python-opencv圆环变矩形矩形变圆环

具体数学原理就很简单了,无非就是坐标变换,从x,y或者r,theta变换一下,cx+r*sin(theta)和cy-r*cos(theta)

当然由于采样原因,导致恢复得结果里有采样不到的点。

import cv2
import numpy as np
from PIL import Image
import math

def get_huan_by_circle(img,circle_center,radius,radius_width):
    black_img = np.zeros((radius_width,int(2*radius*math.pi),3),dtype='uint8')
    for row in range(0,black_img.shape[0]):
        for col in range(0,black_img.shape[1]):
            theta = math.pi*2/black_img.shape[1]*(col+1)#+origin_theta
            rho = radius-row-1
            p_x = int(circle_center[0] + rho*math.sin(theta)+0.5)-1
            p_y = int(circle_center[1] - rho*math.cos(theta)+0.5)-1
            
            black_img[row,col,:] = img[p_y,p_x,:]
    return black_img

img = cv2.imread('timg.jpg')
get_huan_by_circle(img,(305,305),305,131)

def get_circle_by_huan(img):
    cv2.imshow('bk',img)
    cv2.waitKey()
    h,w,_ = img.shape
    radius = w/(math.pi*2)
    black_img = np.zeros((int(2*radius)+1,int(2*radius)+1,3),dtype='uint8')
    circle_center = ((int(2*radius)+1)//2,(int(2*radius)+1)//2)
    for row in range(0,img.shape[0]):
        for col in range(0,img.shape[1]):
            rho = radius-row-1
            theta = (col+1)*(math.pi*2)/img.shape[1]#+origin_theta
            p_x = int(circle_center[0] + rho*math.sin(theta)+0.5)
            p_y = int(circle_center[1] - rho*math.cos(theta)-0.5)
            black_img[p_y,p_x,:] = img[row,col,:]
           # print(col)
    black_img = cv2.blur(black_img,(3,3))
    return black_img

img = cv2.imread('bk.jpg')
get_circle_by_huan(img)

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转载自blog.csdn.net/wi162yyxq/article/details/106980593