4点迫切变换失败 [英] 4 point persective transform failure

问题描述

我一直在尝试进行4点透视变换，以便开始执行一些OCR.

I've been trying to do a 4 point perspective transform in order to start doing some OCR.

从下图开始，我可以检测到车牌

Starting with the following image I can detect the number plate

将其裁剪为绿色框作为边界框，红色点为我要平方的矩形的角.

and crop it out with the green box being the bounding box and the red dots being the corners of the rectangle I want to square up.

这是变换的输出.

乍一看，它看起来像是从里到外进行变换(将零件放在一边而不是在点之间).

At a first look it seams to have done the transform inside out (taking the parts either side rather than between the points).

我正在使用imutils包进行转换，并从此和

I'm using the imutils package to do the transform and working from this and this as a guide. I'm sure it's something relatively simple I'm missing.

#!/usr/bin/python
import numpy as np
import cv2
import imutils
from imutils import contours
from imutils.perspective import four_point_transform

img = cv2.imread("sample7-smaller.jpg")
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
blurred = cv2.bilateralFilter(gray,15,75,75)
v = np.median(blurred)
lower = int(max(0, (1.0 - 0.33) * v))
upper = int(min(255, (1.0 + 0.33) * v))
edged = cv2.Canny(blurred, lower, upper, 255)

conts = cv2.findContours(edged.copy(), cv2.RETR_EXTERNAL,
    cv2.CHAIN_APPROX_SIMPLE)
conts = conts[0] if imutils.is_cv2() else conts[1]
conts = sorted(conts, key=cv2.contourArea, reverse=True)

for cnt in conts:
    approx = cv2.approxPolyDP(cnt,0.01*cv2.arcLength(cnt,True),True)
    if len(approx) == 4:
        x,y,w,h = cv2.boundingRect(cnt)
        cv2.rectangle(img,(x,y),(x+w,y+h),(0,255,0),2)
        for i in approx:
            cv2.circle(img,(i[0][0], i[0][1]),2,(0,0,255), thickness=4)
        warped = four_point_transform(img, approx.reshape(4,2))
        cv2.imshow("crop",img[y:y+h,x:x+w])
        cv2.imshow("warped", warped)
        cv2.waitKey(0)

推荐答案

我建议您使用OpenCV

I would recommend you to use the OpenCV Perspective Transform method, to get the desired results, as per the given image:

首先标记src点的位置:

First mark the position of src points:

src_pts = np.array([[8, 136], [415, 52], [420, 152], [14, 244]], dtype=np.float32)

并且假设您希望将此车牌以50x200形状的矩阵拟合，那么目标点将是:

And suppose you want to fit this number plate in a matrix of shape 50x200, so destination points would be:

dst_pts = np.array([[0, 0],   [200, 0],  [200, 50], [0, 50]], dtype=np.float32)

将透视图变换矩阵"查找为:

Find the perspective Transform Matrix as :

M = cv2.getPerspectiveTransform(src_pts, dst_pts)
warp = cv2.warpPerspective(img, M, (200, 50))

由于您不想对板的最终宽度，高度进行硬编码，因此，为了使计算更加灵活，您可以从4个标记点计算板的宽度和高度，如下所示:

As you didn't wanted to hard code the final width, height of plate, So in order to make the calculations more flexible you can calculate the width and height of the plate from the 4 marker points as:

def get_euler_distance(pt1, pt2):
    return ((pt1[0] - pt2[0])**2 + (pt1[1] - pt2[1])**2)**0.5

src_pts = np.array([[8, 136], [415, 52], [420, 152], [14, 244]], dtype=np.float32)

width = get_euler_distance(src_pts[0][0], src_pts[0][1])
height = get_euler_distance(src_pts[0][0], src_pts[0][3])

dst_pts = np.array([[0, 0],   [width, 0],  [width, height], [0, height]], dtype=np.float32)

M = cv2.getPerspectiveTransform(src_pts, dst_pts)
warp = cv2.warpPerspective(img, M, (width, height))

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