CODEWITHSUNDEEP
pythonnumpy
subtleseeker
subtleseeker

Reputation: 5253

Copying numpy array with '=' operator. Why is it working?

According to this answer, B=A where A is a numpy array, B should be pointing to the same object A.

import cv2
import numpy as np

img = cv2.imread('rose.jpeg')
print("img.shape: ", np.shape(img))

img2 = img
img = cv2.resize(img, (250,100))
print("img.shape: ", img.shape)
print("img2.shape:", img2.shape)

Output:

img.shape:  (331, 500, 3)
img.shape:  (100, 250, 3)
img2.shape: (331, 500, 3)

It seems to be a very basic question, but I have been scratching my head over this. Could someone please explain what's happening behind it?

Upvotes: 0

Views: 66

Answers (1)

Lutz
Lutz

Reputation: 655

The "problem" is that your not using numpy here but opencv and while numpy array.resize() is in-place opencv img.resize() is not.

So your call to 

    img = cv2.resize(img, (250,100))

creates a new object (image) with the given size. So here the img variable will point to a different object then before the call.

    img2 = img

adds a new name for the original object. Here img2 and img are refering to exactly the same object/piece of memory. 

    img = cv2.resize(img, (250,100))

cv2.resize(img, (250,100)) creates a new object and the name img now refers to that new object/piece of memory.

    print("img.shape: ", img.shape)

gets you the size of the new object and 

    print("img2.shape:", img2.shape)

the size of the original object as img2 still refers to the original object.

By the way in numpy the call a = a.resize(...) would be really bad - because a would then by None (return value of resize) instead of the resized array. There you would just do a.resize(...)

Upvotes: 1

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