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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Morphological operations
Morphology is the study of shapes. In image processing, some simple operations can get you a long way. The first things to learn are erosion and dilation. In erosion, we look at a pixel’s local neighborhood and replace the value of that pixel with the minimum value of that neighborhood. In dilation, we instead choose the maximum.
Step1:  The documentation for scikit-image's morphology module is
here.
Importantly, we must use a structuring element, which defines the local
neighborhood of each pixel. To get every neighbor (up, down, left, right, and
diagonals), use morphology.square; to avoid diagonals, use
morphology.diamond
Step2:  The central value of the structuring element represents the pixel being considered, and the surrounding values are the neighbors
Step3:  and
Step4:  and
Step5:  Erosion and dilation can be combined into two slightly more sophisticated operations, opening and closing. Here's an example
Step6:  What happens when run an erosion followed by a dilation of this image?
What about the reverse?
Try to imagine the operations in your head before trying them out below.
Step7:  Exercise
Step8:  Remove the smaller objects to retrieve the large galaxy.
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Python Code:
import numpy as np
from matplotlib import pyplot as plt, cm
import skdemo
plt.rcParams['image.cmap'] = 'cubehelix'
plt.rcParams['image.interpolation'] = 'none'
image = np.array([[0, 0, 0, 0, 0, 0, 0],
                  [0, 0, 0, 0, 0, 0, 0],
                  [0, 0, 1, 1, 1, 0, 0],
                  [0, 0, 1, 1, 1, 0, 0],
                  [0, 0, 1, 1, 1, 0, 0],
                  [0, 0, 0, 0, 0, 0, 0],
                  [0, 0, 0, 0, 0, 0, 0]], dtype=np.uint8)
plt.imshow(image)
Explanation: Morphological operations
Morphology is the study of shapes. In image processing, some simple operations can get you a long way. The first things to learn are erosion and dilation. In erosion, we look at a pixel’s local neighborhood and replace the value of that pixel with the minimum value of that neighborhood. In dilation, we instead choose the maximum.
End of explanation
from skimage import morphology
sq = morphology.square(width=3)
dia = morphology.diamond(radius=1)
disk = morphology.disk(radius=30)
skdemo.imshow_all(sq, dia, disk)
Explanation: The documentation for scikit-image's morphology module is
here.
Importantly, we must use a structuring element, which defines the local
neighborhood of each pixel. To get every neighbor (up, down, left, right, and
diagonals), use morphology.square; to avoid diagonals, use
morphology.diamond:
End of explanation
skdemo.imshow_all(image, morphology.erosion(image, sq), shape=(1, 2))
Explanation: The central value of the structuring element represents the pixel being considered, and the surrounding values are the neighbors: a 1 value means that pixel counts as a neighbor, while a 0 value does not. So:
End of explanation
skdemo.imshow_all(image, morphology.dilation(image, sq))
Explanation: and
End of explanation
skdemo.imshow_all(image, morphology.dilation(image, dia))
Explanation: and
End of explanation
image = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
                  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
                  [0, 0, 1, 1, 1, 1, 1, 1, 0, 0],
                  [0, 0, 1, 1, 1, 1, 1, 1, 0, 0],
                  [0, 0, 1, 1, 1, 1, 1, 1, 0, 0],
                  [0, 0, 1, 1, 1, 0, 0, 1, 0, 0],
                  [0, 0, 1, 1, 1, 0, 0, 1, 0, 0],
                  [0, 0, 1, 1, 1, 0, 0, 1, 0, 0],
                  [0, 0, 1, 1, 1, 1, 1, 1, 0, 0],
                  [0, 0, 1, 1, 1, 1, 1, 1, 0, 0],
                  [0, 0, 1, 1, 1, 1, 1, 1, 0, 0],
                  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
                  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]], np.uint8)
plt.imshow(image)
Explanation: Erosion and dilation can be combined into two slightly more sophisticated operations, opening and closing. Here's an example:
End of explanation
skdemo.imshow_all(image, morphology.opening(image, sq)) # erosion -> dilation
skdemo.imshow_all(image, morphology.closing(image, sq)) # dilation -> erosion
Explanation: What happens when run an erosion followed by a dilation of this image?
What about the reverse?
Try to imagine the operations in your head before trying them out below.
End of explanation
from skimage import data, color
hub = color.rgb2gray(data.hubble_deep_field()[350:450, 90:190])
plt.imshow(hub)
Explanation: Exercise: use morphological operations to remove noise from a binary image.
End of explanation
disk = morphology.disk(radius=8)
gal = morphology.opening(hub, disk)
plt.imshow(gal)
skdemo.imshow_with_histogram(gal)
gal_selector = gal > 50
# use numpy boolean indexing to estimate histogram
# of brightness intensity on isolated galaxy
skdemo.imshow_all(hub, gal_selector)
f = plt.figure()
plt.hist(hub[gal_selector], bins=50);
Explanation: Remove the smaller objects to retrieve the large galaxy.
End of explanation
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