如何使用Python openCV仅在这张表图像中找到左上角框的位置（x，y，宽度，高度）？

Question

我有这张图片，我需要找到only左上方方框的位置及其宽度和高度。如何在openCV中使用Python做到这一点？enter image description here

Answer 1

这是在Python / OpenCV / Numpy中执行此操作的一种方法。

• 读取输入
• 转换为灰色
• 二进制阈值
• 计算每一行每一列的黑色像素之和
• 阈值总和必须大于图像高度和宽度的80％
• 查找这些总和中具有非零值的所有坐标
• 过滤坐标以删除彼此之间不超过10个像素的任何值，以避免与大于1个像素的线重复”>
• 将过滤后的坐标的第一和第二坐标作为左上角矩形的边界
• 在这些边界处裁剪输入图像
• 保存结果

输入：

import cv2
import numpy as np

# read input
img = cv2.imread("table_cells.png")
hh, ww = img.shape[:2]

# convert to gray
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)

# threshold to binary
thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY)[1]

# get sum of black values in rows and columns
row_sums = np.sum(thresh==0, axis=1)
column_sums = np.sum(thresh==0, axis=0)

# threshold sums to counts above 80% of hh and ww
row_sums[np.where(row_sums<0.8*ww)] = 0
column_sums[np.where(column_sums<0.8*hh)] = 0

# find coordinates that have non-zero values
row_coords = np.argwhere(row_sums>0)
column_coords = np.argwhere(column_sums>0)
num_rows = len(row_coords)
num_cols = len(column_coords)

# filter row_coords to avoid duplicates within 10 pixels
row_coords_filt = [row_coords[0]]
for i in range(num_rows-1):
    if (row_coords[i] > row_coords[i-1]+10):
        row_coords_filt.append(row_coords[i])

column_coords_filt = [column_coords[0]]
for i in range(num_cols-1):
    if (column_coords[i] > column_coords[i-1]+10):
        column_coords_filt.append(column_coords[i])

# print row_coords_filt
print('grid row coordinates:')
for c in row_coords_filt:
    print (c)

print('')

# print column_coords_filt
print('grid column coordinates:')
for c in column_coords_filt:
    print (c)

# get left, right, top, bottom of upper left rectangle
left = int(column_coords_filt[0])
right = int(column_coords_filt[1])
top = int(row_coords_filt[0])
bottom = int(row_coords_filt[1])

# crop rectangle
rectangle = img[top:bottom, left:right]

# save output
cv2.imwrite('table_cells_crop.png', rectangle)

cv2.imshow('thresh', thresh)
cv2.imshow('rectangle', rectangle)
cv2.waitKey(0)
cv2.destroyAllWindows()
矩形裁剪：

找到坐标：

grid row coordinates:
[30]
[315]
[599]
[884]

grid column coordinates:
[41]
[790]
[1540]
[2289]