Why datatype has to be 'uint8' in Opencv python wrapper?

Question

I want to develop an image of gray color using the following code:

import numpy as np

import cv2

cv=cv2

s=np.zeros((240, 320, 3), dtype=int)

s[s==0]=128

cv.imshow('3', s)

cv.waitKey()

cv.destroyAllWindows()

When I am executing the code, I am getting a black image. 

fn='example.jpg'

cv.imwrite(fn, s)

Then I have to replace int with uint8 for getting the correct image. I want to know why I need to use uint8 instead of int?

Answer 1

Kindly be informed that the OpenCV package in Python will change its behavior according to the type of array you are using in your code:

Thus, if the image is in the format of 8-bit unsigned, then it will get presented as-is.

But, if the picture is in the structure of a 16-bit unsigned or 32-bit integer, then the number of cells of the picture will get divided by 256 that denotes the value range [0,255*256] is composed to [0,255].

And if the picture is in the form of a 32-bit floating-point, then the pixel of the image will get multiplied by 255. That means the value range [0,1] is drafted to [0,255].

Therefore, the below code will all yield the same grayscale image:

s = np.zeros((240, 320), dtype=np.uint8)

s[s==0] = 128

int32:

s = np.zeros((240, 320), dtype=np.int32)

s[s==0] = 128 * 256

float32:

s = np.zeros((240, 320), dtype=np.float32)

s[s==0] = 128 / 256.0

If you are a beginner and want to know more about Python, then do check out the below Python tutorial video that will help you in understanding the topic in a better way: