[英]Extract an arbitrary rectangular patch from an image given 4 point coordinates
给定图像中四个任意点的坐标(保证形成矩形),我想提取它们所代表的补丁并获得相同的矢量化(平面)表示。 我怎样才能做到这一点?
我看到了这个问题的答案并使用它,我能够达到我需要的补丁。 例如,给定此图像中绿色矩形的4个角的图像坐标:
我能够得到补丁并得到类似的东西:
使用以下代码:
p1 = (334,128)
p2 = (438,189)
p3 = (396,261)
p4 = (292,200)
pts = np.array([p1, p2, p3, p4])
mask = np.zeros((img.shape[0], img.shape[1]))
cv2.fillConvexPoly(mask, pts, 1)
mask = mask.astype(np.bool)
out = np.zeros_like(img)
out[mask] = img[mask]
patch = img[mask]
cv2.imwrite(img_name, out)
然而,问题是我获得的patch
变量只是属于补丁的图像的所有像素的数组,当图像被读作行主要顺序的矩阵时。
我想要的是patch
变量应该按照它们可以形成真实图像的顺序包含像素,以便我可以对其执行操作。 是否有一个我应该知道的opencv函数可以帮助我这样做?
谢谢!
这是你如何实现这个:
码:
# create a subimage with the outer limits of the points
subimg = out[128:261,292:438]
# calculate the angle between the 2 'lowest' points, the 'bottom' line
myradians = math.atan2(p3[0]-p4[0], p3[1]-p4[1])
# convert to degrees
mydegrees = 90-math.degrees(myradians)
# create rotationmatrix
h,w = subimg.shape[:2]
center = (h/2,w/2)
M = cv2.getRotationMatrix2D(center, mydegrees, 1)
# rotate subimage
rotatedImg = cv2.warpAffine(subimg, M, (h, w))
接下来,通过移除100%黑色的所有行/列,可以容易地裁剪图像中的黑色区域。
最后结果:
码:
# converto image to grayscale
img = cv2.cvtColor(rotatedImg, cv2.COLOR_BGR2GRAY)
# sum each row and each volumn of the image
sumOfCols = np.sum(img, axis=0)
sumOfRows = np.sum(img, axis=1)
# Find the first and last row / column that has a sum value greater than zero,
# which means its not all black. Store the found values in variables
for i in range(len(sumOfCols)):
if sumOfCols[i] > 0:
x1 = i
print('First col: ' + str(i))
break
for i in range(len(sumOfCols)-1,-1,-1):
if sumOfCols[i] > 0:
x2 = i
print('Last col: ' + str(i))
break
for i in range(len(sumOfRows)):
if sumOfRows[i] > 0:
y1 = i
print('First row: ' + str(i))
break
for i in range(len(sumOfRows)-1,-1,-1):
if sumOfRows[i] > 0:
y2 = i
print('Last row: ' + str(i))
break
# create a new image based on the found values
finalImage = rotatedImg[y1:y2,x1:x2]
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.