计算正方形周围的框

Question

我想检测图像中的正方形，如下所示：

我想通过在正方形角周围绘制一种 3 维 框来突出正方形，如下图所示：

我如何精确计算所有线坐标以便稍后绘制“3 维”框？ (给出的是黑色方块的4个角点)

注意：您可以在此处找到我想要实现的目标的视频https://www.youtube.com/watch?v=oSq9V2b5AZ8。

如果你愿意帮助我，如果你分享一些代码行如何计算 4 个缺失点以及如何知道哪些点匹配在一起以从 startPoint(x,y) 画一条线，我会非常高兴到端点（x，y）。例如 js 中的某些行会有很大帮助 :)

Answer 1

首先找到轮廓，然后 select 极值点。然后指定新的 3D 角并使用 cv2.line().

绘制它们

示例：

import cv2
import numpy as np
import imutils

image = cv2.imread('3d2.png')

gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
gray = cv2.GaussianBlur(gray, (5, 5), 0)
thresh = cv2.threshold(gray, 190, 255, cv2.THRESH_BINARY)[1]
cv2.bitwise_not(thresh, thresh)

cnts = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL,
        cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if imutils.is_cv2() else cnts[1]
c = max(cnts, key=cv2.contourArea)

extLeft = tuple(c[c[:, :, 0].argmin()][0])
extRight = tuple(c[c[:, :, 0].argmax()][0])
extTop = tuple(c[c[:, :, 1].argmin()][0])
extBot = tuple(c[c[:, :, 1].argmax()][0])

leftx = int(extLeft[0])
lefty = int(extLeft[1]) - 90

rightx = int(extRight[0])
righty = int(extRight[1]) -90

topx = int(extTop[0])
topy = int(extTop[1]) -90

bottomx = int(extBot[0])
bottomy = int(extBot[1]) -90

leftc = (leftx, lefty)
rightc = (rightx, righty)
topc = (topx, topy)
bottomc = (bottomx, bottomy)

line = cv2.line(image, extLeft, leftc, (0,255,0), 2)
line = cv2.line(image, extRight, rightc, (0,255,0), 2)
line = cv2.line(image, extTop, topc, (0,255,0), 2)
line = cv2.line(image, extBot, bottomc, (0,255,0), 2)
line = cv2.line(image, bottomc, leftc, (0,255,0), 2)
line = cv2.line(image, rightc, topc, (0,255,0), 2)
line = cv2.line(image, leftc, topc, (0,255,0), 2)
line = cv2.line(image, rightc, topc, (0,255,0), 2)
line = cv2.line(image, bottomc, rightc, (0,255,0), 2)
cv2.drawContours(image, [c], -1, (0,255,0), 2)

cv2.imshow("Image", image)
cv2.imwrite('3Dbox1.png', image)
cv2.waitKey(0)

结果：

您当然可以按照自己的意愿制作新的点（例如，如果您想要与图片中的相同，请给出 x+50 和 y-150）：

编辑：

要使框旋转，请尝试使用可以从 cv2.minAreaRect() 函数获得的角度，如下所示：

import cv2
import numpy as np
import imutils

cap = cv2.VideoCapture(0)

while True:

    try:
        ret, image = cap.read()

        gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
        gray = cv2.GaussianBlur(gray, (5, 5), 0)
        thresh = cv2.threshold(gray, 190, 255, cv2.THRESH_BINARY)[1]
        #cv2.bitwise_not(thresh, thresh)

        cnts = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL,
                cv2.CHAIN_APPROX_SIMPLE)
        cnts = cnts[0] if imutils.is_cv2() else cnts[1]
        c = max(cnts, key=cv2.contourArea)

        rect = cv2.minAreaRect(c)
        angle = rect[2]


        extLeft = tuple(c[c[:, :, 0].argmin()][0])
        extRight = tuple(c[c[:, :, 0].argmax()][0])
        extTop = tuple(c[c[:, :, 1].argmin()][0])
        extBot = tuple(c[c[:, :, 1].argmax()][0])

        if angle < 0:

            leftx = int(extLeft[0]) - int(angle)
            lefty = int(extLeft[1]) - 50 + int(angle)

            rightx = int(extRight[0]) - int(angle)
            righty = int(extRight[1]) -50 + int(angle)

            topx = int(extTop[0]) - int(angle)
            topy = int(extTop[1]) -50 + int(angle) 

            bottomx = int(extBot[0]) - int(angle)
            bottomy = int(extBot[1]) -50 + int(angle)

            leftc = (leftx, lefty)
            rightc = (rightx, righty)
            topc = (topx, topy)
            bottomc = (bottomx, bottomy)

            line = cv2.line(image, extLeft, leftc, (0,255,0), 2)
            line = cv2.line(image, extRight, rightc, (0,255,0), 2)
            line = cv2.line(image, extTop, topc, (0,255,0), 2)
            line = cv2.line(image, extBot, bottomc, (0,255,0), 2)
            line = cv2.line(image, bottomc, leftc, (0,255,0), 2)
            line = cv2.line(image, rightc, topc, (0,255,0), 2)
            line = cv2.line(image, leftc, topc, (0,255,0), 2)
            line = cv2.line(image, rightc, topc, (0,255,0), 2)
            line = cv2.line(image, bottomc, rightc, (0,255,0), 2)
            cv2.drawContours(image, [c], -1, (0,255,0), 2)

        elif angle > 0:

            leftx = int(extLeft[0]) + int(angle)
            lefty = int(extLeft[1]) + 50 + int(angle)

            rightx = int(extRight[0]) + int(angle)
            righty = int(extRight[1]) +50 + int(angle)

            topx = int(extTop[0]) + int(angle)
            topy = int(extTop[1]) +50 + int(angle) 

            bottomx = int(extBot[0]) + int(angle)
            bottomy = int(extBot[1]) +50 + int(angle)

            leftc = (leftx, lefty)
            rightc = (rightx, righty)
            topc = (topx, topy)
            bottomc = (bottomx, bottomy)

            line = cv2.line(image, extLeft, leftc, (0,255,0), 2)
            line = cv2.line(image, extRight, rightc, (0,255,0), 2)
            line = cv2.line(image, extTop, topc, (0,255,0), 2)
            line = cv2.line(image, extBot, bottomc, (0,255,0), 2)
            line = cv2.line(image, bottomc, leftc, (0,255,0), 2)
            line = cv2.line(image, rightc, topc, (0,255,0), 2)
            line = cv2.line(image, leftc, topc, (0,255,0), 2)
            line = cv2.line(image, rightc, topc, (0,255,0), 2)
            line = cv2.line(image, bottomc, rightc, (0,255,0), 2)
            cv2.drawContours(image, [c], -1, (0,255,0), 2)

    except:
        pass

    cv2.imshow("Image", image)
    if cv2.waitKey(1) & 0xFF == ord('q'):
        break

Answer 2

如果您的照片是等轴测视图，这将简化为以下问题：

您想找到对应于立方体的三个正交边的三个向量的测量值。

边 1：（X1，Y1，Z1）
Edge2: (X2, Y2, Z2)
Edge3: (X3, Y3, Z3)

根据您的图像，您可以测量其中两个向量的 X 和 Y 值，留下 5 个未知值。

由于所有三个边都是正交的，因此您也知道它们的点积为零。

最后，由于您处理的是立方体，所以您知道每个向量都具有相同的大小。

这给了你五个方程来求解五个未知变量，它可以唯一标识一个解。

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