Issue
If I plot a 2D array and contour it, I can get the access to the segmentation map, via cs = plt.contour(...); cs.allsegs
but it's parameterized as a line. I'd like a segmap boolean mask of what's interior to the line, so I can, say, quickly sum everything within that contour.
Many thanks!
Solution
I dont think there is a really easy way, mainly because you want to mix raster and vector data. Matplotlib paths fortunately have a way to check if a point is within the path, doing this for all pixels will make a mask, but i think this method can get very slow for large datasets.
import matplotlib.patches as patches
from matplotlib.nxutils import points_inside_poly
import matplotlib.pyplot as plt
import numpy as np
# generate some data
X, Y = np.meshgrid(np.arange(-3.0, 3.0, 0.025), np.arange(-3.0, 3.0, 0.025))
Z1 = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
# difference of Gaussians
Z = 10.0 * (Z2 - Z1)
fig, axs = plt.subplots(1,2, figsize=(12,6), subplot_kw={'xticks': [], 'yticks': [], 'frameon': False})
# create a normal contour plot
axs[0].set_title('Standard contour plot')
im = axs[0].imshow(Z, cmap=plt.cm.Greys_r)
cs = axs[0].contour(Z, np.arange(-3, 4, .5), linewidths=2, colors='red', linestyles='solid')
# get the path from 1 of the contour lines
verts = cs.collections[7].get_paths()[0]
# highlight the selected contour with yellow
axs[0].add_patch(patches.PathPatch(verts, facecolor='none', ec='yellow', lw=2, zorder=50))
# make a mask from it with the dimensions of Z
mask = verts.contains_points(list(np.ndindex(Z.shape)))
mask = mask.reshape(Z.shape).T
axs[1].set_title('Mask of everything within one contour line')
axs[1].imshow(mask, cmap=plt.cm.Greys_r, interpolation='none')
# get the sum of everything within the contour
# the mask is inverted because everything within the contour should not be masked
print np.ma.MaskedArray(Z, mask=~mask).sum()
Note that contour lines which 'leave' the plot at different edges by default wont make a path which follows these edges. These lines would need some additional processing.
Answered By - Rutger Kassies
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