numpy mask creation broadcasting
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1
I have n-size of image (n,row,col) and n-size of floats (n,1).
what I want to do is to create mask of 0 and 1 of (row, col) size. 1's in center and 0's in edges. size of 1's is according to weights.
example
>>> immask = np.zeros((2,8,8))
[[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
>>> multiplier = np.array([16./64,32./64])
[0.25 0.5 ]
#**insert magic here**
# expected result :
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
is there any way to do this with broadcasting? not using loop. thanks in advance.
python numpy image-processing
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
add a comment |
1
I have n-size of image (n,row,col) and n-size of floats (n,1).
what I want to do is to create mask of 0 and 1 of (row, col) size. 1's in center and 0's in edges. size of 1's is according to weights.
example
>>> immask = np.zeros((2,8,8))
[[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
>>> multiplier = np.array([16./64,32./64])
[0.25 0.5 ]
#**insert magic here**
# expected result :
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
is there any way to do this with broadcasting? not using loop. thanks in advance.
python numpy image-processing
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
1
What have you tried so far? And how should your weights affect the amount of 1's? Is the factor in multiplier the ratio of 1's per row/colum?
– Scotty1-
Jan 4 at 15:11
so far much like @roland-smith answer, using looping on range(n), using half of image row and col, substracted by it's ratio, slicing through like it's from (+)left/up : (-)right/down, assign them by 1. immask[n,(col-(multiplier[n]*col)//2):-(col-(multiplier[n]*col)//2),(row-(multiplier[n]*row)//2):-(row-(multiplier[n]*row)//2)] (forgetting the details, but something like that)
– itonia.x.i
Jan 4 at 16:41
add a comment |
1
1
1
I have n-size of image (n,row,col) and n-size of floats (n,1).
what I want to do is to create mask of 0 and 1 of (row, col) size. 1's in center and 0's in edges. size of 1's is according to weights.
example
>>> immask = np.zeros((2,8,8))
[[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
>>> multiplier = np.array([16./64,32./64])
[0.25 0.5 ]
#**insert magic here**
# expected result :
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
is there any way to do this with broadcasting? not using loop. thanks in advance.
python numpy image-processing
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
I have n-size of image (n,row,col) and n-size of floats (n,1).
what I want to do is to create mask of 0 and 1 of (row, col) size. 1's in center and 0's in edges. size of 1's is according to weights.
example
>>> immask = np.zeros((2,8,8))
[[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]
[[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
>>> multiplier = np.array([16./64,32./64])
[0.25 0.5 ]
#**insert magic here**
# expected result :
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
is there any way to do this with broadcasting? not using loop. thanks in advance.
python numpy image-processing
python numpy image-processing
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
asked Jan 4 at 15:03
itonia.x.iitonia.x.i
103
103
1
What have you tried so far? And how should your weights affect the amount of 1's? Is the factor in multiplier the ratio of 1's per row/colum?
– Scotty1-
Jan 4 at 15:11
so far much like @roland-smith answer, using looping on range(n), using half of image row and col, substracted by it's ratio, slicing through like it's from (+)left/up : (-)right/down, assign them by 1. immask[n,(col-(multiplier[n]*col)//2):-(col-(multiplier[n]*col)//2),(row-(multiplier[n]*row)//2):-(row-(multiplier[n]*row)//2)] (forgetting the details, but something like that)
– itonia.x.i
Jan 4 at 16:41
add a comment |
1
What have you tried so far? And how should your weights affect the amount of 1's? Is the factor in multiplier the ratio of 1's per row/colum?
– Scotty1-
Jan 4 at 15:11
so far much like @roland-smith answer, using looping on range(n), using half of image row and col, substracted by it's ratio, slicing through like it's from (+)left/up : (-)right/down, assign them by 1. immask[n,(col-(multiplier[n]*col)//2):-(col-(multiplier[n]*col)//2),(row-(multiplier[n]*row)//2):-(row-(multiplier[n]*row)//2)] (forgetting the details, but something like that)
– itonia.x.i
Jan 4 at 16:41
1
1
What have you tried so far? And how should your weights affect the amount of 1's? Is the factor in multiplier the ratio of 1's per row/colum?
– Scotty1-
Jan 4 at 15:11
What have you tried so far? And how should your weights affect the amount of 1's? Is the factor in multiplier the ratio of 1's per row/colum?
– Scotty1-
Jan 4 at 15:11
so far much like @roland-smith answer, using looping on range(n), using half of image row and col, substracted by it's ratio, slicing through like it's from (+)left/up : (-)right/down, assign them by 1. immask[n,(col-(multiplier[n]*col)//2):-(col-(multiplier[n]*col)//2),(row-(multiplier[n]*row)//2):-(row-(multiplier[n]*row)//2)] (forgetting the details, but something like that)
– itonia.x.i
Jan 4 at 16:41
so far much like @roland-smith answer, using looping on range(n), using half of image row and col, substracted by it's ratio, slicing through like it's from (+)left/up : (-)right/down, assign them by 1. immask[n,(col-(multiplier[n]*col)//2):-(col-(multiplier[n]*col)//2),(row-(multiplier[n]*row)//2):-(row-(multiplier[n]*row)//2)] (forgetting the details, but something like that)
– itonia.x.i
Jan 4 at 16:41
add a comment |
3 Answers
3
active
oldest
votes
0
This function does something like that:
import numpy as np
def make_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Make masks where criteria is met
return (np.abs(r) <= w) & (np.abs(c) <= w)
# Test
masks = make_masks([0.25, 0.5], 8)
# Convert type as needed
masks_f = masks.astype(np.float32)
print(masks_f)
Output:
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
EDIT
To make circle-shaped masks, the function could be as follows:
import numpy as np
def make_circle_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Arrays with distances to centre
d = r * r + c * c
# Make masks where criteria is met
return d <= w * w
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
thank you, this is what I'm looking for. can 1's be shaped to circle?
– itonia.x.i
Jan 4 at 17:08
@itonia.x.i I have added a version for circle-shaped masks.
– jdehesa
Jan 4 at 17:14
add a comment |
0
I think if you want to use some kind of matrix to be applied over a picture this:
from scipy.ndimage.filters import convolve
could work for you. You could take a deeper look here how this works.
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
noted. what I really want to do is to divide a image into 8x8 images, then only get center of image according to it's weights/percents, so it looks like images but dotted-like and vary on radius of the dot-like.
– itonia.x.i
Jan 4 at 16:49
add a comment |
0
Try this:
In [1]: import numpy as np
In [3]: immask = np.zeros((2,8,8))
In [6]: immask[0, 3:5, 3:5] = 1
In [7]: immask[1, 2:6, 2:6] = 1
In [8]: immask
Out[8]:
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., 1., 1., 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., 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., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.]]])
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
nice, I can use loop on 'n' (i.e. 2), and using it's negative instead immask[0, 3:-3, 3:-3] = 1. it is easy to read in loop but it will take long since I have many 'n'.
– itonia.x.i
Jan 4 at 16:44
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
This function does something like that:
import numpy as np
def make_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Make masks where criteria is met
return (np.abs(r) <= w) & (np.abs(c) <= w)
# Test
masks = make_masks([0.25, 0.5], 8)
# Convert type as needed
masks_f = masks.astype(np.float32)
print(masks_f)
Output:
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
EDIT
To make circle-shaped masks, the function could be as follows:
import numpy as np
def make_circle_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Arrays with distances to centre
d = r * r + c * c
# Make masks where criteria is met
return d <= w * w
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
thank you, this is what I'm looking for. can 1's be shaped to circle?
– itonia.x.i
Jan 4 at 17:08
@itonia.x.i I have added a version for circle-shaped masks.
– jdehesa
Jan 4 at 17:14
add a comment |
0
This function does something like that:
import numpy as np
def make_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Make masks where criteria is met
return (np.abs(r) <= w) & (np.abs(c) <= w)
# Test
masks = make_masks([0.25, 0.5], 8)
# Convert type as needed
masks_f = masks.astype(np.float32)
print(masks_f)
Output:
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
EDIT
To make circle-shaped masks, the function could be as follows:
import numpy as np
def make_circle_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Arrays with distances to centre
d = r * r + c * c
# Make masks where criteria is met
return d <= w * w
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
thank you, this is what I'm looking for. can 1's be shaped to circle?
– itonia.x.i
Jan 4 at 17:08
@itonia.x.i I have added a version for circle-shaped masks.
– jdehesa
Jan 4 at 17:14
add a comment |
0
0
0
This function does something like that:
import numpy as np
def make_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Make masks where criteria is met
return (np.abs(r) <= w) & (np.abs(c) <= w)
# Test
masks = make_masks([0.25, 0.5], 8)
# Convert type as needed
masks_f = masks.astype(np.float32)
print(masks_f)
Output:
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
EDIT
To make circle-shaped masks, the function could be as follows:
import numpy as np
def make_circle_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Arrays with distances to centre
d = r * r + c * c
# Make masks where criteria is met
return d <= w * w
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
This function does something like that:
import numpy as np
def make_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Make masks where criteria is met
return (np.abs(r) <= w) & (np.abs(c) <= w)
# Test
masks = make_masks([0.25, 0.5], 8)
# Convert type as needed
masks_f = masks.astype(np.float32)
print(masks_f)
Output:
[[[0. 0. 0. 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. 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. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 1. 1. 1. 1. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0.]]]
EDIT
To make circle-shaped masks, the function could be as follows:
import numpy as np
def make_circle_masks(weights, rows, cols=None):
if cols is None:
cols = rows
# Add two dimensions to weights
w = np.asarray(weights)[:, np.newaxis, np.newaxis]
# Open grid of row and column coordinates
r, c = np.ogrid[-1:1:rows * 1j, -1:1:cols * 1j]
# Arrays with distances to centre
d = r * r + c * c
# Make masks where criteria is met
return d <= w * w
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
edited Jan 4 at 17:13
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
answered Jan 4 at 15:37
jdehesajdehesa
27.8k43759
27.8k43759
thank you, this is what I'm looking for. can 1's be shaped to circle?
– itonia.x.i
Jan 4 at 17:08
@itonia.x.i I have added a version for circle-shaped masks.
– jdehesa
Jan 4 at 17:14
add a comment |
thank you, this is what I'm looking for. can 1's be shaped to circle?
– itonia.x.i
Jan 4 at 17:08
@itonia.x.i I have added a version for circle-shaped masks.
– jdehesa
Jan 4 at 17:14
thank you, this is what I'm looking for. can 1's be shaped to circle?
– itonia.x.i
Jan 4 at 17:08
thank you, this is what I'm looking for. can 1's be shaped to circle?
– itonia.x.i
Jan 4 at 17:08
@itonia.x.i I have added a version for circle-shaped masks.
– jdehesa
Jan 4 at 17:14
@itonia.x.i I have added a version for circle-shaped masks.
– jdehesa
Jan 4 at 17:14
add a comment |
0
I think if you want to use some kind of matrix to be applied over a picture this:
from scipy.ndimage.filters import convolve
could work for you. You could take a deeper look here how this works.
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
noted. what I really want to do is to divide a image into 8x8 images, then only get center of image according to it's weights/percents, so it looks like images but dotted-like and vary on radius of the dot-like.
– itonia.x.i
Jan 4 at 16:49
add a comment |
0
I think if you want to use some kind of matrix to be applied over a picture this:
from scipy.ndimage.filters import convolve
could work for you. You could take a deeper look here how this works.
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
noted. what I really want to do is to divide a image into 8x8 images, then only get center of image according to it's weights/percents, so it looks like images but dotted-like and vary on radius of the dot-like.
– itonia.x.i
Jan 4 at 16:49
add a comment |
0
0
0
I think if you want to use some kind of matrix to be applied over a picture this:
from scipy.ndimage.filters import convolve
could work for you. You could take a deeper look here how this works.
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
I think if you want to use some kind of matrix to be applied over a picture this:
from scipy.ndimage.filters import convolve
could work for you. You could take a deeper look here how this works.
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
answered Jan 4 at 15:18
default_settingsdefault_settings
5517
5517
noted. what I really want to do is to divide a image into 8x8 images, then only get center of image according to it's weights/percents, so it looks like images but dotted-like and vary on radius of the dot-like.
– itonia.x.i
Jan 4 at 16:49
add a comment |
noted. what I really want to do is to divide a image into 8x8 images, then only get center of image according to it's weights/percents, so it looks like images but dotted-like and vary on radius of the dot-like.
– itonia.x.i
Jan 4 at 16:49
noted. what I really want to do is to divide a image into 8x8 images, then only get center of image according to it's weights/percents, so it looks like images but dotted-like and vary on radius of the dot-like.
– itonia.x.i
Jan 4 at 16:49
noted. what I really want to do is to divide a image into 8x8 images, then only get center of image according to it's weights/percents, so it looks like images but dotted-like and vary on radius of the dot-like.
– itonia.x.i
Jan 4 at 16:49
add a comment |
0
Try this:
In [1]: import numpy as np
In [3]: immask = np.zeros((2,8,8))
In [6]: immask[0, 3:5, 3:5] = 1
In [7]: immask[1, 2:6, 2:6] = 1
In [8]: immask
Out[8]:
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., 1., 1., 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., 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., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.]]])
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
nice, I can use loop on 'n' (i.e. 2), and using it's negative instead immask[0, 3:-3, 3:-3] = 1. it is easy to read in loop but it will take long since I have many 'n'.
– itonia.x.i
Jan 4 at 16:44
add a comment |
0
Try this:
In [1]: import numpy as np
In [3]: immask = np.zeros((2,8,8))
In [6]: immask[0, 3:5, 3:5] = 1
In [7]: immask[1, 2:6, 2:6] = 1
In [8]: immask
Out[8]:
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., 1., 1., 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., 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., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.]]])
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
nice, I can use loop on 'n' (i.e. 2), and using it's negative instead immask[0, 3:-3, 3:-3] = 1. it is easy to read in loop but it will take long since I have many 'n'.
– itonia.x.i
Jan 4 at 16:44
add a comment |
0
0
0
Try this:
In [1]: import numpy as np
In [3]: immask = np.zeros((2,8,8))
In [6]: immask[0, 3:5, 3:5] = 1
In [7]: immask[1, 2:6, 2:6] = 1
In [8]: immask
Out[8]:
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., 1., 1., 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., 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., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.]]])
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
Try this:
In [1]: import numpy as np
In [3]: immask = np.zeros((2,8,8))
In [6]: immask[0, 3:5, 3:5] = 1
In [7]: immask[1, 2:6, 2:6] = 1
In [8]: immask
Out[8]:
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., 1., 1., 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., 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., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 1., 1., 1., 1., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.]]])
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
answered Jan 4 at 15:24
Roland SmithRoland Smith
27.1k33257
27.1k33257
nice, I can use loop on 'n' (i.e. 2), and using it's negative instead immask[0, 3:-3, 3:-3] = 1. it is easy to read in loop but it will take long since I have many 'n'.
– itonia.x.i
Jan 4 at 16:44
add a comment |
nice, I can use loop on 'n' (i.e. 2), and using it's negative instead immask[0, 3:-3, 3:-3] = 1. it is easy to read in loop but it will take long since I have many 'n'.
– itonia.x.i
Jan 4 at 16:44
nice, I can use loop on 'n' (i.e. 2), and using it's negative instead immask[0, 3:-3, 3:-3] = 1. it is easy to read in loop but it will take long since I have many 'n'.
– itonia.x.i
Jan 4 at 16:44
nice, I can use loop on 'n' (i.e. 2), and using it's negative instead immask[0, 3:-3, 3:-3] = 1. it is easy to read in loop but it will take long since I have many 'n'.
– itonia.x.i
Jan 4 at 16:44
add a comment |
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1
What have you tried so far? And how should your weights affect the amount of 1's? Is the factor in multiplier the ratio of 1's per row/colum?
– Scotty1-
Jan 4 at 15:11
so far much like @roland-smith answer, using looping on range(n), using half of image row and col, substracted by it's ratio, slicing through like it's from (+)left/up : (-)right/down, assign them by 1. immask[n,(col-(multiplier[n]*col)//2):-(col-(multiplier[n]*col)//2),(row-(multiplier[n]*row)//2):-(row-(multiplier[n]*row)//2)] (forgetting the details, but something like that)
– itonia.x.i
Jan 4 at 16:41