Question

我正在尝试对倾斜的矩形（信用卡）进行透视校正，该矩形在所有4个方向上倾斜。我可以找到它的四个角和倾斜的相应角度，但我找不到坐标的确切位置，它必须在哪里投影。我正在使用cv2.getPerspectiveTransform进行转换。

我有实际卡片的宽高比（非倾斜的卡片），我想要这样的坐标，以保持原始的宽高比。我尝试过使用边界矩形，但这会增加卡的大小。

任何帮助都将不胜感激。 test image

Bounding Rectangle

Answer 1

以下是您需要遵循的方式...

为了方便起见，我已将图像尺寸调整为较小尺寸，

enter image description here

计算源图像的四边形顶点，这里我手动找到，你可以选择边缘检测，hough线等。

  Q1=manual calculation;
  Q2=manual calculation;
  Q3=manual calculation;
  Q4=manual calculation;

通过保持纵横比来计算目标图像中的四边形顶点，这里您可以从源的四边形顶点上方获取卡的宽度，并通过乘以纵横比来计算高度。

   // compute the size of the card by keeping aspect ratio.
    double ratio=1.6;
    double cardH=sqrt((Q3.x-Q2.x)*(Q3.x-Q2.x)+(Q3.y-Q2.y)*(Q3.y-Q2.y)); //Or you can give your own height
    double cardW=ratio*cardH;
    Rect R(Q1.x,Q1.y,cardW,cardH);

现在您获得了源和目标的四边形顶点，然后应用warpPerspective。

enter image description here

您可以参考下面的C ++代码，

   //Compute quad point for edge
    Point Q1=Point2f(90,11);
    Point Q2=Point2f(596,135);
    Point Q3=Point2f(632,452);
    Point Q4=Point2f(90,513);

    // compute the size of the card by keeping aspect ratio.
    double ratio=1.6;
    double cardH=sqrt((Q3.x-Q2.x)*(Q3.x-Q2.x)+(Q3.y-Q2.y)*(Q3.y-Q2.y));//Or you can give your own height
    double cardW=ratio*cardH;
    Rect R(Q1.x,Q1.y,cardW,cardH);

    Point R1=Point2f(R.x,R.y);
    Point R2=Point2f(R.x+R.width,R.y);
    Point R3=Point2f(Point2f(R.x+R.width,R.y+R.height));
    Point R4=Point2f(Point2f(R.x,R.y+R.height));

    std::vector<Point2f> quad_pts;
    std::vector<Point2f> squre_pts;

    quad_pts.push_back(Q1);
    quad_pts.push_back(Q2);
    quad_pts.push_back(Q3);
    quad_pts.push_back(Q4);

    squre_pts.push_back(R1);
    squre_pts.push_back(R2);
    squre_pts.push_back(R3);
    squre_pts.push_back(R4);


    Mat transmtx = getPerspectiveTransform(quad_pts,squre_pts);
    int offsetSize=150;
    Mat transformed = Mat::zeros(R.height+offsetSize, R.width+offsetSize, CV_8UC3);
    warpPerspective(src, transformed, transmtx, transformed.size());

    //rectangle(src, R, Scalar(0,255,0),1,8,0);

    line(src,Q1,Q2, Scalar(0,0,255),1,CV_AA,0);
    line(src,Q2,Q3, Scalar(0,0,255),1,CV_AA,0);
    line(src,Q3,Q4, Scalar(0,0,255),1,CV_AA,0);
    line(src,Q4,Q1, Scalar(0,0,255),1,CV_AA,0);

    imshow("quadrilateral", transformed);
    imshow("src",src);
    waitKey();

Answer 2

我有一个更好的解决方案，这很简单： - 原始图像上的红色矩形和矩形的角点是源点
- 我们使用cv2.warpPerspective()将源点和目标点作为参数，并返回转换矩阵，将矩阵转换为目标图像，如图所示 - 我们在import cv2 import matplotlib.pyplot as plt import numpy as np def unwarp(img, src, dst, testing): h, w = img.shape[:2] # use cv2.getPerspectiveTransform() to get M, the transform matrix, and Minv, the inverse M = cv2.getPerspectiveTransform(src, dst) # use cv2.warpPerspective() to warp your image to a top-down view warped = cv2.warpPerspective(img, M, (w, h), flags=cv2.INTER_LINEAR) if testing: f, (ax1, ax2) = plt.subplots(1, 2, figsize=(20, 10)) f.subplots_adjust(hspace=.2, wspace=.05) ax1.imshow(img) x = [src[0][0], src[2][0], src[3][0], src[1][0], src[0][0]] y = [src[0][1], src[2][1], src[3][1], src[1][1], src[0][1]] ax1.plot(x, y, color='red', alpha=0.4, linewidth=3, solid_capstyle='round', zorder=2) ax1.set_ylim([h, 0]) ax1.set_xlim([0, w]) ax1.set_title('Original Image', fontsize=30) ax2.imshow(cv2.flip(warped, 1)) ax2.set_title('Unwarped Image', fontsize=30) plt.show() else: return warped, M im = cv2.imread("so.JPG") w, h = im.shape[0], im.shape[1] # We will first manually select the source points # we will select the destination point which will map the source points in # original image to destination points in unwarped image src = np.float32([(20, 1), (540, 130), (20, 520), (570, 450)]) dst = np.float32([(600, 0), (0, 0), (600, 531), (0, 531)]) unwarp(im, src, dst, True) cv2.imshow("so", im) cv2.waitKey(0)[![enter image description here][1]][1] cv2.destroyAllWindows()中使用此转换矩阵 - 你可以看到结果更好。你会得到一个非常漂亮的鸟瞰图像

$("#submit_click").on("click",function(){

alert('Yepeee it is working ');
});

Answer 3

我正在用Python写@Haris提供的答案。

import cv2
import math
import numpy as np
import matplotlib.pyplot as plt

img = cv2.imread('test.jpg')
rows,cols,ch = img.shape

pts1 = np.float32([[360,50],[2122,470],[2264, 1616],[328,1820]])

ratio=1.6
cardH=math.sqrt((pts1[2][0]-pts1[1][0])*(pts1[2][0]-pts1[1][0])+(pts1[2][1]-pts1[1][1])*(pts1[2][1]-pts1[1][1]))
cardW=ratio*cardH;
pts2 = np.float32([[pts1[0][0],pts1[0][1]], [pts1[0][0]+cardW, pts1[0][1]], [pts1[0][0]+cardW, pts1[0][1]+cardH], [pts1[0][0], pts1[0][1]+cardH]])

M = cv2.getPerspectiveTransform(pts1,pts2)

offsetSize=500
transformed = np.zeros((int(cardW+offsetSize), int(cardH+offsetSize)), dtype=np.uint8);
dst = cv2.warpPerspective(img, M, transformed.shape)

plt.subplot(121),plt.imshow(img),plt.title('Input')
plt.subplot(122),plt.imshow(dst),plt.title('Output')
plt.show()

使用python在OpenCV中进行透视校正

3 个答案: