OpenCV：形状检测

Question

我有不同形状的.jpg文件。我试图检测每个图形的形状，并将其打印在旁边。我正在使用下面的代码创建和绘制轮廓

contours, _ = cv2.findContours(threshold, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
for con in contours:
    approx = cv2.approxPolyDP(con, 0.01*cv2.arcLength(con, True), True)
    cv2.drawContours(img, [approx], 0, (0,0,0), 5)

if len(approx) == 4:
        cv2.putText(img, "Quadrilateral", (x,y), font, 1, (0))

现在我已经计算出了cv2.approxPolyDp()连接轮廓点并创建了一个确定形状的边界，我如何确定确切的形状，即它是正方形还是矩形？如上述代码所示，如果 len(approx) == 4

这是图像：

Answer 1

可以使用aspect ratio来区分正方形和矩形。通过观察，正方形的长度和宽度相等，而矩形的一侧更长。同样的逻辑也可用于识别圆形与椭圆形。结果如下：

import cv2

def detect_shape(c):
    shape = ""
    peri = cv2.arcLength(c, True)
    approx = cv2.approxPolyDP(c, 0.04 * peri, True)

    # Triangle
    if len(approx) == 3:
        shape = "triangle"

    # Square or rectangle
    elif len(approx) == 4:
        (x, y, w, h) = cv2.boundingRect(approx)
        ar = w / float(h)

        # A square will have an aspect ratio that is approximately
        # equal to one, otherwise, the shape is a rectangle
        shape = "square" if ar >= 0.95 and ar <= 1.05 else "rectangle"

    # Pentagon
    elif len(approx) == 5:
        shape = "pentagon"

    # Otherwise assume as circle or oval
    else:
        (x, y, w, h) = cv2.boundingRect(approx)
        ar = w / float(h)
        shape = "circle" if ar >= 0.95 and ar <= 1.05 else "oval"

    return shape

image = cv2.imread('1.png')
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
thresh = cv2.adaptiveThreshold(gray,255,cv2.ADAPTIVE_THRESH_GAUSSIAN_C, cv2.THRESH_BINARY_INV,51,7)

cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
for c in cnts:
    shape = detect_shape(c)
    x,y,w,h = cv2.boundingRect(c)
    cv2.putText(image, shape, (x, y - 5), cv2.FONT_HERSHEY_SIMPLEX, 0.9, (36,255,12), 2)

cv2.imshow('thresh', thresh)
cv2.imshow('image', image)
cv2.waitKey()