从skimage轮廓创建蒙版 [英] Create mask from skimage contour
问题描述
我有一幅图像,该图像是通过 skimage.measure.find_contours()
在其上找到轮廓的,但是现在我想为完全在最大闭合轮廓之外的像素创建一个遮罩.任何想法如何做到这一点?
I have an image that I found contours on with skimage.measure.find_contours()
but now I want to create a mask for the pixels fully outside the largest closed contour. Any idea how to do this?
修改文档中的示例:
import numpy as np
import matplotlib.pyplot as plt
from skimage import measure
# Construct some test data
x, y = np.ogrid[-np.pi:np.pi:100j, -np.pi:np.pi:100j]
r = np.sin(np.exp((np.sin(x)**2 + np.cos(y)**2)))
# Find contours at a constant value of 0.8
contours = measure.find_contours(r, 0.8)
# Select the largest contiguous contour
contour = sorted(contours, key=lambda x: len(x))[-1]
# Display the image and plot the contour
fig, ax = plt.subplots()
ax.imshow(r, interpolation='nearest', cmap=plt.cm.gray)
X, Y = ax.get_xlim(), ax.get_ylim()
ax.step(contour.T[1], contour.T[0], linewidth=2, c='r')
ax.set_xlim(X), ax.set_ylim(Y)
plt.show()
这是红色的轮廓:
但是,如果放大,请注意轮廓不在像素的分辨率上.
But if you zoom in, notice the contour is not at the resolution of the pixels.
如何创建与原始尺寸相同尺寸的图像,且像素完全位于外部(即,轮廓线未与之交叉)?例如
How can I create an image of the same dimensions as the original with the pixels fully outside (i.e. not crossed by the contour line) masked? E.g.
from numpy import ma
masked_image = ma.array(r.copy(), mask=False)
masked_image.mask[pixels_outside_contour] = True
谢谢!
推荐答案
好,我可以通过将轮廓转换为路径然后选择其中的像素来完成这项工作:
Ok, I was able to make this work by converting the contour to a path and then selecting the pixels inside:
# Convert the contour into a closed path
from matplotlib import path
closed_path = path.Path(contour.T)
# Get the points that lie within the closed path
idx = np.array([[(i,j) for i in range(r.shape[0])] for j in range(r.shape[1])]).reshape(np.prod(r.shape),2)
mask = closed_path.contains_points(idx).reshape(r.shape)
# Invert the mask and apply to the image
mask = np.invert(mask)
masked_data = ma.array(r.copy(), mask=mask)
但是,这是一种缓慢的测试,需要对 N = r.shape [0] * r.shape [1]
个像素进行遏制.有人有更快的算法吗?谢谢!
However, this is kind of slow testing N = r.shape[0]*r.shape[1]
pixels for containment. Anyone have a faster algorithm? Thanks!
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