Opencv学习笔记 透视变换(perspective transform)

        拉伸、收缩、扭曲、旋转是图像的几何变换，在三维视觉技术中大量应用到这些变换，又分为仿射变换和透视变换。仿射变换通常用单应性建模，利用cvWarpAffine解决密集映射，用cvTransform解决稀疏映射。仿射变换可以将矩形转换成平行四边形，它可以将矩形的边压扁但必须保持边是平行的，也可以将矩形旋转或者按比例变化。透视变换提供了更大的灵活性，一个透视变换可以将矩阵转变成梯形。当然，平行四边形也是梯形，所以仿射变换是透视变换的子集。

        opencv中的函数主要是：

        对图像进行透视变换

        void cvWarpPerspective( const CvArr* src, CvArr* dst,const CvMat* map_matrix,

                       int flags=CV_INTER_LINEAR+CV_WARP_FILL_OUTLIERS, CvScalar fillval=cvScalarAll(0) );

        由四对点计算透射变换

        CvMat* cvGetPerspectiveTransform( const CvPoint2D32f*src, const CvPoint2D32f* dst, CvMat*map_matrix );

        透视变换(Perspective Transformation)是将成像投影到一个新的视平面(Viewing Plane)，也称作投影映射(Projective Mapping)。如图1，通过透视变换ABC变换到A'B'C'。

        参考代码如下：

# 透视变换
# import the necessary packages
import numpy as np
import argparse
import cv2
def order_points(pts):
	# initialzie a list of coordinates that will be ordered
	# such that the first entry in the list is the top-left,
	# the second entry is the top-right, the third is the
	# bottom-right, and the fourth is the bottom-left
	rect = np.zeros((4, 2), dtype = "float32")
	# the top-left point will have the smallest sum, whereas
	# the bottom-right point will have the largest sum
	s = pts.sum(axis = 1)
	rect[0] = pts[np.argmin(s)]
	rect[2] = pts[np.argmax(s)]
	# now, compute the difference between the points, the
	# top-right point will have the smallest difference,
	# whereas the bottom-left will have the largest difference
	diff = np.diff(pts, axis = 1)
	rect[1] = pts[np.argmin(diff)]
	rect[3] = pts[np.argmax(diff)]
	# return the ordered coordinates
	return rect

def four_point_transform(image, pts):
	# obtain a consistent order of the points and unpack them
	# individually
	rect = order_points(pts)
	(tl, tr, br, bl) = rect
	# compute the width of the new image, which will be the
	# maximum distance between bottom-right and bottom-left
	# x-coordiates or the top-right and top-left x-coordinates
	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))
	# compute the height of the new image, which will be the
	# maximum distance between the top-right and bottom-right
	# y-coordinates or the top-left and bottom-left y-coordinates
	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))
	# now that we have the dimensions of the new image, construct
	# the set of destination points to obtain a "birds eye view",
	# (i.e. top-down view) of the image, again specifying points
	# in the top-left, top-right, bottom-right, and bottom-left
	# order
	dst = np.array([
		[0, 0],
		[maxWidth - 1, 0],
		[maxWidth - 1, maxHeight - 1],
		[0, maxHeight - 1]], dtype = "float32")
	# compute the perspective transform matrix and then apply it
	M = cv2.getPerspectiveTransform(rect, dst)
	warped = cv2.warpPerspective(image, M, (maxWidth, maxHeight))
	# return the warped image
	return warped


# load the image and grab the source coordinates (i.e. the list of
# of (x, y) points)
# NOTE: using the 'eval' function is bad form, but for this example
# let's just roll with it -- in future posts I'll show you how to
# automatically determine the coordinates without pre-supplying them
image = cv2.imread("C:/Users/zyh/Desktop/20201004114816.png")
pts = np.array([[73, 239], [356, 117], [475, 265], [187, 443]], dtype = "float32")
# apply the four point tranform to obtain a "birds eye view" of
# the image
warped = four_point_transform(image, pts)
# show the original and warped images
cv2.imshow("Original", image)
cv2.imshow("Warped", warped)
cv2.waitKey(0)

原图

变换后的图

        参考如下： 

        perspective transform（透视变换）的实现过程_一只安静的大白的博客-CSDN博客_perspectivetransform

        【OpenCV3】透视变换——cv::getPerspectiveTransform()与cv::warpPerspective()详解 - 程序员大本营