How to use cv2.imread() with an image type object instead of "image path"?

Question

I wonder if there’s any way to do something like this:

import cv2, pyautogui

screenshot = pyautogui.screenshot()
cv2.imread(screenshot, 0)

Instead:

import cv2, pyautogui
screenshot = pyautogui.screenshot("imagem.png")
cv2.imread("imagem.png", 0)

Answer 1

I didn’t quite understand the purpose, but if you want to simply save the image to use it again with Opencv, you can use the method save() of the image in this way:

pil_image = pyautogui.screenshot()
pil_image.save("<Nome_do_arquivo>")

image = cv2.imread("<Nome_do_arquivo>")

If what you want is to actually convert the image directly without saving to use with Opencv, the process will be a little more complicated and you will need to use the library numpy for that reason.

First, you must turn your image into an object of numpy.array and after that you should convert it from RGB to BGR. See below:

pil_image = pyautogui.screenshot()
image = numpy.array(pil_image)

# Converte de RGB para BGR
image = image[:, :, ::-1].copy()

cv2.imshow("Imagem", image)

Answer 2

Actually you don’t have to do anything.

take this test:

import cv2, pyautogui

screenshot = pyautogui.screenshot()
####### editado
img_opencv = cv2.imencode('.jpg', screenshot )[1].tostring()
####### editado
cv2.imshow("teste",img_opencv )
cv2.waitKey(0)

it will open a window with the print you took.