How to change an new array without changing the initial array that the new array was created?

Question

img_1 = np.arange(25).reshape(5,5)

img_2 = img_1

locs = np.where(img_2 >= 0)

img_2[locs]  = [65000]

Wondering is there an way to have img_1 with initial array and only img_2 will will be modified?

Both img_1 and img_2 array is changing.

You can use `img2 = img1.copy()`. `img2 = img1` makes `img2` point to the same data as `img1`. — B Remmelzwaal, Feb 07 '23 at 02:09
This isn't unique to numpy arrays. `img2 = img1` will make `img2` reference the exact same object as `img1`, regardless of what `img1` is. For more on this, see [what does it mean by 'passed by assignment'?](https://stackoverflow.com/q/50534394/11082165) — Brian61354270, Feb 07 '23 at 02:14
for a numpy array or a pandas dataframe.. ```img2=img1.copy()``` will give you a copy .. for other objects which donot have a copy() method inplemented ... `import copy` and `obj2=copy.deepcopy(obj1)` will give you true copy and not a reference — geekay, Feb 07 '23 at 02:34
In addition, there's also a shorter way to do the operation you did the code. ```img_2[img_2>=0]``` produces the same result. The argument produces a Boolean for each element of img_2, i.e., accordingly the array img_2 takes the values. — Pixel_Bear, Feb 07 '23 at 04:14

score 1 · Answer 1 · answered Feb 07 '23 at 05:39

1

All you need is just the np.where() alone. Do like this.

import numpy as np

img_1 = np.arange(25).reshape(5,5)
img_2 = np.where(img_1 >= 0, 65000, img_1)

answered Feb 07 '23 at 05:39

relent95

1 Answers1