Question

我正在尝试在openCV中对图片进行旋转矩形选择。

我想从这个图像开始（选择用红色突出显示）： Original

获得这个：

enter image description here

问题是我只能通过使用Mat结构找到如何提取Rect的区域。 Rect结构始终是垂直的。

我知道原始图像中矩形选区的点坐标，但我不知道如何获得一个矩阵，可以通过Rect选择我的片段，我怎么去过度计算Rect的坐标和大小。

这是否有共同的方法？

Answer 1

你走了。以下代码是一行到一行的副本 http://www.pyimagesearch.com/2014/03/24/building-pokedex-python-scraping-pokemon-sprites-step-2-6/和 https://www.pyimagesearch.com/2014/04/21/building-pokedex-python-finding-game-boy-screen-step-4-6/。代码基本上尝试使用轮廓查找给定图像中的最大矩形。

import cv2
import numpy as np

def biggestRect(image):
    ratio = image.shape[0] / 300.0
    orig = image.copy()
    image = aspectResize(image, rows = 300)

# convert the image to grayscale, blur it, and find edges 
# in the image
    gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
    gray = cv2.bilateralFilter(gray, 11, 17, 17)
    edged = cv2.Canny(gray, 30, 200)
    # find contours in the edged image, keep only the largest
    # ones, and initialize our screen contour
    (cnts, _) = cv2.findContours(edged.copy(), cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
    cnts = sorted(cnts, key = cv2.contourArea, reverse = True)[:10]
    screenCnt = None
    # loop over our contours
    for c in cnts:
        # approximate the contour
        peri = cv2.arcLength(c, True)
        approx = cv2.approxPolyDP(c, 0.02 * peri, True)

        # if our approximated contour has four points, then
        # we can assume that we have found our screen
        if len(approx) == 4:
            screenCnt = approx
            break

    pts = screenCnt.reshape(4, 2)
    rect = np.zeros((4, 2), dtype = "float32")

    # the top-left point has the smallest sum whereas the
    # bottom-right has the largest sum
    s = pts.sum(axis = 1)
    rect[0] = pts[np.argmin(s)]
    rect[2] = pts[np.argmax(s)]

    # compute the difference between the points -- the top-right
   # will have the minumum difference and the bottom-left will
    # have the maximum difference
    diff = np.diff(pts, axis = 1)
    rect[1] = pts[np.argmin(diff)]
    rect[3] = pts[np.argmax(diff)]

    # multiply the rectangle by the original ratio
    rect *= ratio
    # now that we have our rectangle of points, let's compute
    # the width of our new image
    (tl, tr, br, bl) = rect
    widthA = np.sqrt(((br[0] - bl[0]) ** 2) + ((br[0] - bl[0]) ** 2))
    widthB = np.sqrt(((tr[0] - tl[0]) ** 2) + ((tr[0] - tl[0]) ** 2))

    # ...and now for the height of our new image
    heightA = np.sqrt(((tr[1] - br[1]) ** 2) + ((tr[1] - br[1]) ** 2))
    heightB = np.sqrt(((tl[1] - bl[1]) ** 2) + ((tl[1] - bl[1]) ** 2))

    # take the maximum of the width and height values to reach
    # our final dimensions
    maxWidth = max(int(widthA), int(widthB))
    maxHeight = max(int(heightA), int(heightB))

    # construct our destination points which will be used to
    # map the screen to a top-down, "birds eye" view
    dst = np.array([
            [0, 0],
            [maxWidth - 1, 0],
            [maxWidth - 1, maxHeight - 1],
            [0, maxHeight - 1]], dtype = "float32")

    # calculate the perspective transform matrix and warp
    # the perspective to grab the screen
    M = cv2.getPerspectiveTransform(rect, dst)
    warp = cv2.warpPerspective(orig, M, (maxWidth, maxHeight))

    return warp


def aspectResize(img,rows=None,cols=None):
    if rows==None and cols ==None:
        return img
    if rows==None:
        ratio=float(img.shape[1])/cols
    if cols==None:
        ratio=float(img.shape[0])/rows
    img=cv2.resize(img,(int(img.shape[1]/ratio),int(img.shape[0]/ratio)))
    return img

我在您的图像上尝试了上述代码并获得了所需的输出。

如何在OpenCV中进行对角线选择

1 个答案: