Reducing matrix size in 2D using KNN
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I have a large binary matrix. I want to reduce the size of this matrix by using knn-approximation. What my idea is to cluster the matrix in groups of 4 neighbors and replace the group with a 1, if the number of 1s in the group is greater than or equal to the number of zeros.
To be concrete, let the matrix be
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0
First I want to create neighborhood group as
1 0 |0 1| 0|
0 1 |1 0| 0|
------------
1 1 |0 0| 0|
0 1 |1 1| 0|
------------
0 0 |1 1| 0|
------------
and then the final matrix I want to generate is
1 1 0
1 1 0
0 1 0
by replacing the group with the majority score. How can I efficiently do this is MATLAB?
matlab matrix binary-data knn downsampling
edited Nov 11 at 9:10
Dev-iL
16.3k64074
asked Nov 11 at 8:25
Shew
4971516
add a comment |
up vote
0
down vote
favorite
I have a large binary matrix. I want to reduce the size of this matrix by using knn-approximation. What my idea is to cluster the matrix in groups of 4 neighbors and replace the group with a 1, if the number of 1s in the group is greater than or equal to the number of zeros.
To be concrete, let the matrix be
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0
First I want to create neighborhood group as
1 0 |0 1| 0|
0 1 |1 0| 0|
------------
1 1 |0 0| 0|
0 1 |1 1| 0|
------------
0 0 |1 1| 0|
------------
and then the final matrix I want to generate is
1 1 0
1 1 0
0 1 0
by replacing the group with the majority score. How can I efficiently do this is MATLAB?
matlab matrix binary-data knn downsampling
edited Nov 11 at 9:10
Dev-iL
16.3k64074
asked Nov 11 at 8:25
Shew
4971516
Several questions: 1) Will your matrix always divide with a remainder into blocks? 2) Which MATLAB version are you using? 3) Do you have the image processing toolbox? 4) Did you try something already (e.g.mat2cell)?
– Dev-iL
Nov 11 at 8:37
1) nope. as in figure, the remainder is grouped into the remaining size. 2) I am using version 2016b. 3) Yes, I have image processing toolbox. 4) nope. i did not try anything other than splitting using loop.
– Shew
Nov 11 at 8:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a large binary matrix. I want to reduce the size of this matrix by using knn-approximation. What my idea is to cluster the matrix in groups of 4 neighbors and replace the group with a 1, if the number of 1s in the group is greater than or equal to the number of zeros.
To be concrete, let the matrix be
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0
First I want to create neighborhood group as
1 0 |0 1| 0|
0 1 |1 0| 0|
------------
1 1 |0 0| 0|
0 1 |1 1| 0|
------------
0 0 |1 1| 0|
------------
and then the final matrix I want to generate is
1 1 0
1 1 0
0 1 0
by replacing the group with the majority score. How can I efficiently do this is MATLAB?
matlab matrix binary-data knn downsampling
edited Nov 11 at 9:10
Dev-iL
16.3k64074
asked Nov 11 at 8:25
Shew
4971516
I have a large binary matrix. I want to reduce the size of this matrix by using knn-approximation. What my idea is to cluster the matrix in groups of 4 neighbors and replace the group with a 1, if the number of 1s in the group is greater than or equal to the number of zeros.
To be concrete, let the matrix be
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0
First I want to create neighborhood group as
1 0 |0 1| 0|
0 1 |1 0| 0|
------------
1 1 |0 0| 0|
0 1 |1 1| 0|
------------
0 0 |1 1| 0|
------------
and then the final matrix I want to generate is
1 1 0
1 1 0
0 1 0
by replacing the group with the majority score. How can I efficiently do this is MATLAB?
matlab matrix binary-data knn downsampling
matlab matrix binary-data knn downsampling
edited Nov 11 at 9:10
Dev-iL
16.3k64074
asked Nov 11 at 8:25
Shew
4971516
edited Nov 11 at 9:10
Dev-iL
16.3k64074
asked Nov 11 at 8:25
Shew
4971516
edited Nov 11 at 9:10
Dev-iL
16.3k64074
edited Nov 11 at 9:10
Dev-iL
16.3k64074
edited Nov 11 at 9:10
Dev-iL
16.3k64074
16.3k64074
asked Nov 11 at 8:25
Shew
4971516
asked Nov 11 at 8:25
Shew
4971516
asked Nov 11 at 8:25
Shew
4971516
4971516
Several questions: 1) Will your matrix always divide with a remainder into blocks? 2) Which MATLAB version are you using? 3) Do you have the image processing toolbox? 4) Did you try something already (e.g.mat2cell)?
– Dev-iL
Nov 11 at 8:37
1) nope. as in figure, the remainder is grouped into the remaining size. 2) I am using version 2016b. 3) Yes, I have image processing toolbox. 4) nope. i did not try anything other than splitting using loop.
– Shew
Nov 11 at 8:40
add a comment |
Several questions: 1) Will your matrix always divide with a remainder into blocks? 2) Which MATLAB version are you using? 3) Do you have the image processing toolbox? 4) Did you try something already (e.g.mat2cell)?
– Dev-iL
Nov 11 at 8:37
1) nope. as in figure, the remainder is grouped into the remaining size. 2) I am using version 2016b. 3) Yes, I have image processing toolbox. 4) nope. i did not try anything other than splitting using loop.
– Shew
Nov 11 at 8:40
Several questions: 1) Will your matrix always divide with a remainder into blocks? 2) Which MATLAB version are you using? 3) Do you have the image processing toolbox? 4) Did you try something already (e.g.
mat2cell)?– Dev-iL
Nov 11 at 8:37
Several questions: 1) Will your matrix always divide with a remainder into blocks? 2) Which MATLAB version are you using? 3) Do you have the image processing toolbox? 4) Did you try something already (e.g.
mat2cell)?– Dev-iL
Nov 11 at 8:37
1) nope. as in figure, the remainder is grouped into the remaining size. 2) I am using version 2016b. 3) Yes, I have image processing toolbox. 4) nope. i did not try anything other than splitting using loop.
– Shew
Nov 11 at 8:40
1) nope. as in figure, the remainder is grouped into the remaining size. 2) I am using version 2016b. 3) Yes, I have image processing toolbox. 4) nope. i did not try anything other than splitting using loop.
– Shew
Nov 11 at 8:40
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Initially I tried getting it to work using imresize but couldn't quite get it without "hacks" (it was off by 1 value in all my "proper" attempts).
imresize(M, ceil(size(M)/2), 'bilinear') >= 0.4 % This works but is hacky and not recommended!
However, I can think of a way to solve this using 2D convolution. Note that I pad the array (which requires a toolbox) in order to simplify the indexing stage at the end:
function C = q53247013(M)
if nargin < 1
M = [
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0];
end
% Constants:
BLK_SZ = 2;
A = ones(BLK_SZ);
% Pad array if needed (note: this WILL require modification if BLK_SZ > 2 ):
padBottom = rem(size(M,BLK_SZ),1);
padRight = rem(size(M,BLK_SZ),2);
M = padarray(M, [padBottom, padRight], 'replicate', 'post');
% Perform convolution:
C = conv2(M, A, 'valid') >= ceil(BLK_SZ^2 / 2);
% Remove every other row and column:
C = C(1:2:end, 1:2:end);
Another alternative is the blockproc function:
C = blockproc(M, [2 2], @(x)sum(x.data(:))) >= 2;
answered Nov 11 at 9:05
Dev-iL
16.3k64074
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Initially I tried getting it to work using imresize but couldn't quite get it without "hacks" (it was off by 1 value in all my "proper" attempts).
imresize(M, ceil(size(M)/2), 'bilinear') >= 0.4 % This works but is hacky and not recommended!
However, I can think of a way to solve this using 2D convolution. Note that I pad the array (which requires a toolbox) in order to simplify the indexing stage at the end:
function C = q53247013(M)
if nargin < 1
M = [
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0];
end
% Constants:
BLK_SZ = 2;
A = ones(BLK_SZ);
% Pad array if needed (note: this WILL require modification if BLK_SZ > 2 ):
padBottom = rem(size(M,BLK_SZ),1);
padRight = rem(size(M,BLK_SZ),2);
M = padarray(M, [padBottom, padRight], 'replicate', 'post');
% Perform convolution:
C = conv2(M, A, 'valid') >= ceil(BLK_SZ^2 / 2);
% Remove every other row and column:
C = C(1:2:end, 1:2:end);
Another alternative is the blockproc function:
C = blockproc(M, [2 2], @(x)sum(x.data(:))) >= 2;
answered Nov 11 at 9:05
Dev-iL
16.3k64074
add a comment |
up vote
1
down vote
accepted
Initially I tried getting it to work using imresize but couldn't quite get it without "hacks" (it was off by 1 value in all my "proper" attempts).
imresize(M, ceil(size(M)/2), 'bilinear') >= 0.4 % This works but is hacky and not recommended!
However, I can think of a way to solve this using 2D convolution. Note that I pad the array (which requires a toolbox) in order to simplify the indexing stage at the end:
function C = q53247013(M)
if nargin < 1
M = [
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0];
end
% Constants:
BLK_SZ = 2;
A = ones(BLK_SZ);
% Pad array if needed (note: this WILL require modification if BLK_SZ > 2 ):
padBottom = rem(size(M,BLK_SZ),1);
padRight = rem(size(M,BLK_SZ),2);
M = padarray(M, [padBottom, padRight], 'replicate', 'post');
% Perform convolution:
C = conv2(M, A, 'valid') >= ceil(BLK_SZ^2 / 2);
% Remove every other row and column:
C = C(1:2:end, 1:2:end);
Another alternative is the blockproc function:
C = blockproc(M, [2 2], @(x)sum(x.data(:))) >= 2;
answered Nov 11 at 9:05
Dev-iL
16.3k64074
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Initially I tried getting it to work using imresize but couldn't quite get it without "hacks" (it was off by 1 value in all my "proper" attempts).
imresize(M, ceil(size(M)/2), 'bilinear') >= 0.4 % This works but is hacky and not recommended!
However, I can think of a way to solve this using 2D convolution. Note that I pad the array (which requires a toolbox) in order to simplify the indexing stage at the end:
function C = q53247013(M)
if nargin < 1
M = [
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0];
end
% Constants:
BLK_SZ = 2;
A = ones(BLK_SZ);
% Pad array if needed (note: this WILL require modification if BLK_SZ > 2 ):
padBottom = rem(size(M,BLK_SZ),1);
padRight = rem(size(M,BLK_SZ),2);
M = padarray(M, [padBottom, padRight], 'replicate', 'post');
% Perform convolution:
C = conv2(M, A, 'valid') >= ceil(BLK_SZ^2 / 2);
% Remove every other row and column:
C = C(1:2:end, 1:2:end);
Another alternative is the blockproc function:
C = blockproc(M, [2 2], @(x)sum(x.data(:))) >= 2;
answered Nov 11 at 9:05
Dev-iL
16.3k64074
Initially I tried getting it to work using imresize but couldn't quite get it without "hacks" (it was off by 1 value in all my "proper" attempts).
imresize(M, ceil(size(M)/2), 'bilinear') >= 0.4 % This works but is hacky and not recommended!
However, I can think of a way to solve this using 2D convolution. Note that I pad the array (which requires a toolbox) in order to simplify the indexing stage at the end:
function C = q53247013(M)
if nargin < 1
M = [
1 0 0 1 0
0 1 1 0 0
1 1 0 0 0
0 1 1 1 0
0 0 1 1 0
1 0 0 1 0];
end
% Constants:
BLK_SZ = 2;
A = ones(BLK_SZ);
% Pad array if needed (note: this WILL require modification if BLK_SZ > 2 ):
padBottom = rem(size(M,BLK_SZ),1);
padRight = rem(size(M,BLK_SZ),2);
M = padarray(M, [padBottom, padRight], 'replicate', 'post');
% Perform convolution:
C = conv2(M, A, 'valid') >= ceil(BLK_SZ^2 / 2);
% Remove every other row and column:
C = C(1:2:end, 1:2:end);
Another alternative is the blockproc function:
C = blockproc(M, [2 2], @(x)sum(x.data(:))) >= 2;
answered Nov 11 at 9:05
Dev-iL
16.3k64074
edited Nov 11 at 9:27
answered Nov 11 at 9:05
Dev-iL
16.3k64074
answered Nov 11 at 9:05
Dev-iL
16.3k64074
answered Nov 11 at 9:05
Dev-iL
16.3k64074
16.3k64074
add a comment |
add a comment |
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Several questions: 1) Will your matrix always divide with a remainder into blocks? 2) Which MATLAB version are you using? 3) Do you have the image processing toolbox? 4) Did you try something already (e.g.
mat2cell)?– Dev-iL
Nov 11 at 8:37
1) nope. as in figure, the remainder is grouped into the remaining size. 2) I am using version 2016b. 3) Yes, I have image processing toolbox. 4) nope. i did not try anything other than splitting using loop.
– Shew
Nov 11 at 8:40