Finding width and height of concave curved shaped blob

Question

I have been scratching my head over this problem of calculating the width and height measurements of a figure like this. The major challenge is I cant use miBoundingrectangl

Answer 1

Here's an OpenCV solution. The main idea is

• Convert image to grayscale and invert image
• Find contour and centroid of the blob
• Use Numpy slicing to grab all row/column pixels
• Count non-zero pixels to determine width/height

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We convert the image to grayscale and invert it. This is the image we will count pixels on since the desired ROI is in white. From here, we find the contour of the blob and the center coordinates using cv2.moments(). This gives us the centroid (i.e., the center (x, y)-coordinates of the object)

M = cv2.moments(c)
cX = int(M["m10"] / M["m00"])
cY = int(M["m01"] / M["m00"])

Next we use Numpy slicing to obtain all the row and column pixels. We use cv2.countNonZero() to find the width/height of the row/column like this

row_pixels = cv2.countNonZero(gray[cY][:])
column_pixels = cv2.countNonZero(gray[:, cX])

Here's a visualization

Here's the result

row 150

column 354

import cv2
import numpy as np

image = cv2.imread('1.png')
inverted = 255 - image.copy()
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
gray = 255 - gray

cnts = cv2.findContours(gray, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]

for c in cnts:
    M = cv2.moments(c)
    cX = int(M["m10"] / M["m00"])
    cY = int(M["m01"] / M["m00"])

    cv2.circle(inverted, (cX, cY), 5, (36, 255, 12), -1)
    inverted[cY][:] = (36, 255, 12)
    inverted[:, cX] = (36, 255, 12)
    row_pixels = cv2.countNonZero(gray[cY][:])
    column_pixels = cv2.countNonZero(gray[:, cX])

print('row', row_pixels)
print('column', column_pixels)
cv2.imshow('inverted', inverted)
cv2.imwrite('inverted.png', image)
cv2.waitKey(0)