you can’t check an equal sign on an image

Is that the result you’re trying to achieve?

enter image description here

Since it seems that your math expressions are written horizontally, as @Micka points out, you have very strong priors on the relationship between the two different components of an equal sign, so there is a straight-forward (but hacky) way to detect when to minus signs are actually an equal:

import cv2
import matplotlib.pyplot as plt
import numpy as np

class Rect:
    def __init__(self,
        self.d=d[(a+c)/2, (b+d)/2])

def merge_rectangles(r_1, r_2):

    a_m= min(r_1.a, r_2.a)
    b_m= min(r_1.b, r_2.b)
    c_M= max(r_1.c, r_2.c)
    d_M= max(r_1.d, r_2.d)

    return Rect(a_m, b_m, c_M, d_M)

def they_form_an_equal_sign(rect_1,
        tol_w=10,#in pixels
    #check if the bounding boxes approximately align
    b0 = (np.abs(rect_1.a - rect_2.a) < tol_w ) and (np.abs(rect_1.c - rect_2.c) <     tol_w)

    #check if the bounding boxes have approximately the same height
    h1 = np.abs(rect_1.d - rect_1.b)
    h2 = np.abs(rect_2.d - rect_2.b)
    b1 = np.abs(h1-h2) < tol_h

    return b0 and b1

image = cv2.imread('/tmp/m.png')
grey = cv2.cvtColor(image.copy(), cv2.COLOR_BGR2GRAY)
ret, thresh = cv2.threshold(grey.copy(), 0, 255, cv2.THRESH_BINARY_INV)
contours, _ = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL,     cv2.CHAIN_APPROX_SIMPLE)

for c in contours:
    x,y,w,h = cv2.boundingRect(c)

for i in range(len(candidate_rectangles)):
    for j in range(i+1,len(candidate_rectangles)):
        b=they_form_an_equal_sign(candidate_rectangles[i], candidate_rectangles[j])
        if b:
            new_rect=merge_rectangles(candidate_rectangles[i], candidate_rectangles[j])

for i in range(len(kept)):
    if kept[i]:
        cv2.rectangle(image, (rect.a, rect.b), (rect.c, rect.d), color=(0, 255, 0), thickness=2)
for rect in new_rectangles:
    cv2.rectangle(image, (rect.a, rect.b), (rect.c, rect.d), color=(0, 255, 0), thickness=2)

plt.imshow(image, cmap="gray")

Basically, this takes the brute-force approach of comparing every two bounding boxes that have been detected in your code, and merging them into a larger one if they meet the requirements from your prior: i.e. if they are horizontally algined (as the top and bottom parts of a minus sign should) and if their height is approximately the same.

However, this is obviously not robust: you need to adjust thresholds and the entire thing will fall apart if your expressions aren’t horizontal and clearly separated. If you want a more robust/useful system, basic ML approaches for character recognition are a better place to start.

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