python :

I get a segmentation graph (

2D numpy array) with the class value (integers 0 to N) for each pixel of the original img, and I want to find the bounding box coordinates of each connected cluster in the segmentation graph.

EDIT: There may be multiple clusters per class in map!

I

think I could use something like skimage.measure.label(seg_map, connectivity=1).

Full answer

There is an existing function scipy.ndimage.measurements.find_objects Allllmost is exactly what you need. Numpy and are required A little help from scipy.ndimage.measurements.label, though:

import numpy as np
import scipy.ndimage.measurements as mnts

A = np.array([
    [0, 0, 0, 0, 0, 0, 0],
    [0, 1, 1, 0, 2, 2, 0],
    [0, 1, 1, 0, 2, 2, 0],
    [0, 0, 0, 0, 0, 0, 0],
    [0, 4, 4, 0, 1, 1, 0],
    [0, 4, 4, 0, 1, 1, 0],
    [4, 0, 0, 0, 0, 0, 1]
])

structure = np.array([
    [1,1,1],
    [1,1,1],
    [1,1,1]
])

bboxSlices = {}
for i in range(1, A.max() + 1):
    B = A.copy()
    B[B != i] = 0

bboxSlices[i] = mnts.find_objects(mnts.label(B, structure=structure)[0])

print(bboxSlices)

Output:

{1: [(slice(1, 3, None), slice(1, 3, None)),
     (slice(4, 7, None), slice(4, 7, None))],
 2: [(slice(1, 3, None), slice(4, 6, None))],
 3: [],
 4: [(slice(4, 7, None), slice(0, 3, None))]}

Each entry in the bboxSlices dictionary is a tuple list. Each tuple contains two slices, a row slice and a column slice, each of which defines a bounding box around the cluster of corresponding classes.

Details

label(...) finds feature clusters and replaces their values with labels (for example, 1 for the first cluster, 2 for the second, and so on). find_objects(...) Then find the bounding box around each label. The problem is that the label treats all non-zero values as “features”. Therefore, for each class value i, we need a copy of A with all non-i values cleared.

structure defines the connectivity of the cluster. If you want a cluster that is not connected diagonally, you can use a different structure:

structure = np.array([
    [0,1,0],
    [1,1,1],
    [0,1,0]
])

Simple answer

This is easy if there is only 1 cluster per class :

import numpy as np

A = np.array([
    [0, 0, 0, 0, 0, 0, 0],
    [0, 1, 1, 0, 2, 2, 0],
    [0, 1, 1, 0, 2, 2, 0],
    [0, 0, 0, 0, 0, 0, 0],
    [0, 4, 4, 0, 3, 3, 0],
    [0, 4, 4, 0, 3, 3, 0],
    [0, 0, 0, 0, 0, 0, 0]
])

bboxCorners = {}
for i in range(1, A.max()+1):
    B = np.argwhere(A==i)
    bboxCorners[i] = B.min(0), B.max(0)

print(bboxCorners)

Output:

{1: (array([1, 1]), array([2, 2])),
 2: (array([1, 4]), array([2, 5])),
 3: (array([4, 4]), array([5, 5])),
 4: (array([4, 1]), array([5, 2]))}