如何扭曲矩形 object 以适应其较大的边界框

Question

给定这张图片：

我想让它旋转和拉伸以完全适合最大矩形框外侧没有空白的边界框。它还应该考虑更糟糕的透视情况，就像我稍后列出的链接一样。

基本上，虽然不明显，但矩形旋转了一点，我想修复这个变形。

但是，我在尝试检索轮廓的四个点时遇到错误。我已经确定并利用轮廓近似来隔离并仅获得相关的外观轮廓，正如您在图像中看到的那样，它是成功的，尽管我不能在其上使用透视扭曲。

我已经尝试过这里的链接：

How to straighten a rotated rectangle area of an image using opencv in python?
https://www.pyimagesearch.com/2014/05/05/building-pokedex-python-opencv-perspective-warping-step-5-6/
https://www.pyimagesearch.com/2014/09/01/build-kick-ass-mobile-document-scanner-just-5-minutes/

紧随其后，只是做了一些小的修改（比如不缩小图像然后再放大）和不同的输入图像。

评论里有一个reader遇到了类似的错误，不过作者只是说用contour approximation。我这样做了，但我仍然收到同样的错误。

我已经检索到轮廓（连同它的边界框，是前面说明的图像），并使用此代码尝试透视扭曲：

def warp_perspective(cnt):
    # reshape cnt to get tl, tr, br, bl points
    pts = cnt.reshape(4, 2)
    rect = np.zeros((4, 2), dtype="float32")

    s = pts.sum(axis=1)
    rect[0] = pts[np.argmin(s)]
    rect[2] = pts[np.argmin(s)]

    diff = np.diff(pts, axis=1)
    rect[1] = pts[np.argmin(diff)]
    rect[2] = pts[np.argmax(diff)]

    # solve for the width of the image
    (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))

    # solve for the height of the image
    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))

    # get the final dimensions
    maxWidth = max(int(widthA), int(widthB))
    maxHeight = max(int(heightA), int(heightB))

    # construct the dst image
    dst = np.array([
        [0, 0],
        [maxWidth - 1, 0],
        [maxWidth - 1, maxHeight - 1],
        [0, maxHeight - 1]], dtype="float32")

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

    cv2.imshow("warped", warp)

    return warp

该函数接受 cnt 作为单个轮廓。

在运行我运行进入我之前提到的这个错误：

in warp_perspective
    pts = cnt.reshape(4, 2)
ValueError: cannot reshape array of size 2090 into shape (4,2)

我完全不明白。我已经成功地分离并检索了正确的轮廓和边界框，我唯一不同的是跳过了降尺度..

Answer 1

试试这个方法：

将图像转换为灰度并使用双边滤波器进行模糊处理
大津的门槛
寻找轮廓
对最大方形等高线进行等高线逼近
透视变换和旋转

结果

import cv2
import numpy as np
import imutils

def perspective_transform(image, corners):
    def order_corner_points(corners):
        # Separate corners into individual points
        # Index 0 - top-right
        #       1 - top-left
        #       2 - bottom-left
        #       3 - bottom-right
        corners = [(corner[0][0], corner[0][1]) for corner in corners]
        top_r, top_l, bottom_l, bottom_r = corners[0], corners[1], corners[2], corners[3]
        return (top_l, top_r, bottom_r, bottom_l)

    # Order points in clockwise order
    ordered_corners = order_corner_points(corners)
    top_l, top_r, bottom_r, bottom_l = ordered_corners

    # Determine width of new image which is the max distance between 
    # (bottom right and bottom left) or (top right and top left) x-coordinates
    width_A = np.sqrt(((bottom_r[0] - bottom_l[0]) ** 2) + ((bottom_r[1] - bottom_l[1]) ** 2))
    width_B = np.sqrt(((top_r[0] - top_l[0]) ** 2) + ((top_r[1] - top_l[1]) ** 2))
    width = max(int(width_A), int(width_B))

    # Determine height of new image which is the max distance between 
    # (top right and bottom right) or (top left and bottom left) y-coordinates
    height_A = np.sqrt(((top_r[0] - bottom_r[0]) ** 2) + ((top_r[1] - bottom_r[1]) ** 2))
    height_B = np.sqrt(((top_l[0] - bottom_l[0]) ** 2) + ((top_l[1] - bottom_l[1]) ** 2))
    height = max(int(height_A), int(height_B))

    # Construct new points to obtain top-down view of image in 
    # top_r, top_l, bottom_l, bottom_r order
    dimensions = np.array([[0, 0], [width - 1, 0], [width - 1, height - 1], 
                    [0, height - 1]], dtype = "float32")

    # Convert to Numpy format
    ordered_corners = np.array(ordered_corners, dtype="float32")

    # Find perspective transform matrix
    matrix = cv2.getPerspectiveTransform(ordered_corners, dimensions)

    # Transform the image
    transformed = cv2.warpPerspective(image, matrix, (width, height))

    # Rotate and return the result
    return imutils.rotate_bound(transformed, angle=-90)

image = cv2.imread('1.png')
original = image.copy()
blur = cv2.bilateralFilter(image,9,75,75)
gray = cv2.cvtColor(blur, cv2.COLOR_BGR2GRAY)
thresh = cv2.threshold(gray,0,255, cv2.THRESH_BINARY_INV + cv2.THRESH_OTSU)[1]

cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]

for c in cnts:
    peri = cv2.arcLength(c, True)
    approx = cv2.approxPolyDP(c, 0.015 * peri, True)

    if len(approx) == 4:
        cv2.drawContours(image,[c], 0, (36,255,12), 3)
        transformed = perspective_transform(original, approx)

cv2.imshow('thresh', thresh)
cv2.imshow('image', image)
cv2.imshow('transformed', transformed)
cv2.waitKey()

如何扭曲矩形 object 以适应其较大的边界框

How to warp a rectangular object to fit its larger bounding box

python

opencv

image

image-processing

contour