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First, my 3 separate scenarios and my input image

enter image description here

Scenario 1: Copying input image onto new variable by " = "

a = cv2.imread("/content/consec2.jpg")
b = a

b [b < 200] = 0 #Some random change to an image

plt.imshow(a, cmap='gray')

Image Displayed: ("changed" image) enter image description here

Scenario 2: Copying input image onto new variable by .copy() method

a = cv2.imread("/content/consec2.jpg")
b = a.copy()

b [b < 100] = 0 #Same random change to the image as in scenario 1

plt.imshow(a, cmap='gray')

Image Displayed: The original input image

Scenario 3: Normal calculation (Disregard the image here)

a = 10
b = a
print (a)

Result: the printed value comes out to 10 (This scenario is just for extra refernce).

QUESTION: Why are scenario 1 and scenario 2 giving differnet results? Shouldn't the " = " and .copy() function the same way in this case?

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1 Answer 1

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In scenario 1, the underlying memory is used for both Mat a and Mat b, so when you change b, you end up changing a as well, i.e. it's just a reference to the underlying Mat that is copied. From the opencv docs: https://docs.opencv.org/4.x/d6/d6d/tutorial_mat_the_basic_image_container.html

"All the above objects, in the end, point to the same single data matrix and making a modification using any of them will affect all the other ones as well."

In scenario 2, you have created a new Mat through the call to a.copy() and initialized the underlying memory, so changes to b won't impact a.

This is the same for all objects in python btw, it has nothing to do specifically with opencv. https://docs.python.org/3/library/copy.html

hth

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