Python 如何使用OpenCV findContours检测相交形状？

在黑白图像中有两个相交的椭圆。我正试图使用OpenCV findContours，使用此代码（以及下面的附图）将单独的形状识别为单独的轮廓

但是，发现了四个轮廓，其中没有一个是原始轮廓：

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如何配置OpenCV.findContours以识别这两个单独的形状？（注意，我已经玩过Hough圆，发现它对于我正在分析的图像不可靠）

从哲学上讲，你想找到两个圆，因为你搜索它们，你需要中心和半径。从图形上看，这些图形是连接在一起的，我们可以看到它们是分开的，因为我们知道“圆”是什么，并推断出坐标，这些坐标与重叠的部分相匹配

那么，如何为每个轮廓（或在某些情况下使用椭圆并使用其参数）找到最小封闭圆呢

然后说，在一张清晰的图像中画一个圆，然后用掩模取不为零的像素坐标，或者通过一步一步地画一个圆来计算它们

然后以一定精度将这些坐标与其他轮廓中的坐标进行比较，并将匹配坐标附加到当前轮廓

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最后：在清晰的画布上绘制延伸轮廓，并将HoughCircles应用于单个不重叠的圆。（或者计算圆心和半径，一个圆的坐标，并与轮廓进行精确比较。）

从哲学上讲，你想找到两个圆，因为你搜索它们，你需要圆心和半径。从图形上看，这些图形是连接在一起的，我们可以看到它们是分开的，因为我们知道“圆”是什么，并推断出坐标，这些坐标与重叠的部分相匹配

那么，如何为每个轮廓（或在某些情况下使用椭圆并使用其参数）找到最小封闭圆呢

然后说，在一张清晰的图像中画一个圆，然后用掩模取不为零的像素坐标，或者通过一步一步地画一个圆来计算它们

然后以一定精度将这些坐标与其他轮廓中的坐标进行比较，并将匹配坐标附加到当前轮廓

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最后：在清晰的画布上绘制延伸轮廓，并将HoughCircles应用于单个不重叠的圆。（或者计算圆心和半径、圆的坐标并与轮廓进行精确比较。）

也许我用这种方法做得过火了，但它可以作为一种工作方法。您可以找到图像上的所有轮廓-您将得到两个像“半圆”的轮廓，交点轮廓和两个连接圆的外部形状轮廓。最小的三条等高线应该是两个半圆和交点。如果从这三个轮廓中画出两个的组合，将得到三个遮罩，其中两个将具有一个半圆和交点的组合。如果在遮罩上执行闭合操作，将获得圆。然后你应该简单地做一个算法来检测哪两个掩模代表一个完整的圆，你就会得到你的结果。以下是示例解决方案：

import numpy as np
import cv2


# Function for returning solidity of contour - ratio of contour area to its 
# convex hull area.
def checkSolidity(cnt):
    area = cv2.contourArea(cnt)
    hull = cv2.convexHull(cnt)
    hull_area = cv2.contourArea(hull)
    solidity = float(area)/hull_area
    return solidity


img_orig = cv2.imread("circles.png")
# Had to dilate the image so the contour was completly connected.
img = cv2.dilate(img_orig, np.ones((3, 3), np.uint8))
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)  # Grayscale transformation.
# Otsu threshold.
thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY_INV+cv2.THRESH_OTSU)[1]
# Search for contours.
contours = cv2.findContours(thresh, cv2.CHAIN_APPROX_NONE, cv2.RETR_TREE)[0]

# Sorting contours from smallest to biggest.
contours.sort(key=lambda cnt: cv2.contourArea(cnt))

# Three contours - two semi circles and the intersection of the circles.
cnt1 = contours[0]
cnt2 = contours[1]
cnt3 = contours[2]

# Create three empty images
h, w = img.shape[:2]
mask1 = np.zeros((h, w), np.uint8)
mask2 = np.zeros((h, w), np.uint8)
mask3 = np.zeros((h, w), np.uint8)

# Draw all combinations of two out of three contours on the masks.
# The goal here is to draw one semicircle and the intersection together.

cv2.drawContours(mask1, [cnt1], 0, (255, 255, 255), -1)
cv2.drawContours(mask1, [cnt3], 0, (255, 255, 255), -1)

cv2.drawContours(mask2, [cnt2], 0, (255, 255, 255), -1)
cv2.drawContours(mask2, [cnt3], 0, (255, 255, 255), -1)

cv2.drawContours(mask3, [cnt1], 0, (255, 255, 255), -1)
cv2.drawContours(mask3, [cnt2], 0, (255, 255, 255), -1)


# Perform closing operation on the masks so that you get uniform contours.
kernel_size = 25
kernel = np.ones((kernel_size, kernel_size), np.uint8)
mask1 = cv2.morphologyEx(mask1, cv2.MORPH_CLOSE, kernel)
mask2 = cv2.morphologyEx(mask2, cv2.MORPH_CLOSE, kernel)
mask3 = cv2.morphologyEx(mask3, cv2.MORPH_CLOSE, kernel)

masks = []  # List for storing all the masks.
masks.append(mask1)
masks.append(mask2)
masks.append(mask3)

# List where you will append solidity of the found biggest contour of every mask.
solidity = []
for mask in masks:
    cnts = cv2.findContours(mask, cv2.CHAIN_APPROX_NONE, cv2.RETR_TREE)[0]
    cnt = max(cnts, key=lambda c: cv2.contourArea(c))
    s = checkSolidity(cnt)
    solidity.append(s)


# Index of the mask with smallest solidity.
min_solidity = solidity.index(min(solidity))
# The mask with the contour that has smallest solidity should be the one that
# has two semicirles drawn instead of one semicircle and the intersection. 
#You could build a better function to check which mask is the one with   
# two semicircles... like maybe the contour with the largest 
# height and width of the bounding box etc.
# I chose solidity because it is enough for this example.

# Selection of colors.
colors = {
    0: (0, 0, 255),
    1: (0, 255, 0),
    2: (255, 0, 0),
}

# Draw contours of the mask other two masks - those two that have the        
# semicircle and the intersection.
for i, s in enumerate(solidity):
    if s != solidity[min_solidity]:
        cnts = cv2.findContours(
            masks[i], cv2.CHAIN_APPROX_NONE, cv2.RETR_TREE)[0]
        cnt = max(cnts, key=lambda c: cv2.contourArea(c))
        cv2.drawContours(img_orig, [cnt], 0, colors[i], 1)

# Display result
cv2.imshow("img", img_orig)
cv2.waitKey(0)
cv2.destroyAllWindows()

结果:

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也许我用这种方法做得过火了，但它可以作为一种有效的方法。您可以找到图像上的所有轮廓-您将得到两个像“半圆”的轮廓，交点轮廓和两个连接圆的外部形状轮廓。最小的三条等高线应该是两个半圆和交点。如果从这三个轮廓中画出两个的组合，将得到三个遮罩，其中两个将具有一个半圆和交点的组合。如果在遮罩上执行闭合操作，将获得圆。然后你应该简单地做一个算法来检测哪两个掩模代表一个完整的圆，你就会得到你的结果。以下是示例解决方案：

import numpy as np
import cv2


# Function for returning solidity of contour - ratio of contour area to its 
# convex hull area.
def checkSolidity(cnt):
    area = cv2.contourArea(cnt)
    hull = cv2.convexHull(cnt)
    hull_area = cv2.contourArea(hull)
    solidity = float(area)/hull_area
    return solidity


img_orig = cv2.imread("circles.png")
# Had to dilate the image so the contour was completly connected.
img = cv2.dilate(img_orig, np.ones((3, 3), np.uint8))
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)  # Grayscale transformation.
# Otsu threshold.
thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY_INV+cv2.THRESH_OTSU)[1]
# Search for contours.
contours = cv2.findContours(thresh, cv2.CHAIN_APPROX_NONE, cv2.RETR_TREE)[0]

# Sorting contours from smallest to biggest.
contours.sort(key=lambda cnt: cv2.contourArea(cnt))

# Three contours - two semi circles and the intersection of the circles.
cnt1 = contours[0]
cnt2 = contours[1]
cnt3 = contours[2]

# Create three empty images
h, w = img.shape[:2]
mask1 = np.zeros((h, w), np.uint8)
mask2 = np.zeros((h, w), np.uint8)
mask3 = np.zeros((h, w), np.uint8)

# Draw all combinations of two out of three contours on the masks.
# The goal here is to draw one semicircle and the intersection together.

cv2.drawContours(mask1, [cnt1], 0, (255, 255, 255), -1)
cv2.drawContours(mask1, [cnt3], 0, (255, 255, 255), -1)

cv2.drawContours(mask2, [cnt2], 0, (255, 255, 255), -1)
cv2.drawContours(mask2, [cnt3], 0, (255, 255, 255), -1)

cv2.drawContours(mask3, [cnt1], 0, (255, 255, 255), -1)
cv2.drawContours(mask3, [cnt2], 0, (255, 255, 255), -1)


# Perform closing operation on the masks so that you get uniform contours.
kernel_size = 25
kernel = np.ones((kernel_size, kernel_size), np.uint8)
mask1 = cv2.morphologyEx(mask1, cv2.MORPH_CLOSE, kernel)
mask2 = cv2.morphologyEx(mask2, cv2.MORPH_CLOSE, kernel)
mask3 = cv2.morphologyEx(mask3, cv2.MORPH_CLOSE, kernel)

masks = []  # List for storing all the masks.
masks.append(mask1)
masks.append(mask2)
masks.append(mask3)

# List where you will append solidity of the found biggest contour of every mask.
solidity = []
for mask in masks:
    cnts = cv2.findContours(mask, cv2.CHAIN_APPROX_NONE, cv2.RETR_TREE)[0]
    cnt = max(cnts, key=lambda c: cv2.contourArea(c))
    s = checkSolidity(cnt)
    solidity.append(s)


# Index of the mask with smallest solidity.
min_solidity = solidity.index(min(solidity))
# The mask with the contour that has smallest solidity should be the one that
# has two semicirles drawn instead of one semicircle and the intersection. 
#You could build a better function to check which mask is the one with   
# two semicircles... like maybe the contour with the largest 
# height and width of the bounding box etc.
# I chose solidity because it is enough for this example.

# Selection of colors.
colors = {
    0: (0, 0, 255),
    1: (0, 255, 0),
    2: (255, 0, 0),
}

# Draw contours of the mask other two masks - those two that have the        
# semicircle and the intersection.
for i, s in enumerate(solidity):
    if s != solidity[min_solidity]:
        cnts = cv2.findContours(
            masks[i], cv2.CHAIN_APPROX_NONE, cv2.RETR_TREE)[0]
        cnt = max(cnts, key=lambda c: cv2.contourArea(c))
        cv2.drawContours(img_orig, [cnt], 0, colors[i], 1)

# Display result
cv2.imshow("img", img_orig)
cv2.waitKey(0)
cv2.destroyAllWindows()

结果:

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为了便于参考，我将在这里发布我根据一些想法提出的解决方案以及其他一些想法。该解决方案具有99.9%的有效性，可以从图像中恢复椭圆，这些图像通常包含许多重叠、包含以及其他图像噪声，如线条、文本等

代码太长，分布太广，无法在此处发布，但伪代码如下所示

分割图像

• 使用RETR_EXTERNAL运行cv2 findContours以获取图像中的单独区域
• 对于每个图像，填充内部，应用遮罩，并独立于其他区域提取要处理的区域
• 其余步骤分别针对每个区域执行

使用RETR_列表运行cv2 findContours以获取所有内部和外部轮廓

对于找到的每个轮廓，应用多边形平滑以减少像素化的影响

对于每个平滑轮廓，识别轮廓内具有相同曲率符号的连续段，即完全向右弯曲或向左弯曲的段（仅计算角度和符号变化）

在每个分段内，用最小二乘法拟合椭圆模型（scikit学习椭圆模型）

对原始图像执行Lee算法，计算每个像素到白色像素的最小距离

对于每个模型，执行贪婪的局部邻域搜索以改进对原始模型的拟合-拟合是从lee算法输出到白色像素的拟合椭圆最大距离

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不简单也不优雅，但对于我正在处理的内容来说非常准确，这是通过对大量图像的手动查看确认的。

为了便于参考，我将在这里发布我根据一些想法提出的解决方案，以及其他一些想法。该解决方案具有99.9%的有效性，可从通常包含许多重叠的图像中恢复椭圆，