numpy.where on 2d or 3d matrix

Question

I want to get the index where the target is located in background image using numpy. the background color of each image is actually varible
so except for the color of the sqaure, other colors (in this case, black painted) including the inside of the square will vary.
i have no idea how to approach this at all as im not familiar with numpy.

import numpy as np

#changing to grayscale to have 2d array 
output = cv2.imread('backgroundimage.png', cv2.IMREAD_GRAYSCALE)
output1 = cv2.imread('target.png', cv2.IMREAD_GRAYSCALE)

i tried to change the images to 2d array because i thought it might be easier to approach.

a = np.where(output==output1)

apparently this doesnt work for 2d or 3d.
my desired output will be something like this

desired output = (108, 23) (x and y coordination of where its found)

so how would i able to do what i want?

Asked By: kerz

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Source

Answer 1

You have to use a sliding window approach, and make sure that all pixels are equal for each window you compare with the target. You can do that using sliding_window_view and all. You can mask out any values inside the target that you do not want to match by setting them to True:

import cv2
import numpy as np

output = cv2.imread('backgroundimage.png', cv2.IMREAD_GRAYSCALE)
output1 = cv2.imread('target.png', cv2.IMREAD_GRAYSCALE)

# Apply sliding window view
windows = np.lib.stride_tricks.sliding_window_view(output, output1.shape)

# Only check the 1 pixel border by making a mask
mask = np.full_like(output1, False)
mask[1:-1, 1:-1] = True

# Apply mask and check match everywhere
masked = (windows == output1) | mask
matches = masked.all(axis=(2, 3))

locations = np.where(matches)

locations:

(array([24], dtype=int64), array([108], dtype=int64))

Answered By: Chrysophylaxs

numpy.where on 2d or 3d matrix

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