高级方形检测(连接区域)

Question

如果方块在图像中连接了区域,我该如何检测它们.

我测试了OpenCV C++/Obj-C中提到的方法 :高级方形检测

它运作不佳.

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.

Answer 1

应用基于距离变换的分水岭变换将分离对象:

处理边界处的对象总是有问题的,并且经常被丢弃,因此左上角的粉红色矩形不是分开的根本不是问题.

给定二值图像,我们可以应用距离变换(DT)并从中获取分水岭的标记.理想情况下,会有一个用于查找区域最小值/最大值的就绪函数,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.现在您可以担心区分矩形组件和非矩形组件.

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)

从上面的图像中,您可以考虑在每个组件中拟合椭圆来确定矩形.然后,您可以使用某些测量来定义组件是否为矩形.这种方法更有可能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.

# 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)