图像子部分的扭曲透视变换不对齐
Warp Perspective transform of sub sections of image do not align
我想扭曲图像的部分以将其投影到不均匀的表面上。最终,我想像 HERE, kinda like was is done in HERE from THIS 项目那样扭曲图像。
我的问题是当我将变换应用到图像的每个部分时,事情就是不对齐
这是我实现转换然后拼接(裁剪并将它们粘贴到最终图像上)的过程。
- 获取所有点的列表
- 从一组 4 个点创建一个四边形感兴趣区域 (ROI)
这4个点是用对应的原始4个点对图像进行Transform。这是使用我的函数 perspective_transform()
完成的
一个。我取 2 组 4 个点并将它们传递给 M = cv2.getPerspectiveTransform(corners, newCorners)
b。然后我调用:warped = cv2.warpPerspective(roi, M, (width, height))
获得新的变形图像后,我使用蒙版根据与之关联的 ROI 将所有内容拼接在一起:
一个。这是由函数 quadr_croped()
完成的
屏幕初始化以从屏幕获取原始像素,将其保存到 Numpy 数组
img0 = np.array(sct.grab(monitor))
clone = img0.copy()
total_height, total_width, channels = img0.shape
xSub =int (input("How many columns would you like to divide the screen in to? (integers only)"))
ySub =int (input("How many rows would you like to divide the screen in to? (integers only)"))
roi_width = float(total_width/xSub)
roi_height = float(total_height/ySub)
point_list = []
第三:使用2组4点来扭曲图像的透视
def perspective_transform(图像, roi, 角点, newCorners, i = -1 ):
corners = list (corners)
newCorners = list (newCorners)
height, width, pixType = image.shape
corners = np.array([[corners[0][0],corners[0][1],corners[0][2],corners[0][3]]],np.float32)
newCorners = np.array([[newCorners[0][0],newCorners[0][1],newCorners[0][2],newCorners[0][3]]],np.float32)
M = cv2.getPerspectiveTransform(corners, newCorners)
#warped = cv2.warpPerspective(roi, M, (width, height), flags=cv2.INTER_LINEAR)
warped = cv2.warpPerspective(roi, M, (width, height))
return warped
其次:将四边形剪切并粘贴到主图中
def quadr_croped (mainImg,image, pts, i):
# example
# mask defaulting to black for 3-channel and transparent for 4-channel
# (of course replace corners with yours)
mask = np.zeros(image.shape, dtype=np.uint8)
roi_corners = pts #np.array([[(10,10), (300,300), (10,300)]], dtype=np.int32)
# fill the ROI so it doesn't get wiped out when the mask is applied
channel_count = image.shape[2] # i.e. 3 or 4 depending on your image
ignore_mask_color = (255,)*channel_count
cv2.fillConvexPoly(mask, roi_corners, ignore_mask_color)
# apply the mask
masked_image = cv2.bitwise_and(image, mask)
mainImg = cv2.bitwise_or(mainImg, mask)
mainImg = mainImg + masked_image
# cv2.imshow("debug: image, mainImg: " +str(i), mainImg)
return mainImg
第一:启动函数
def draw_quadr(img1):
#set up list for ROIquadrilateral == polygon with 4 sides
numb_ROI = xSub * ySub
skips =int((numb_ROI-1)/xSub)
numb_ROI = skips + numb_ROI
quadrilateral_list.clear()
for i in range(numb_ROI):
if not point_list[i][0] <= point_list[(i+xSub+2)][0]:
continue
vert_poly = np.array([[
point_list[i],
point_list[i+1],
point_list[i+xSub+2],
point_list[i+xSub+1]
]], dtype=np.int32)
verticesPoly_old = np.array([[
H_points_list[i],
H_points_list[i+1],
H_points_list[i+xSub+2],
H_points_list[i+xSub+1]
]], dtype=np.int32)
roi = img0.copy()
# cv2.imshow("debug: roi"+str(i), roi)
overlay = perspective_transform(
img1,
roi,
verticesPoly_old,
vert_poly,
i)
img1 = quadr_croped(img1,overlay,vert_poly,i)
cv2.polylines(img1,vert_poly,True,(255,255,0))
quadrilateral_list.append(vert_poly)
pt1 = point_list[i]
pt2 = point_list[i+xSub+2]
cntPt = (int( (pt1[0]+pt2[0])/2),int((pt1[1]+pt2[1])/2) )
cv2.putText(img1,str(len(quadrilateral_list)-1),cntPt,cv2.FONT_HERSHEY_SIMPLEX, 1,(0,255,0),2,cv2.LINE_AA)
#cv2.imshow(str(i), img1)
return img1
图片结果链接
请查看这些,因为它们很好地说明了问题。
没有失真的原始图像
这张图片从中心向左偏移(没有 y 方向移动)
x方向畸变图像结果
此图像从中心向上偏移(没有 x 方向移动)
y方向畸变图像结果
这张图片从中心向上和向左偏移
x和y方向失真图像的结果
我是计算机视觉和 Whosebug 的新手,我希望我已经包含了帮助描述问题的所有内容,如果您需要了解其他任何帮助,请告诉我
代码中肯定存在一些错误,因为输出图像看起来不像它们应该的那样(或可能不是)。但是由于透视变换的数学性质,您永远无法准确地得到您想要的东西。即,因为它们是非线性的。你可以让矩形的角点重合,但是角点之间的图像缩放不均匀,你不能让这些不均匀度在分界线的两侧相同。
但是您可以使用仿射变换来统一缩放图像。这保证了如果一条线上的两点重合,则所有其他点也重合。这里唯一的问题是仿射变换是使用三角形确定的,因此您需要将四边形拆分为三角形。例如。在下面的代码中,每个四边形都被分成 4 个三角形,使用四边形的中心作为附加顶点。
import numpy as np
import matplotlib.pyplot as plt
import cv2
# generate a test image
im = np.full((400,600), 255, 'u1')
h, w = im.shape
for i in range(1, w//20):
im = cv2.line(im, (i*20, 0), (i*20, h), i*8)
for i in range(1, h//20):
im = cv2.line(im, (0, i*20), (w, i*20), i*10)
plt.figure(figsize=(w/30, h/30))
plt.imshow(im, 'gray')
plt.show()
# Number of grid cells
nx, ny = 3, 2
p0 = np.meshgrid(np.linspace(0, w-1, nx+1, dtype='f'), np.linspace(0, h-1, ny+1, dtype='f'))
print(np.vstack(p0))
p1 = [v.copy() for v in p0]
# Move the central points
p1[0][1,1] -= 30; p1[1][1,1] -= 40
p1[0][1,2] += 20; p1[1][1,2] += 10
print(np.vstack(p1))
# Set perspective = True to see what happens if we use perspective transform
perspective = False
im1 = np.zeros_like(im)
for i in range(nx):
for j in range(ny):
x0, y0 = p0[0][j,i], p0[1][j,i]
c0 = np.stack((p0[0][j:(j+2),i:(i+2)].ravel() - x0, p0[1][j:(j+2),i:(i+2)].ravel() - y0))
c1 = np.stack((p1[0][j:(j+2),i:(i+2)].ravel(), p1[1][j:(j+2),i:(i+2)].ravel()))
if perspective:
ic0 = np.round(c0).astype('i')
ic1 = np.round(c1).astype('i')
M = cv2.getPerspectiveTransform(c0.T, c1.T)
imw = cv2.warpPerspective(im[ic0[1,0]:ic0[1,3], ic0[0,0]:ic0[0,3]], M, (w, h))
im1 |= cv2.fillConvexPoly(np.zeros_like(im), ic1[:,[0,1,3,2]].T, 255) & imw
else:
c0 = np.append(c0, np.mean(c0, axis=1, keepdims=True), 1)
c1 = np.append(c1, np.mean(c1, axis=1, keepdims=True), 1)
ic0 = np.round(c0).astype('i')
ic1 = np.round(c1).astype('i')
for ind in ([0,1,4], [1,3,4], [3,2,4], [2,0,4]):
M = cv2.getAffineTransform(c0[:,ind].T, c1[:,ind].T)
imw = cv2.warpAffine(im[ic0[1,0]:ic0[1,3], ic0[0,0]:ic0[0,3]], M, (w, h))
im1 |= cv2.fillConvexPoly(np.zeros_like(im), ic1[:,ind].T, 255) & imw
plt.figure(figsize=(w/30, h/30))
plt.imshow(im1, 'gray')
plt.show()
我想扭曲图像的部分以将其投影到不均匀的表面上。最终,我想像 HERE, kinda like was is done in HERE from THIS 项目那样扭曲图像。
我的问题是当我将变换应用到图像的每个部分时,事情就是不对齐
这是我实现转换然后拼接(裁剪并将它们粘贴到最终图像上)的过程。
- 获取所有点的列表
- 从一组 4 个点创建一个四边形感兴趣区域 (ROI)
这4个点是用对应的原始4个点对图像进行Transform。这是使用我的函数 perspective_transform()
完成的一个。我取 2 组 4 个点并将它们传递给 M = cv2.getPerspectiveTransform(corners, newCorners)
b。然后我调用:warped = cv2.warpPerspective(roi, M, (width, height))
获得新的变形图像后,我使用蒙版根据与之关联的 ROI 将所有内容拼接在一起:
一个。这是由函数 quadr_croped()
完成的屏幕初始化以从屏幕获取原始像素,将其保存到 Numpy 数组
img0 = np.array(sct.grab(monitor)) clone = img0.copy() total_height, total_width, channels = img0.shape xSub =int (input("How many columns would you like to divide the screen in to? (integers only)")) ySub =int (input("How many rows would you like to divide the screen in to? (integers only)")) roi_width = float(total_width/xSub) roi_height = float(total_height/ySub) point_list = []
第三:使用2组4点来扭曲图像的透视
def perspective_transform(图像, roi, 角点, newCorners, i = -1 ):
corners = list (corners) newCorners = list (newCorners) height, width, pixType = image.shape corners = np.array([[corners[0][0],corners[0][1],corners[0][2],corners[0][3]]],np.float32) newCorners = np.array([[newCorners[0][0],newCorners[0][1],newCorners[0][2],newCorners[0][3]]],np.float32) M = cv2.getPerspectiveTransform(corners, newCorners) #warped = cv2.warpPerspective(roi, M, (width, height), flags=cv2.INTER_LINEAR) warped = cv2.warpPerspective(roi, M, (width, height)) return warped
其次:将四边形剪切并粘贴到主图中
def quadr_croped (mainImg,image, pts, i): # example
# mask defaulting to black for 3-channel and transparent for 4-channel # (of course replace corners with yours) mask = np.zeros(image.shape, dtype=np.uint8) roi_corners = pts #np.array([[(10,10), (300,300), (10,300)]], dtype=np.int32) # fill the ROI so it doesn't get wiped out when the mask is applied channel_count = image.shape[2] # i.e. 3 or 4 depending on your image ignore_mask_color = (255,)*channel_count cv2.fillConvexPoly(mask, roi_corners, ignore_mask_color) # apply the mask masked_image = cv2.bitwise_and(image, mask) mainImg = cv2.bitwise_or(mainImg, mask) mainImg = mainImg + masked_image # cv2.imshow("debug: image, mainImg: " +str(i), mainImg) return mainImg
第一:启动函数
def draw_quadr(img1):
#set up list for ROIquadrilateral == polygon with 4 sides numb_ROI = xSub * ySub skips =int((numb_ROI-1)/xSub) numb_ROI = skips + numb_ROI quadrilateral_list.clear() for i in range(numb_ROI): if not point_list[i][0] <= point_list[(i+xSub+2)][0]: continue vert_poly = np.array([[ point_list[i], point_list[i+1], point_list[i+xSub+2], point_list[i+xSub+1] ]], dtype=np.int32) verticesPoly_old = np.array([[ H_points_list[i], H_points_list[i+1], H_points_list[i+xSub+2], H_points_list[i+xSub+1] ]], dtype=np.int32) roi = img0.copy() # cv2.imshow("debug: roi"+str(i), roi) overlay = perspective_transform( img1, roi, verticesPoly_old, vert_poly, i) img1 = quadr_croped(img1,overlay,vert_poly,i) cv2.polylines(img1,vert_poly,True,(255,255,0)) quadrilateral_list.append(vert_poly) pt1 = point_list[i] pt2 = point_list[i+xSub+2] cntPt = (int( (pt1[0]+pt2[0])/2),int((pt1[1]+pt2[1])/2) ) cv2.putText(img1,str(len(quadrilateral_list)-1),cntPt,cv2.FONT_HERSHEY_SIMPLEX, 1,(0,255,0),2,cv2.LINE_AA) #cv2.imshow(str(i), img1) return img1
图片结果链接
请查看这些,因为它们很好地说明了问题。
没有失真的原始图像
这张图片从中心向左偏移(没有 y 方向移动)
x方向畸变图像结果
此图像从中心向上偏移(没有 x 方向移动)
y方向畸变图像结果
这张图片从中心向上和向左偏移
x和y方向失真图像的结果
我是计算机视觉和 Whosebug 的新手,我希望我已经包含了帮助描述问题的所有内容,如果您需要了解其他任何帮助,请告诉我
代码中肯定存在一些错误,因为输出图像看起来不像它们应该的那样(或可能不是)。但是由于透视变换的数学性质,您永远无法准确地得到您想要的东西。即,因为它们是非线性的。你可以让矩形的角点重合,但是角点之间的图像缩放不均匀,你不能让这些不均匀度在分界线的两侧相同。
但是您可以使用仿射变换来统一缩放图像。这保证了如果一条线上的两点重合,则所有其他点也重合。这里唯一的问题是仿射变换是使用三角形确定的,因此您需要将四边形拆分为三角形。例如。在下面的代码中,每个四边形都被分成 4 个三角形,使用四边形的中心作为附加顶点。
import numpy as np
import matplotlib.pyplot as plt
import cv2
# generate a test image
im = np.full((400,600), 255, 'u1')
h, w = im.shape
for i in range(1, w//20):
im = cv2.line(im, (i*20, 0), (i*20, h), i*8)
for i in range(1, h//20):
im = cv2.line(im, (0, i*20), (w, i*20), i*10)
plt.figure(figsize=(w/30, h/30))
plt.imshow(im, 'gray')
plt.show()
# Number of grid cells
nx, ny = 3, 2
p0 = np.meshgrid(np.linspace(0, w-1, nx+1, dtype='f'), np.linspace(0, h-1, ny+1, dtype='f'))
print(np.vstack(p0))
p1 = [v.copy() for v in p0]
# Move the central points
p1[0][1,1] -= 30; p1[1][1,1] -= 40
p1[0][1,2] += 20; p1[1][1,2] += 10
print(np.vstack(p1))
# Set perspective = True to see what happens if we use perspective transform
perspective = False
im1 = np.zeros_like(im)
for i in range(nx):
for j in range(ny):
x0, y0 = p0[0][j,i], p0[1][j,i]
c0 = np.stack((p0[0][j:(j+2),i:(i+2)].ravel() - x0, p0[1][j:(j+2),i:(i+2)].ravel() - y0))
c1 = np.stack((p1[0][j:(j+2),i:(i+2)].ravel(), p1[1][j:(j+2),i:(i+2)].ravel()))
if perspective:
ic0 = np.round(c0).astype('i')
ic1 = np.round(c1).astype('i')
M = cv2.getPerspectiveTransform(c0.T, c1.T)
imw = cv2.warpPerspective(im[ic0[1,0]:ic0[1,3], ic0[0,0]:ic0[0,3]], M, (w, h))
im1 |= cv2.fillConvexPoly(np.zeros_like(im), ic1[:,[0,1,3,2]].T, 255) & imw
else:
c0 = np.append(c0, np.mean(c0, axis=1, keepdims=True), 1)
c1 = np.append(c1, np.mean(c1, axis=1, keepdims=True), 1)
ic0 = np.round(c0).astype('i')
ic1 = np.round(c1).astype('i')
for ind in ([0,1,4], [1,3,4], [3,2,4], [2,0,4]):
M = cv2.getAffineTransform(c0[:,ind].T, c1[:,ind].T)
imw = cv2.warpAffine(im[ic0[1,0]:ic0[1,3], ic0[0,0]:ic0[0,3]], M, (w, h))
im1 |= cv2.fillConvexPoly(np.zeros_like(im), ic1[:,ind].T, 255) & imw
plt.figure(figsize=(w/30, h/30))
plt.imshow(im1, 'gray')
plt.show()