计算正方形周围的框

Question

我想检测如下图像中的正方形：

我想通过在正方形角周围绘制一种 3 维 框来突出显示正方形，如下图所示：

如何准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四个角点）

注意：你可以在这里https://www.youtube.com/watch?v=oSq9V2b5AZ8找到我想要实现的视频。

如果您愿意帮助我，如果您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绘制从 startPoint(x,y) 到 endPoint 的线，我将非常高兴(x,y)。例如 js 中的一些行会有很大帮助 :)

Answer 1

首先找到你的轮廓，然后选择极值点。然后指定新的 3D 角并使用 cv2.line() 绘制它们。

例子：

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：

您可以按照自己的意愿提出新的观点（例如，如果您希望与图片中的相同，则给出 x+50 和 y-150）：

编辑：

要使盒子旋转，请尝试使用您可以从 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

如果您的照片是等距视图，则这会简化为以下问题：

您想要找到对应于立方体三个正交边的三个向量的测量值。

Edge1：（X1，Y1，Z1）
边 2：（X2，Y2，Z2）
Edge3：（X3，Y3，Z3）

从您的图像中，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，由于您正在处理一个立方体，您知道每个向量具有相同的大小。

这为您提供了五个方程来求解五个未知变量，它们可以唯一地标识一个解。