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```{code-cell} ipython3
%matplotlib inline
```

# 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.

```{code-cell} ipython3
import numpy as np
from matplotlib import pyplot as plt, cm
import skdemo
```

```{code-cell} ipython3
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);
```

The documentation for scikit-image's morphology module is
[here](http://scikit-image.org/docs/0.10.x/api/skimage.morphology.html).

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`:

```{code-cell} ipython3
from skimage import morphology
sq = morphology.square(width=3)
dia = morphology.diamond(radius=1)
print(sq)
print(dia)
```

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:

```{code-cell} ipython3
skdemo.imshow_all(image, morphology.erosion(image, sq), shape=(1, 2))
```

and

```{code-cell} ipython3
skdemo.imshow_all(image, morphology.dilation(image, sq))
```

and

```{code-cell} ipython3
skdemo.imshow_all(image, morphology.dilation(image, dia))
```

Erosion and dilation can be combined into two slightly more sophisticated operations, *opening* and *closing*. Here's an example:

```{code-cell} ipython3
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);
```

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.

```{code-cell} ipython3
skdemo.imshow_all(image, morphology.opening(image, sq)) # erosion -> dilation
```

```{code-cell} ipython3
skdemo.imshow_all(image, morphology.closing(image, sq)) # dilation -> erosion
```

**Exercise**: use morphological operations to remove noise from a binary image.

```{code-cell} ipython3
from skimage import data, color
hub = color.rgb2gray(data.hubble_deep_field()[350:450, 90:190])
plt.imshow(hub);
```

Remove the smaller objects to retrieve the large galaxy.