how to change the particuler elements of an array
Question:
I have an array for an example:
import numpy as np
data=np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,1],
[3,0,1,0,1,1,1,1,1,1,1,1,1,1,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,1,0],
[2,0,0,0,1,1,1,0,1,0,1,1,1,0,0],
[2,0,1,0,1,1,1,0,1,0,1,0,1,0,0]])
Requirement :
In the data array, if element 1’s are consecutive as the square size
of ((3,3)) and more than square size no changes. Otherwise, replace
element value 1 with zero except the square size.
Expected output :
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 0]
[3 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 0 0 0 0 0]]
current code:
k = np.ones((3,3))
print(k)
jk=binary_dilation(binary_erosion(data==1, k), k)
print(jk)
current output:
[[False False False False True True True True True True False
False False False False]
[False False False False True True True True True True True
True True False False]
[False False False False True True True True True True True
True True False False]
[False False False False True True True False False False True
True True False False]
[False False False False True True True False False False False
False False False False]]
Answers:
The below proposed solution uses skimage.util.view_as_windows in order to get all of the possible 3×3 views onto the larger array and then looping using them with help of Python loops to set the found hits back to ones after zeroing all the ones in the first step.
It seems that OpenCV, scimage, numpy don’t provide a method able to label areas in which a 3×3 square can move around having all the values beneath set to 1’s, but I am not sure here. So if you read this and know much about dilations, convolutions, etc. please leave a note pointing me in the right direction allowing to label the area of image by a methond implemented in C-code for better speed on huge arrays.
# https://stackoverflow.com/questions/73649612/how-to-change-the-particuler-elements-of-an-array
# Example array:
import numpy as np
D_in =np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,1],
[3,0,1,0,1,1,1,1,1,1,1,1,1,1,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,1,0],
[2,0,0,0,1,1,1,0,1,0,1,1,1,0,0],
[2,0,1,0,1,1,1,0,1,0,1,0,1,0,0]])
# Requirements: If in the D_in array at least one 3x3 square with 1's
# is found, replace in D_in all 1's with 0's except the square size and more than square size no changes.
# Otherwise.
# Expected output :
D_tgt=np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,0],
[3,0,0,0,1,1,1,1,1,1,1,1,1,0,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,0,0],
[2,0,0,0,1,1,1,0,0,0,1,1,1,0,0],
[2,0,0,0,1,1,1,0,0,0,0,0,0,0,0]])
# ----------------------------------------------------------------------
import numpy as np
D_out = np.copy(D_in) # D_out for restoring hit 3x3 ONEs
D_out[D_out==1] = 0 # set all ONEs to zero
needle = np.ones((3,3)) # not necessary here, except documentation
# Create all possible 3x3 needle views on larger D_in heystack:
from skimage.util import view_as_windows as winview
# vvvvvvvvvvvvvvvvvvvvvvvvvvv
all3x3 = winview(D_in,(3,3)) # the CORE of the algorithm
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^
all3x3_rows, all3x3_cols, needle_rows, needle_cols = all3x3.shape
print(f'{all3x3_rows=}, {all3x3_cols=}, {needle_rows=}, {needle_cols=}')
noOfHits = 0 # used also to decide about the output if not found Hits
for row in range(all3x3_rows):
for col in range(all3x3_cols):
if np.all(all3x3[row,col,:,:]):
noOfHits += 1
# print(f'3x3 Ones at: {row=}, {col=}')
D_out[row:row+3, col:col+3] = 1
print('--------------------------')
print(D_in)
print('--------------------------')
if noOfHits > 0:
print(D_out)
assert D_out.all() == D_tgt.all() # make sure the result is correct
else:
print(D_in)
gives on output:
all3x3_rows=3, all3x3_cols=13, needle_rows=3, needle_cols=3
--------------------------
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 1]
[3 0 1 0 1 1 1 1 1 1 1 1 1 1 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 1 0]
[2 0 0 0 1 1 1 0 1 0 1 1 1 0 0]
[2 0 1 0 1 1 1 0 1 0 1 0 1 0 0]]
--------------------------
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 0]
[3 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 0 0 0 0 0]]
The same output can be achieved with the one-liner from the answer by mozway. Here some preliminary code:
import numpy as np
data =np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,1],
[3,0,1,0,1,1,1,1,1,1,1,1,1,1,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,1,0],
[2,0,0,0,1,1,1,0,1,0,1,1,1,0,0],
[2,0,1,0,1,1,1,0,1,0,1,0,1,0,0]])
print(data)
print(" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^")
from scipy.signal import convolve2d
from scipy.ndimage import binary_dilation
b = binary_dilation
c = convolve2d
d = data
k = (3,3)
m = data==1 # ; print( m.astype(int) )
n = np.ones
s = 'same'
w = np.where
than
out=w(m,b(c(d==1,n(k),mode=s)==9,n(k)),d)
and finally
print(out)
You can use a 2D convolution on the 1s with a 3×3 kernel of 1s to identify the centers of the 3×3 squares, then dilate them and restore the non 1 numbers
from scipy.signal import convolve2d
from scipy.ndimage import binary_dilation
# get 1s (as boolean)
m = data==1
kernel = np.ones((3, 3))
# get centers
conv = convolve2d(m, kernel, mode='same')
# dilate and restore the other numbers
out = np.where(m, binary_dilation(conv == 9, kernel).astype(int), data)
print(out)
Alternative, erosion and dilation, similarly to my previous answer:
from scipy.ndimage import binary_dilation, binary_erosion
m = data==1
kernel = np.ones((3, 3))
out = np.where(m, binary_dilation(binary_erosion(m, kernel), kernel), data)
Output:
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 0]
[3 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 0 0 0 0 0]]
I have an array for an example:
import numpy as np
data=np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,1],
[3,0,1,0,1,1,1,1,1,1,1,1,1,1,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,1,0],
[2,0,0,0,1,1,1,0,1,0,1,1,1,0,0],
[2,0,1,0,1,1,1,0,1,0,1,0,1,0,0]])
Requirement :
In the data array, if element 1’s are consecutive as the square size
of ((3,3)) and more than square size no changes. Otherwise, replace
element value 1 with zero except the square size.
Expected output :
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 0]
[3 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 0 0 0 0 0]]
current code:
k = np.ones((3,3))
print(k)
jk=binary_dilation(binary_erosion(data==1, k), k)
print(jk)
current output:
[[False False False False True True True True True True False
False False False False]
[False False False False True True True True True True True
True True False False]
[False False False False True True True True True True True
True True False False]
[False False False False True True True False False False True
True True False False]
[False False False False True True True False False False False
False False False False]]
The below proposed solution uses skimage.util.view_as_windows in order to get all of the possible 3×3 views onto the larger array and then looping using them with help of Python loops to set the found hits back to ones after zeroing all the ones in the first step.
It seems that OpenCV, scimage, numpy don’t provide a method able to label areas in which a 3×3 square can move around having all the values beneath set to 1’s, but I am not sure here. So if you read this and know much about dilations, convolutions, etc. please leave a note pointing me in the right direction allowing to label the area of image by a methond implemented in C-code for better speed on huge arrays.
# https://stackoverflow.com/questions/73649612/how-to-change-the-particuler-elements-of-an-array
# Example array:
import numpy as np
D_in =np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,1],
[3,0,1,0,1,1,1,1,1,1,1,1,1,1,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,1,0],
[2,0,0,0,1,1,1,0,1,0,1,1,1,0,0],
[2,0,1,0,1,1,1,0,1,0,1,0,1,0,0]])
# Requirements: If in the D_in array at least one 3x3 square with 1's
# is found, replace in D_in all 1's with 0's except the square size and more than square size no changes.
# Otherwise.
# Expected output :
D_tgt=np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,0],
[3,0,0,0,1,1,1,1,1,1,1,1,1,0,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,0,0],
[2,0,0,0,1,1,1,0,0,0,1,1,1,0,0],
[2,0,0,0,1,1,1,0,0,0,0,0,0,0,0]])
# ----------------------------------------------------------------------
import numpy as np
D_out = np.copy(D_in) # D_out for restoring hit 3x3 ONEs
D_out[D_out==1] = 0 # set all ONEs to zero
needle = np.ones((3,3)) # not necessary here, except documentation
# Create all possible 3x3 needle views on larger D_in heystack:
from skimage.util import view_as_windows as winview
# vvvvvvvvvvvvvvvvvvvvvvvvvvv
all3x3 = winview(D_in,(3,3)) # the CORE of the algorithm
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^
all3x3_rows, all3x3_cols, needle_rows, needle_cols = all3x3.shape
print(f'{all3x3_rows=}, {all3x3_cols=}, {needle_rows=}, {needle_cols=}')
noOfHits = 0 # used also to decide about the output if not found Hits
for row in range(all3x3_rows):
for col in range(all3x3_cols):
if np.all(all3x3[row,col,:,:]):
noOfHits += 1
# print(f'3x3 Ones at: {row=}, {col=}')
D_out[row:row+3, col:col+3] = 1
print('--------------------------')
print(D_in)
print('--------------------------')
if noOfHits > 0:
print(D_out)
assert D_out.all() == D_tgt.all() # make sure the result is correct
else:
print(D_in)
gives on output:
all3x3_rows=3, all3x3_cols=13, needle_rows=3, needle_cols=3
--------------------------
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 1]
[3 0 1 0 1 1 1 1 1 1 1 1 1 1 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 1 0]
[2 0 0 0 1 1 1 0 1 0 1 1 1 0 0]
[2 0 1 0 1 1 1 0 1 0 1 0 1 0 0]]
--------------------------
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 0]
[3 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 0 0 0 0 0]]
The same output can be achieved with the one-liner from the answer by mozway. Here some preliminary code:
import numpy as np
data =np.array([[4,4,4,0,1,1,1,1,1,1,0,0,0,0,1],
[3,0,1,0,1,1,1,1,1,1,1,1,1,1,0],
[6,0,0,0,1,1,1,1,1,1,1,1,1,1,0],
[2,0,0,0,1,1,1,0,1,0,1,1,1,0,0],
[2,0,1,0,1,1,1,0,1,0,1,0,1,0,0]])
print(data)
print(" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^")
from scipy.signal import convolve2d
from scipy.ndimage import binary_dilation
b = binary_dilation
c = convolve2d
d = data
k = (3,3)
m = data==1 # ; print( m.astype(int) )
n = np.ones
s = 'same'
w = np.where
than
out=w(m,b(c(d==1,n(k),mode=s)==9,n(k)),d)
and finally
print(out)
You can use a 2D convolution on the 1s with a 3×3 kernel of 1s to identify the centers of the 3×3 squares, then dilate them and restore the non 1 numbers
from scipy.signal import convolve2d
from scipy.ndimage import binary_dilation
# get 1s (as boolean)
m = data==1
kernel = np.ones((3, 3))
# get centers
conv = convolve2d(m, kernel, mode='same')
# dilate and restore the other numbers
out = np.where(m, binary_dilation(conv == 9, kernel).astype(int), data)
print(out)
Alternative, erosion and dilation, similarly to my previous answer:
from scipy.ndimage import binary_dilation, binary_erosion
m = data==1
kernel = np.ones((3, 3))
out = np.where(m, binary_dilation(binary_erosion(m, kernel), kernel), data)
Output:
[[4 4 4 0 1 1 1 1 1 1 0 0 0 0 0]
[3 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[6 0 0 0 1 1 1 1 1 1 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 1 1 1 0 0]
[2 0 0 0 1 1 1 0 0 0 0 0 0 0 0]]