我想创建一个二进制图像掩码,在 python 中只包含 1 和 0。感兴趣区域(白色)是非矩形的,由 4 个角点定义,如下所示:
在我的方法中,我首先计算 ROI 上下边界的线方程,然后我检查每个掩码元素,它是否小于或大于边界。该代码正在运行,但速度很慢。一个 2000x1000 的蒙版最多需要 4 秒的时间来处理我的机器。
from matplotlib import pyplot as plt
import cv2
import numpy as np
import time
def line_eq(line):
"""input:
2 points of a line
returns:
slope and intersection of the line
"""
(x1, y1), (x2, y2) = line
slope = (y2 - y1) / float((x2 - x1))
intersect = int(slope * (-x1) + y1)
return slope,intersect
def maskByROI(mask,ROI):
"""
input:
ROI: with 4 corner points e.g. ((x0,y0),(x1,y1),(x2,y2),(x3,y3))
mask:
output:
mask with roi set to 1, rest to 0
"""
line1 = line_eq((ROI[0],ROI[1]))
line2 = line_eq((ROI[2],ROI[3]))
slope1 = line1[0]
intersect1 = line1[1]
#upper line
if slope1>0:
for (x,y), value in np.ndenumerate(mask):
if y > slope1*x +intersect1:
mask[x,y] = 0
else:
for (x,y), value in np.ndenumerate(mask):
if y < slope1*x +intersect1:
mask[x,y] = 0
#lower line
slope2 = line2[0]
intersect2 = line2[1]
if slope2<0:
for (x,y), value in np.ndenumerate(mask):
if y > slope2*x +intersect2:
mask[x,y] = 0
else:
for (x,y), value in np.ndenumerate(mask):
if y < slope2*x +intersect2:
mask[x,y] = 0
return mask
mask = np.ones((2000,1000))
myROI = ((750,0),(900,1000),(1000,1000),(1500,0))
t1 = time.time()
mask = maskByROI(mask,myROI)
t2 = time.time()
print "execution time: ", t2-t1
plt.imshow(mask,cmap='Greys_r')
plt.show()
创建这样的 mask 的更有效方法是什么?
对于非矩形形状是否有类似的解决方案 numpy、OpenCV 或类似的库?
最佳答案
使用 fillPoly
绘制 mask :
mask = np.ones((1000, 2000)) # (height, width)
myROI = [(750, 0), (900, 1000), (1000, 1000), (1500, 0)] # (x, y)
cv2.fillPoly(mask, [np.array(myROI)], 0)
这应该需要 ~1 毫秒。
关于python - 在 OpenCV 中选择图像的非矩形 ROI 的最有效方法是什么?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37624509/