问题
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?
Modifying the example in the documentation:
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()
Here is the contour in red:
But if you zoom in, notice the contour is not at the resolution of the pixels.
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
Thanks!
回答1:
A bit late but you know the saying. Here is how I would accomplish this.
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
回答2:
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)
However, this is kind of slow testing N = r.shape[0]*r.shape[1]
pixels for containment. Anyone have a faster algorithm? Thanks!
回答3:
If you're still looking for a faster way to achieve that, I would recommend using skimage.draw.polygon, I'm kind of new to this but it seems to be built in to do exactly what you are trying to accomplish:
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
So in your case, a 'closed contour' is a 'poly', we are creating a blank image with the shape of your contours filled with the value 1:
mask = np.zeros(r.shape)
rr, cc = polygon(contour[:, 0], contour[:, 1], mask.shape)
mask[rr, cc] = 1
And now you can apply your mask to the original image to mask out everything outside the contours:
masked = ma.array(r.copy(), mask=mask)
Documented in scikit image - Shapes
来源:https://stackoverflow.com/questions/39642680/create-mask-from-skimage-contour