我正在使用的图像:

我正在尝试查找此图像中的每个框。结果不必是 100% 准确，只要找到的框的位置/大小大致正确即可。通过使用正方形检测示例，我设法获得了轮廓、边界框、角和框的中心。

我在这里遇到了一些问题:

1. 检测绘制线的内部和外部的边界矩形。
2. 检测到一些无关的角/中心。
3. 我不确定如何将角/中心与相关的轮廓/边界框匹配，尤其是在考虑嵌套框时。

来自代码的图像:

这是我用来生成上图的代码:

import numpy as np
import cv2
from operator import itemgetter
from glob import glob
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 makebin(gray):
    bin = cv2.adaptiveThreshold(gray, 255, cv2.ADAPTIVE_THRESH_MEAN_C, cv2.THRESH_BINARY, 5, 2)
    return cv2.bitwise_not(bin)
def find_squares(img):
    img = cv2.GaussianBlur(img, (11, 11), 0)
    squares = []
    points = []`
    for gray in cv2.split(img):
        bin = makebin(gray)
        contours, hierarchy = cv2.findContours(bin, cv2.RETR_LIST, cv2.CHAIN_APPROX_SIMPLE)
        corners = cv2.goodFeaturesToTrack(gray,len(contours)*4,0.2,15)
        cv2.cornerSubPix(gray,corners,(6,6),(-1,-1),(cv2.TERM_CRITERIA_MAX_ITER | cv2.TERM_CRITERIA_EPS,10, 0.1))
        for cnt in contours:
            cnt_len = cv2.arcLength(cnt, True)
            if len(cnt) >= 4 and cv2.contourArea(cnt) > 200:
                rect = cv2.boundingRect(cnt)
                if rect not in squares:
                    squares.append(rect)
    return squares, corners, contours
if __name__ == '__main__':
    for fn in glob('../1 (Small).jpg'):
        img = cv2.imread(fn)
        squares, corners, contours = find_squares(img)
        for p in corners:
            cv2.circle(img, (p[0][0],p[0][3]), 3, (0,0,255),2)
        squares = sorted(squares,key=itemgetter(1,0,2,3))
        areas = []
        moments = []
        centers = []
        for s in squares:
            areas.append(s[2]*s[3])
            cv2.rectangle( img, (s[0],s[1]),(s[0]+s[2],s[1]+s[3]),(0,255,0),1)
        for c in contours:
            moments.append(cv2.moments(np.array(c)))
        for m in moments:
            centers.append((int(m["m10"] // m["m00"]), int(m["m01"] // m["m00"])))
        for cent in centers:
            print cent
            cv2.circle(img, (cent[0],cent[1]), 3, (0,255,0),2)
        cv2.imshow('squares', img)
        ch = 0xFF & cv2.waitKey()
        if ch == 27:
            break
    cv2.destroyAllWindows()             

最佳答案

我建议以更简单的方法作为起点。例如，形态梯度可以作为强边缘的良好局部检测器，其阈值往往很简单。然后，您可以删除太小的组件，这对于您的问题也相对容易。在您的示例中，每个剩余的连接组件都是一个盒子，因此问题在本例中得到解决。

通过这个简单的过程，您将获得以下结果:

红点代表组件的质心，因此如果黄色对您不利，您可以从那里再长出另一个包含在黄色中的盒子。

以下是实现该目标的代码:

import sys
import numpy
from PIL import Image, ImageOps, ImageDraw
from scipy.ndimage import morphology, label

def boxes(orig):
    img = ImageOps.grayscale(orig)
    im = numpy.array(img)

    # Inner morphological gradient.
    im = morphology.grey_dilation(im, (3, 3)) - im

    # Binarize.
    mean, std = im.mean(), im.std()
    t = mean + std
    im[im < t] = 0
    im[im >= t] = 1

    # Connected components.
    lbl, numcc = label(im)
    # Size threshold.
    min_size = 200 # pixels
    box = []
    for i in range(1, numcc + 1):
        py, px = numpy.nonzero(lbl == i)
        if len(py) < min_size:
            im[lbl == i] = 0
            continue

        xmin, xmax, ymin, ymax = px.min(), px.max(), py.min(), py.max()
        # Four corners and centroid.
        box.append([
            [(xmin, ymin), (xmax, ymin), (xmax, ymax), (xmin, ymax)],
            (numpy.mean(px), numpy.mean(py))])

    return im.astype(numpy.uint8) * 255, box


orig = Image.open(sys.argv[1])
im, box = boxes(orig)

# Boxes found.
Image.fromarray(im).save(sys.argv[2])

# Draw perfect rectangles and the component centroid.
img = Image.fromarray(im)
visual = img.convert('RGB')
draw = ImageDraw.Draw(visual)
for b, centroid in box:
    draw.line(b + [b[0]], fill='yellow')
    cx, cy = centroid
    draw.ellipse((cx - 2, cy - 2, cx + 2, cy + 2), fill='red')
visual.save(sys.argv[3])

关于python - 查找草率手绘矩形的属性，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/14248571/