如何使用opencv检测图像中的矩形(白板)? [英] How to detect a rectangle (whiteboard) in an image using opencv?

问题描述

我有以下图片.我想检测和透视变换矩形白板.

我想检测这 4 个边界/角并对其应用透视变换.看看下面的图片:

我无法检测矩形的边界.这是我尝试过的:

导入 cv2, os将 numpy 导入为 np从 google.colab.patches 导入 cv2_imshowimage = cv2.imread(img.jpg")orig1 = image.copy()# 1) 灰度图像灰色 = cv2.cvtColor(图像，cv2.COLOR_BGR2GRAY)# cv2_imshow(灰色)# 2) 侵蚀内核 = np.ones((5, 5), np.uint8)侵蚀= cv2.erode(灰色，内核，迭代= 1)# cv2_imshow(侵蚀)# 3)阈值(OTSU)模糊 = cv2.GaussianBlur(erosion, (5,5),0)ret3, thresh = cv2.threshold(blur,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)# cv2_imshow(thresh)# 4) 轮廓复制 = 脱粒；原点 = 图像；cnts = cv2.findContours(复制，cv2.RETR_EXTERNAL，cv2.CHAIN_APPROX_SIMPLE)cnts = cnts[0] 如果 len(cnts) == 2 else cnts[1]面积 = -1;c1 = 0对于 cnts 中的 c:如果面积＜cv2.contourArea(c):面积 = cv2.contourArea(c)c1 = ccv2.drawContours(orig,[c1], 0, (0,255,0), 3)epsilon = 0.09 * cv2.arcLength(c1,True)近似 = cv2.approxPolyDP(c1,epsilon,True)如果 len(approx) != 4:# 那么这里会失败.经过酷 = []对于范围内的 i (0, len(approx)):cood.append([approx[i][0][0],approx[i][0][1]])# 5) 透视变换定义重新排序(myPoints):myPoints = np.array(myPoints).reshape((4, 2))myPointsNew = np.zeros((4, 1, 2), dtype=np.int32)添加 = myPoints.sum(1)myPointsNew[0] = myPoints[np.argmin(add)]myPointsNew[3] =myPoints[np.argmax(add)]diff = np.diff(myPoints, 轴=1)myPointsNew[1] =myPoints[np.argmin(diff)]myPointsNew[2] = myPoints[np.argmax(diff)]返回我的新积分pts1 = np.float32(reorder(cood))w = 1000；h = 1000；m1 = 1000；平方米 = 1000pts2 = np.float32([[0, 0], [w, 0], [0, h], [w, h]])矩阵 = cv2.getPerspectiveTransform(pts1, pts2)结果 = cv2.warpPerspective(orig1, matrix, (m1, m2))cv2_imshow(结果)

我也经历过

# 寻找轮廓并应用透视尝试:复制 = thresh.copy();原件 = img.copy()cnts = cv2.findContours(复制，cv2.RETR_EXTERNAL，cv2.CHAIN_APPROX_SIMPLE)cnts = cnts[0] 如果 len(cnts) == 2 else cnts[1]面积 = -1;c1 = 0对于 cnts 中的 c:epsilon = 0.01 * cv2.arcLength(c,True)近似 = cv2.approxPolyDP(c,epsilon,True)如果 len(approx) == 4 并且 area 

这是扭曲的图像:

I have the following image. I want to detect and perspective transform the rectangular whiteboard.

I want to detect these 4 boundaries/corners and apply a perspective transformation to it. Have a look at the below image:

I am not able to detect the boundaries of the rectangle. Here's what I have tried:

import cv2, os
import numpy as np
from google.colab.patches import cv2_imshow


image = cv2.imread("img.jpg")
orig1 = image.copy()
# 1) Grayscale image
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
# cv2_imshow(gray)
# 2) Erosion
kernel = np.ones((5, 5), np.uint8)
erosion = cv2.erode(gray, kernel, iterations = 1)
# cv2_imshow(erosion)

# 3) Thresholding (OTSU)
blur = cv2.GaussianBlur(erosion, (5,5),0)
ret3, thresh = cv2.threshold(blur,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
# cv2_imshow(thresh)

# 4) Contours
copy = thresh; orig = image; 
cnts = cv2.findContours(copy, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
area = -1; c1 = 0
for c in cnts:
    if area < cv2.contourArea(c):
        area = cv2.contourArea(c)
        c1 = c
cv2.drawContours(orig,[c1], 0, (0,255,0), 3)    

epsilon = 0.09 * cv2.arcLength(c1,True)
approx = cv2.approxPolyDP(c1,epsilon,True)

if len(approx) != 4:
    # Then it will fail here.
    pass 
cood = []
for i in range(0, len(approx)):
    cood.append([approx[i][0][0], approx[i][0][1]])

# 5) Perspective Transformation
def reorder(myPoints):
    myPoints = np.array(myPoints).reshape((4, 2))
    myPointsNew = np.zeros((4, 1, 2), dtype=np.int32)
    add = myPoints.sum(1)
    myPointsNew[0] = myPoints[np.argmin(add)]
    myPointsNew[3] =myPoints[np.argmax(add)]
    diff = np.diff(myPoints, axis=1)
    myPointsNew[1] =myPoints[np.argmin(diff)]
    myPointsNew[2] = myPoints[np.argmax(diff)] 
    return myPointsNew

pts1 = np.float32(reorder(cood))
w = 1000; h = 1000; m1 = 1000; m2 = 1000
pts2 = np.float32([[0, 0], [w, 0], [0, h], [w, h]])
matrix = cv2.getPerspectiveTransform(pts1, pts2)
result = cv2.warpPerspective(orig1, matrix, (m1, m2)) 
cv2_imshow(result)

I have also gone through Microsoft's research, but not sure how to implement it. I am not able to detect and perspective transform the board. It would be great if anyone of you can help me out. Also, do let me know if my question requires more details.

解决方案

I manage to get the 4 coordinates of the whiteboard. I have used adaptive thresholding to detect the edges rather than canny-edge detection, not sure whether the methodology is correct or not, but it is giving the required results. Here's the code for the same.

import ...
img = cv2.imread("path-to-image")
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
thresh = cv2.adaptiveThreshold(gray, 255, cv2.ADAPTIVE_THRESH_MEAN_C, cv2.THRESH_BINARY, 199, 5)
cv2_imshow(thresh)

# finding contours and applying perspective
try:
    copy = thresh.copy(); orig = img.copy()
    cnts = cv2.findContours(copy, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
    cnts = cnts[0] if len(cnts) == 2 else cnts[1]
    area = -1; c1 = 0

    for c in cnts:
        epsilon = 0.01 * cv2.arcLength(c,True)
        approx = cv2.approxPolyDP(c,epsilon,True)
        if len(approx) == 4 and area < cv2.contourArea(c):
            area = cv2.contourArea(c)
            c1 = c; approx1 = approx

    warped = four_point_transform(orig, approx1.reshape(4, 2))
    cv2_imshow(warped)
except:
    print("Image cannot be transformed!!\n")

# four point transform
def order_points(pts):
    # https://www.pyimagesearch.com/2016/03/21/ordering-coordinates-clockwise-with-python-and-opencv/
    xSorted = pts[np.argsort(pts[:, 0]), :]
    leftMost = xSorted[:2, :]
    rightMost = xSorted[2:, :]
    leftMost = leftMost[np.argsort(leftMost[:, 1]), :]
    (tl, bl) = leftMost
    D = dist.cdist(tl[np.newaxis], rightMost, "euclidean")[0]
    (br, tr) = rightMost[np.argsort(D)[::-1], :]
    return np.array([tl, tr, br, bl], dtype="float32")

def four_point_transform(image, pts):
    # https://www.pyimagesearch.com/2014/08/25/4-point-opencv-getperspective-transform-example/
    rect = order_points(pts)
    (tl, tr, br, bl) = rect  
    widthA = np.sqrt(((br[0] - bl[0]) ** 2) + ((br[1] - bl[1]) ** 2))
    widthB = np.sqrt(((tr[0] - tl[0]) ** 2) + ((tr[1] - tl[1]) ** 2))
    maxWidth = max(int(widthA), int(widthB))
    heightA = np.sqrt(((tr[0] - br[0]) ** 2) + ((tr[1] - br[1]) ** 2))
    heightB = np.sqrt(((tl[0] - bl[0]) ** 2) + ((tl[1] - bl[1]) ** 2))
    maxHeight = max(int(heightA), int(heightB))
    dst = np.array([
        [0, 0],
        [maxWidth - 1, 0],
        [maxWidth - 1, maxHeight - 1],
        [0, maxHeight - 1]], dtype = "float32")
    M = cv2.getPerspectiveTransform(rect, dst)
    warped = cv2.warpPerspective(image, M, (maxWidth, maxHeight))
    return warped

Here's the warped image:

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