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有没有办法在opencv中计算粗糙度轮廓？

图片示例：https ://docs.google.com/file/d/0ByS6Z5WRz-h2NDgySmJ6NnpId0U/edit?usp=sharing

更新

我计算粗糙度的代码：

周长轮廓/凸包周长

nomeimg = 'Riscalate2/JPEG/e (5).jpg'

img = cv2.imread(nomeimg)

gray = cv2.imread(nomeimg,0)#convert grayscale adn binarize

element = cv2.getStructuringElement(cv2.MORPH_CROSS,(6,6)) 
graydilate = cv2.erode(gray, element) #imgbnbin

cv2.imshow('image',graydilate)
cv2.waitKey(0)

ret,thresh = cv2.threshold(graydilate,127,255,cv2.THRESH_BINARY_INV)   # binarize

imgbnbin = thresh
cv2.imshow('bn',thresh)
cv2.waitKey()

#element = cv2.getStructuringElement(cv2.MORPH_CROSS,(2,2))
#element = np.ones((11,11),'uint8')


contours, hierarchy = cv2.findContours(imgbnbin, cv2.RETR_TREE ,cv2.CHAIN_APPROX_SIMPLE)
print(len(contours))


# Take only biggest contour basing on area
Areacontours = list()
calcarea = 0.0
unicocnt = 0.0
for i in range (0, len(contours)):
    area = cv2.contourArea(contours[i])
    #print("area")
    #print(area)
    if (area > 90 ):  #con 90 trova i segni e togli puntini
        if (calcarea<area):
            calcarea = area
            unicocnt = contours[i]

#ROUGHNESS
perimeter = cv2.arcLength(unicocnt,True)
hull = cv2.convexHull(unicocnt,returnPoints = False)
hullperimeter = cv2.arcLength(hull,True)

print("perimeter")
print(perimeter)
print("hullperimeter")
print(hullperimeter)


roughness = perimeter/hullperimeter
print("roughness")
print(roughness)

错误：

Traceback (most recent call last):
  File "C:\Python27\nuovefeature.py", line 417, in <module>
    hullperimeter = cv2.arcLength(hull,True)
error: ..\..\..\src\opencv\modules\imgproc\src\contours.cpp:1886: error: (-215) curve.checkVector(2) >= 0 && (curve.depth() == CV_32F || curve.depth() == CV_32S)
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1 回答 1

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错误在convexHull 查找线中。

hull = cv2.convexHull(unicocnt,returnPoints = False)

默认情况下，标志returnPoints是True. 然后返回的船体包含船体点或轮廓点的像素坐标。它通常用于绘制船体，找到它的面积，周长等。

但是，如果将其指定为False，则返回的值是轮廓中这些坐标的索引。它通常用于查找凸性缺陷。

所以你的代码应该如下所示：

hull = cv2.convexHull(unicocnt)

有关示例的更详细说明，请阅读 中的convexHull 部分this article。

于 2013-02-28T17:30:50.070 回答