CODEWITHSUNDEEP
pythonnumpyplotlyclassification
Sergey Vladimirovich
Sergey Vladimirovich

Reputation: 489

numpy detection of borders in 2-d array

I have matrix (below), which represents classes (e.x. 0,1,2). I am plottling it with plotly (python) using heatmap, and I can't find any function which will give me the coordinates of classes' borders.

array([[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 1, 2, 2],
       [1, 1, 1, 2, 2, 2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2, 2, 2, 2, 0, 0],
       [2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
       [2, 2, 2, 2, 2, 0, 0, 0, 0, 0],
       [2, 2, 2, 2, 0, 0, 0, 0, 0, 0],
       [2, 2, 2, 0, 0, 0, 0, 0, 0, 0],
       [2, 2, 2, 0, 0, 0, 0, 0, 0, 0]])

The black lines below are borders, could you give some tips to calculate it in python efficient way? Checking every element in array and its neighbour works very slow.

enter image description here

upd. I also tried to look in plotly contour plot, but the shape of lines has interpolation so it looks no as in the exmaple above...

Upvotes: 0

Views: 730

Answers (1)

yatu
yatu

Reputation: 88226

You can use skimage.measure to find the components in the array. Since 0 is considered to be "background", you'll have to change the label to another one, the maximum value +1 for instance. This will give you tuples of slices with the coordinates.

Labeling is necessary if blocks with the same label are not necessarily "attached", or in the same component. Otherwise you can directly use regionprops to find the slices.

from skimage.measure import label, regionprops

a[a==0]=a.max()+1
l = label(a)
for s in regionprops(l):
    print(s.slice)
(slice(0, 4, None), slice(0, 10, None))
(slice(2, 10, None), slice(0, 10, None))
(slice(4, 10, None), slice(3, 10, None))

Input data:

a = np.array([[1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 1, 2, 2],
       [1, 1, 1, 2, 2, 2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2, 2, 2, 2, 0, 0],
       [2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
       [2, 2, 2, 2, 2, 0, 0, 0, 0, 0],
       [2, 2, 2, 2, 0, 0, 0, 0, 0, 0],
       [2, 2, 2, 0, 0, 0, 0, 0, 0, 0],
       [2, 2, 2, 0, 0, 0, 0, 0, 0, 0]])

Upvotes: 1

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