ndarray object, but is a view. So yes you want a copy, but it doesn't need to a deep copy (that's more applicable to lists).

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