高级方形检测（连接区域） [英] Advanced square detection (with connected region)

问题描述

如果方块在图像中有连接区域，我该如何检测它们。

if the squares has connected region in image, how can I detect them.

我测试了
中提到的方法 OpenCV C ++ / Obj-C：高级方形检测

效果不佳。

有什么好主意吗？

import cv2
import numpy as np

def angle_cos(p0, p1, p2):
    d1, d2 = (p0-p1).astype('float'), (p2-p1).astype('float')
    return abs( np.dot(d1, d2) / np.sqrt( np.dot(d1, d1)*np.dot(d2, d2) ) )

def find_squares(img):
    squares = []
    gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
    # cv2.imshow("gray", gray)

    gaussian = cv2.GaussianBlur(gray, (5, 5), 0)

    temp,bin = cv2.threshold(gaussian, 80, 255, cv2.THRESH_BINARY)
    # cv2.imshow("bin", bin)

    contours, hierarchy = cv2.findContours(bin, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)

    cv2.drawContours( gray, contours, -1, (0, 255, 0), 3 )

    #cv2.imshow('contours', gray)
    for cnt in contours:
        cnt_len = cv2.arcLength(cnt, True)
        cnt = cv2.approxPolyDP(cnt, 0.02*cnt_len, True)
        if len(cnt) == 4 and cv2.contourArea(cnt) > 1000 and cv2.isContourConvex(cnt):
            cnt = cnt.reshape(-1, 2)
            max_cos = np.max([angle_cos( cnt[i], cnt[(i+1) % 4], cnt[(i+2) % 4] ) for i in xrange(4)])
            if max_cos < 0.1:
                squares.append(cnt)
    return squares

if __name__ == '__main__':
    img = cv2.imread('123.bmp')

    #cv2.imshow("origin", img)

    squares = find_squares(img)  
    print "Find %d squres" % len(squares)
    cv2.drawContours( img, squares, -1, (0, 255, 0), 3 )
    cv2.imshow('squares', img)

    cv2.waitKey()

我在opencv示例中使用了一些方法，但结果并不好。

I use some method in the opencv example, but the result is not good.

推荐答案

根据距离变换应用分水岭变换将分离对象：

Applying a Watershed Transform based on the Distance Transform will separate the objects:

处理边境处的物体总是有问题，经常被丢弃，所以左上方没有分开的粉红色矩形根本不是问题。

Handling objects at the border is always problematic, and often discarded, so that pink rectangle at top left not separated is not a problem at all.

给定二进制图像，我们可以应用距离Tra nsform（DT）并从中获取Watershed的标记。理想情况下，会有一个用于查找区域最小值/最大值的现成函数，但由于它不存在，我们可以对如何阈值DT进行合理的猜测。根据我们可以使用Watershed进行分段的标记，问题就解决了。现在你可以担心区分矩形和非矩形的组件。

Given a binary image, we can apply the Distance Transform (DT) and from it obtain markers for the Watershed. Ideally there would be a ready function for finding regional minima/maxima, but since it isn't there, we can make a decent guess on how we can threshold DT. Based on the markers we can segment using Watershed, and the problem is solved. Now you can worry about distinguishing components that are rectangles from those that are not.

import sys
import cv2
import numpy
import random
from scipy.ndimage import label

def segment_on_dt(img):
    dt = cv2.distanceTransform(img, 2, 3) # L2 norm, 3x3 mask
    dt = ((dt - dt.min()) / (dt.max() - dt.min()) * 255).astype(numpy.uint8)
    dt = cv2.threshold(dt, 100, 255, cv2.THRESH_BINARY)[1]
    lbl, ncc = label(dt)

    lbl[img == 0] = lbl.max() + 1
    lbl = lbl.astype(numpy.int32)
    cv2.watershed(cv2.cvtColor(img, cv2.COLOR_GRAY2BGR), lbl)
    lbl[lbl == -1] = 0
    return lbl


img = cv2.cvtColor(cv2.imread(sys.argv[1]), cv2.COLOR_BGR2GRAY)
img = cv2.threshold(img, 0, 255, cv2.THRESH_OTSU)[1]
img = 255 - img # White: objects; Black: background

ws_result = segment_on_dt(img)
# Colorize
height, width = ws_result.shape
ws_color = numpy.zeros((height, width, 3), dtype=numpy.uint8)
lbl, ncc = label(ws_result)
for l in xrange(1, ncc + 1):
    a, b = numpy.nonzero(lbl == l)
    if img[a[0], b[0]] == 0: # Do not color background.
        continue
    rgb = [random.randint(0, 255) for _ in xrange(3)]
    ws_color[lbl == l] = tuple(rgb)

cv2.imwrite(sys.argv[2], ws_color)

从上图中你可以考虑在每个组件中拟合椭圆来确定矩形。然后，您可以使用一些测量来定义组件是否为矩形。这种方法有更大的机会适用于完全可见的矩形，并且可能会对部分可见的矩形产生不良结果。下图显示了这种方法的结果，如果拟合椭圆中的矩形在组件面积的10％范围内，则组件为矩形。

From the above image you can consider fitting ellipses in each component to determine rectangles. Then you can use some measurement to define whether the component is a rectangle or not. This approach has a greater chance to work for rectangles that are fully visible, and will likely produce bad results for partially visible ones. The following image shows the result of such approach considering that a component is a rectangle if the rectangle from the fitted ellipse is within 10% of component's area.

# Fit ellipse to determine the rectangles.
wsbin = numpy.zeros((height, width), dtype=numpy.uint8)
wsbin[cv2.cvtColor(ws_color, cv2.COLOR_BGR2GRAY) != 0] = 255

ws_bincolor = cv2.cvtColor(255 - wsbin, cv2.COLOR_GRAY2BGR)
lbl, ncc = label(wsbin)
for l in xrange(1, ncc + 1):
    yx = numpy.dstack(numpy.nonzero(lbl == l)).astype(numpy.int64)
    xy = numpy.roll(numpy.swapaxes(yx, 0, 1), 1, 2)
    if len(xy) < 100: # Too small.
        continue

    ellipse = cv2.fitEllipse(xy)
    center, axes, angle = ellipse
    rect_area = axes[0] * axes[1]
    if 0.9 < rect_area / float(len(xy)) < 1.1:
        rect = numpy.round(numpy.float64(
                cv2.cv.BoxPoints(ellipse))).astype(numpy.int64)
        color = [random.randint(60, 255) for _ in xrange(3)]
        cv2.drawContours(ws_bincolor, [rect], 0, color, 2)

cv2.imwrite(sys.argv[3], ws_bincolor)

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