Joe*_*lip 6 python contour scikit-image masked-array
我有一张我发现轮廓的图像,skimage.measure.find_contours()但现在我想为完全在最大闭合轮廓之外的像素创建一个蒙版。知道如何做到这一点吗?
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()
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这是红色的轮廓:
但是如果放大,请注意轮廓不是像素的分辨率。
如何创建与原始图像尺寸相同且像素完全在外部(即没有与轮廓线交叉)被屏蔽的图像?例如
from numpy import ma
masked_image = ma.array(r.copy(), mask=False)
masked_image.mask[pixels_outside_contour] = True
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谢谢!
有点晚了,但你知道这句话。这是我将如何实现这一点。
import scipy.ndimage as ndimage
# Create an empty image to store the masked array
r_mask = np.zeros_like(r, dtype='bool')
# Create a contour image by using the contour coordinates rounded to their nearest integer value
r_mask[np.round(contour[:, 0]).astype('int'), np.round(contour[:, 1]).astype('int')] = 1
# Fill in the hole created by the contour boundary
r_mask = ndimage.binary_fill_holes(r_mask)
# Invert the mask since you want pixels outside of the region
r_mask = ~r_mask
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如果您仍在寻找更快的方法来实现这一目标,我建议使用skimage.draw.polygon,我对此有点陌生,但它似乎是内置的,可以完全完成您想要完成的任务:
import numpy as np
from skimage.draw import polygon
# fill polygon
poly = np.array((
(300, 300),
(480, 320),
(380, 430),
(220, 590),
(300, 300),
))
rr, cc = polygon(poly[:, 0], poly[:, 1], img.shape)
img[rr, cc, 1] = 1
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因此,在您的情况下,“闭合轮廓”是“多边形”,我们正在创建一个空白图像,其中轮廓的形状填充值 1:
mask = np.zeros(r.shape)
rr, cc = polygon(contour[:, 0], contour[:, 1], mask.shape)
mask[rr, cc] = 1
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现在,您可以将蒙版应用到原始图像,以遮盖轮廓之外的所有内容:
masked = ma.array(r.copy(), mask=mask)
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好的,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:
# 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)
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然而,这是一种缓慢的测试N = r.shape[0]*r.shape[1]像素遏制的方法。有人有更快的算法吗?谢谢!