MultiLabelBinarizer() with inverse_transform() - python

I have multilabel labels. Elements in a label means voting. Here is how labels look:
array([[ 4, 0, 0, 1, 3, 2, 0, 0],
[ 6, 0, 1, 1, 0, 0, 0, 0],
[ 5, 0, 0, 3, 1, 0, 0, 0],
[ 4, 0, 0, 4, 1, 0, 0, 0],
[ 9, 0, 0, 1, 0, 0, 0, 0],
[ 6, 0, 0, 1, 0, 0, 1, 1],
[ 2, 0, 0, 8, 0, 0, 0, 0],
[ 0, 10, 0, 0, 0, 0, 0, 0],
[ 0, 10, 0, 0, 0, 0, 0, 0],
[ 0, 0, 6, 0, 0, 0, 4, 0]])
And here is what I tried:
from sklearn.preprocessing import MultiLabelBinarizer
mlb = MultiLabelBinarizer()
nn = mlb.fit_transform(labels_train)
nn[:10]
Output:
array([[1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0],
[1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0]])
And when I tried inverse_transform():
zet = mlb.inverse_transform(nn)
zet[:10]
out:
[(0, 1, 2, 3, 4),
(0, 1, 6),
(0, 1, 3, 5),
(0, 1, 4),
(0, 1, 9),
(0, 1, 6),
(0, 2, 8),
(0, 10),
(0, 10),
(0, 4, 6)]
What am I doing wrong? Why is it showing unique values in ascending order?

Related

Filtering elements of a numpy array depending on their occurrences

i have the following 2D numpy array M
M = np.array([[1,1,1,0,0,0,0,0,0,0,0],
[1,1,1,0,0,0,0,0,0,1,1],
[1,1,1,0,0,0,0,0,0,1,1],
[0,0,0,0,0,1,1,1,0,0,0],
[0,0,0,0,0,1,1,1,0,0,0],
[1,1,1,0,1,1,1,1,0,0,0],
[1,1,1,0,0,1,1,1,0,0,0],
[1,1,1,0,0,1,1,1,0,0,0]])
which I want to identify its spots (Pixels with value==1 and connected to each other).
Thanks to the function 'label' from scipy, I can identify all of my spots in the matrix. The output should seem like this:
Output, Nbr= label(M)
#Output= array([[1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2],
# [0, 0, 0, 0, 0, 3, 3, 3, 0, 0, 0],
# [0, 0, 0, 0, 0, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 3, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 0, 3, 3, 3, 0, 0, 0],
# [4, 4, 4, 0, 0, 3, 3, 3, 0, 0, 0]])
I want only to have spots with 9 elements, that means the first and fourth spot.
using a for loop like this works fine:
for i in range(Nbr+1):
Spot= np.argwhere(components[:,:]== i)
if len(Spot)!=9:
M[Spot[:, 0], Spot[:, 1]]=0
#M= array([[1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
# [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, 0, 0, 0, 0, 0, 0, 0, 0],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
# [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0]])
The porblem is when my Spots are more than 4, my code is slower.
Is there any faster alternative that can do the job of the for loop?
Thanks.

Finding groups of same value inside a 2D list

I'm currently working on something were I need to find groups of values in a 2d list that are surrounded by another value and then change the values of the surrounded elements. Start/end of a sub-list counts as surrounded.
For exemple if I have this list :
[[1, 2, 1, 1, 1, 0, 0, 0, 0],
[2, 2, 2, 2, 2, 1, 0, 0, 0],
[1, 2, 2, 1, 1, 0, 0, 0, 0],
[0, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 1, 0],
[0, 1, 2, 1, 0, 0, 1, 1, 0],
[0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]]
I want the function to change it to this:
[[1, 3, 1, 1, 1, 0, 0, 0, 0],
[3, 3, 3, 3, 3, 1, 0, 0, 0],
[1, 3, 3, 1, 1, 0, 0, 0, 0],
[0, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 1, 0],
[0, 1, 3, 1, 0, 0, 1, 1, 0],
[0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]]
I have no idea how to do it. Can anyone please help me ?

lst = [
[1, 2, 1, 1, 1, 0, 0, 0, 0],
[2, 2, 2, 2, 2, 1, 0, 0, 0],
[1, 2, 2, 1, 1, 0, 0, 0, 0],
[0, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 1, 0],
[0, 1, 2, 1, 0, 0, 1, 1, 0],
[0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
]
for row in range(len(lst)):
for col in range(len(lst[row])):
if lst[row][col] == 1:
continue
v1 = lst[row][col - 1] >= 1 if col - 1 >= 0 else 1
v2 = lst[row - 1][col] >= 1 if row - 1 >= 0 else 1
v3 = lst[row][col + 1] >= 1 if col + 1 < len(lst[row]) else 1
v4 = lst[row + 1][col] >= 1 if row + 1 < len(lst) else 1
if v1 + v2 + v3 + v4 == 4:
lst[row][col] += 1
from pprint import pprint
pprint(lst)
Prints:
[[1, 3, 1, 1, 1, 0, 0, 0, 0],
[3, 3, 3, 3, 3, 1, 0, 0, 0],
[1, 3, 3, 1, 1, 0, 0, 0, 0],
[0, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 1, 0],
[0, 1, 3, 1, 0, 0, 1, 1, 0],
[0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0]]

can't multiply matrix and list which type 'float'

I want to multiply matrix (transpose_test_feature) with a list (log_train_probs_nonspam_words). Matrix has a size of (2500 * 260) and list has a size of 2500 items. I want to multiply them.
But I have this type error:
TypeError: can't multiply sequence by non-int of type 'float'
Is there anybody who can quickly help me?
multiply_nonspam_test = []
for i in range(0, len(transpose_test_feature)):
x = log_train_probs_nonspam_words[i] *transpose_test_feature[i]
multiply_nonspam_test.append(x)
print multiply_nonspam_test
Input:
log_train_probs_nonspam_words (sample):
[-5.259440519499674, -5.5014525551182665, -6.226860597896433, -4.000908730923304, -6.970438632083271, -6.082121752251521, -6.7823864005803305, -4.22184233658671, -5.805134968916488, -6.28594951426644, -5.396092039460441, -4.935057080197465, -6.321351441317356, -6.054147900209116, -7.698677132454486, -6.71339352909338, -5.403260528939053, -5.454932539483374, -5.837924791739479, -9.239122173401634, -6.405908829345418, -8.83365706529347, -6.9200077784563785, -5.406864136442351, -6.089239220020385, -5.794439679799741, -5.6556032349455245, -7.8528278122817445, -4.863365151741348, -7.3673199965000435, -5.588463932107897, -7.293212024346322, -6.509093065580649, -5.605931625148287, -4.93280736046345, -6.674172815940098, -7.663585812643216, -5.918893854273146, -6.7542155236136345, -7.534374081163209, -6.6241623953654365, -5.095987447010102, -8.140509884733525, -6.2350910970329485, -5.287878454820207]
transpose_test_feature (originally it is a 2500 * 260 matrix, I gave the first row of a matrix):
[0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 6, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 3, 1, 0, 2, 3, 0, 0, 2, 2, 0, 5, 1, 3, 0, 0, 3, 3, 0, 0, 1, 0, 1, 1, 1, 3, 0, 0, 3, 4, 1, 8, 0, 0, 0, 1, 3, 2, 0, 2, 2, 0, 0, 0, 1, 2, 0, 3, 2, 1, 0, 0, 1, 0, 3, 0, 0, 3, 0, 0, 0, 5, 0, 0, 0, 1, 0, 0, 2, 1, 1, 0, 18, 0, 10, 13, 3, 0, 2, 1, 7, 7, 0, 11, 8, 1, 0, 7, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 9, 0, 4, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 42, 11, 5, 0, 19, 3, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 5, 3, 1, 8, 0, 0, 2, 3, 14, 0, 2, 0, 13, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 20, 2, 29, 29, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0]

You are multiplying a list by a float. You should multiply each row of the matrix with words. Something like this will work.
multiply_nonspam_test = []
for row in transpose_test_feature:
multiply_nonspam_test.append([x*y for x,y in zip(row, log_train_probs_nonspam_words)])
print multiply_nonspam_test

You can use NumPy to speed things up:
import numpy as np
a1 = np.array(log_train_probs_nonspam_words)
a1 = a1.reshape(a1.shape[0], 1)
a2 = np.array(transpose_test_feature)
multiply_nonspam_test = a1 * a2

Split contiguous regions within numpy array into smaller contiguous regions

I am trying to identify the center of mass of each contiguous regions of a certain size (i.e. 30 total pixels) within a binary numpy array.
I've used scipy.ndimage.label to identify any contiguous regions of pixels within my array and it worked great. However, in some cases there are contiguous regions that are larger than the size I am looking for (i.e. 60 pixels, 75 pixels, 90 pixels, etc.) In these cases I need to split the large contiguous region into several contiguous regions of my desired size.
For example:
Imagine the following array and I need to find all contiguous areas within the array.
>>> x
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0],
[0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 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, 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, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 1, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
Labels
>>> labels, numLabels = scipy.ndimage.label(x)
>>> labels
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0],
[0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 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, 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, 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, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
Let's say I'm only interested in contiguous areas of 9 pixels. This is an example of the output I would be looking for.
>>> contiguous_regions
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 5, 5, 5, 0, 0, 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, 5, 5, 5, 0, 0, 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, 5, 5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 6, 6, 6, 7, 7, 7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 6, 6, 6, 7, 7, 7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 6, 6, 6, 7, 7, 7, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 8, 8, 8, 9, 9, 9, 19, 19, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 8, 8, 8, 9, 9, 9, 19, 19, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 3, 8, 8, 8, 9, 9, 9, 19, 19, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 10, 11, 11, 11, 12, 12, 12, 19, 19, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 10, 11, 11, 11, 12, 12, 12, 19, 20, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 10, 11, 11, 11, 12, 12, 12, 20, 20, 0, 4, 4, 4, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 13, 13, 14, 14, 14, 15, 15, 15, 20, 20, 0, 4, 4, 4, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 13, 13, 14, 14, 14, 15, 15, 15, 20, 20, 0, 4, 4, 4, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 13, 13, 14, 14, 14, 15, 15, 15, 20, 20, 0, 22, 22, 22, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 16, 16, 17, 17, 17, 18, 18, 18, 0, 0, 0, 22, 22, 22, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 16, 16, 17, 17, 17, 18, 18, 18, 0, 0, 0, 22, 22, 22, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 16, 16, 17, 17, 17, 18, 18, 18, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 21, 21, 21, 21, 21, 21, 21, 21, 21, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
It's ok if edges get cutoff and the contiguous regions do not all need to be the same shape necessarily, but the more square the better. Eventually, once I have the smaller regions of interest, I need to get the center of mass of each one using scipy.ndimage.measurements.center_of_mass. If there is a way I could reduce these contiguous areas into individual pixels x distance from one another, that would work too.
Any ideas on how to accomplish this with numpy and scipy?
Thanks in advance

If your labels are just 0 and 1 you could mess around with the following, though I think there are some cases where it might not work. To get you started, here's how you can create an index grid:
igrid = np.repeat(np.arange(4), 5)[..., None] + np.repeat(np.arange(4), 5)[None, ...]
array([[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3],
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4],
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4],
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4],
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4],
[1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4],
[2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
[3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6],
[3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6],
[3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6],
[3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6],
[3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6]])
So in this case, your maximum allowed size is 25 but you could change this by changing the second parameter of repeat. To get the right size to match your original array you could just generate a large enough igrid and slice out the region you want.
You can then do:
result = np.where(contiguous_regions, contiguous_regions, contiguous_regions + igrid)

Scipy label erosion

How can I keep a ring of pixels around labeled regions in a numpy array?
In a simple case, I'd subtract the erosion. That approach doesn't work when the labels touch. How can I get get B from A?
A = array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 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, 0, 0, 0, 0]])
B = array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0],
[0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0],
[0, 0, 2, 0, 0, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 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, 0, 0, 0, 0]])
I'm working with large arrays with many labels, so separate erosions on each label isn't an option.

New Answer
Actually, I just thought of a better way:
B = A * (np.abs(scipy.ndimage.laplace(A)) > 0)
As a full example:
import numpy as np
import scipy.ndimage
A = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 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, 0, 0, 0, 0]])
B = A * (np.abs(scipy.ndimage.laplace(A)) > 0)
I think this should work in all cases (of "labeled" arrays like A, at any rate...).
If you're worried about performance, you can split this into a few pieces to reduce memory overhead:
B = scipy.ndimage.laplace(A)
B = np.abs(B, B) # Preform abs in-place
B /= B # This will produce a divide by zero warning that you can safely ignore
B *= A
This version is a lot more verbose, but should use much less memory.
Old Answer
I can't think of a good way to do it in one step with the usual scipy.ndimage functions. (I feel like a tophat filter should do what you want, but I can't quite figure it out.)
However, doing several separate erosions is an option, as you mentioned.
You should get reasonable performance even on very large arrays if you use find_objects to extract the subregion of each label, and then just do the erosion on the subregion.
For example:
import numpy as np
import scipy.ndimage
A = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 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, 0, 0, 0, 0]])
regions = scipy.ndimage.find_objects(A)
mask = np.zeros_like(A).astype(np.bool)
for val, region in enumerate(regions, start=1):
if region is not None:
subregion = A[region]
mask[region] = scipy.ndimage.binary_erosion(subregion == val)
B = A.copy()
B[mask] = 0
This yields:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0],
[0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0],
[0, 0, 2, 0, 0, 2, 2, 2, 2, 0, 0, 0],
[0, 0, 2, 2, 2, 1, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 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, 0, 0, 0, 0]])
The performance should be reasonable for large arrays, but it's going to depend strongly on how large of an area the different labeled objects span and the number of labeled objects that you have....

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