计算机视觉：如何在发票中找到多少行（行）边界框？

此解决方案对阈值敏感，您可能需要根据每行中的文本量进行调整！

• 首先，根据存在和文本量（黑色像素）分割线条。
• 其次，找到要与边界框进行比较的线条边界。
• 最后，在边界框索引和分段线索引之间进行比较。

输出：

[17,40,53,79,95,117]
box [4,43]  belongs to line 1
box [9,47]  belongs to line 1
box [76,122]  belongs to line 3
box [10,47]  belongs to line 1
box [81,125]  belongs to line 3
box [84,124]  belongs to line 3
box [47,90]  belongs to line 2

代码：

import cv2

# Read the image
orig = cv2.imread('input.jpg',0)[:,15:]

# The detected boxes
boxes = [[4,43],[9,47],[76,122],[30,74],[10,[81,125],[84,124],[47,90],[1,38]]

# make a deep copy
img = orig.copy()

# quantify the black pixels in each line
summ = img.sum(axis=1)

# Threshold
th = summ.mean()

img[summ>th,:] = 0
img[summ<=th,:] = 1

rows = []
for y in range(img.shape[0]-1):
    if img[y,0]>img[y+1,0] or img[y,0]<img[y+1,0]:
        rows.append(y)

# sort lines indices. 
rows.sort()

print(rows)

# compare the indices
for box in boxes:
    for idx in range(0,len(rows),2):
        if box[0] < rows[idx] and box[1] > rows[idx+1]:
            print("box [{},{}]".format(box[0],box[1])," belongs to line {}".format(idx//2+1))