Find sorrunding elements and element from an array

Question

Life is Wonderful 2020-10-26

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/626663-find-sorrunding-elements-and-element-from-an-array

评论： Life is Wonderful 2021-1-28

I have an array

y = [
 0   0   0   0   0   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   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   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]

where Index 7,41,75 are the locations where 1 is found .

My requirement is

create a block around true(1) with a size of 5
get the indices like 5,6,7,8,9 and data 0 0 1 0 0

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Life is Wonderful 2020-10-27

编辑：Life is Wonderful 2020-10-27

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No no.. this is wrong explanation for edge detection

As per your logic ,

start = max(1, position-2);
stop = start + 4;
we don't know what is the error value at start but position define error.If you train
a edge detection polynomial with leading edge with non error bit. Then the logic won't make 
sense

Below is an example

Leading bit can be 0.

a = [ 1 0 0 0 1 ]; sys = idpoly(a)
sys =
Discrete-time AR model: A(z)y(t) = e(t)
  A(z) = 1 + z^-4                      
                                       
Sample time: unspecified
  
Parameterization:
   Polynomial orders:   na=4
   Number of free coefficients: 4
   Use "polydata", "getpvec", "getcov" for parameters and their uncertainties.
Status:                                                         
Created by direct construction or transformation. Not estimated.
K>> a = [ 0 0 0 0 1 ]; sys = idpoly(a)
Error using idpoly (line 384)
The leading coefficient of the "A" polynomial must be 1.

But for a polynomial leading position must 1 and tail bits can be 0/1. What if block started with 0. So edge detection will not proceed at the initial state.

For the end block , it is Ok to process 1-4 to additional bits to go pass the edge unless we have a conving logic

as I proposed

bit reverse
padd to sufice the block which is a polynomial
short polynomial

for a block to process for edge detection. The problem with Short polynomial size than a block size then we have size error...

Please have re look into below code and suggest solution for ( I prefer to have offset elements to have 0 filled , then we don't alter edge detection error bit and it will be safe)

    if bit_position>= (siz_y-blksize)
        blockBits(:,:,i)  = y(bit_position:1:siz_y);
        blockindcs(:,:,i) =  (bit_position:1:siz_y);
    end

Current implementation with a bad size for end block

y = [
    0   0   0   1   0   0   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   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   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   0]
siz_y = size(y,2);
[Val,idx] = find(y>0);
blksize    = 5;
blockBits  = zeros(1,blksize,size(idx,2));
blockindcs = zeros(1,blksize,size(idx,2));
statelock  = 0;
yMax = movmax(y, blksize);
props = regionprops(yMax > 0, 'PixelIdxList'); % Requires the Image Processing Toolbox.
for i = 1:size(idx,2)
    bit_position = idx(i);
    %     blockBits(:,:,i)  = y(props(i).PixelIdxList);
    %     blockindcs(:,:,i) = props(i).PixelIdxList;
    %     start = max(1, bit_position-2)
    %     stop = start + 4
    %     blockBits(:,:,i)  = y(start:1:stop)
    %     blockindcs(:,:,i) =   (start:1:stop)
    
    if (bit_position>blksize) && bit_position< (siz_y-blksize)
        blockBits(:,:,i)  = y(bit_position:1:bit_position+4);
        blockindcs(:,:,i) =   bit_position:1:bit_position+4;
    end
    if bit_position<=blksize
        blockBits(:,:,i)  = y(bit_position:1:bit_position+4);
        blockindcs(:,:,i) =   bit_position:1:bit_position+4;
    end
    if bit_position>= (siz_y-blksize)
                              % bit_position = 100, 
                              % siz_y        = 101
                              % offset       = 3
                              % Fill blockindcs offset & substitute with 0's
                              % 97    98    99   100   101
                              % blockindcs = 100  101 99 98  97
                              % blockBits  = 1     0   0   0  0
        blockBits(:,:,i)  = y(bit_position:1:siz_y); % fill offset with 0's
        blockindcs(:,:,i) =  (bit_position:1:siz_y);
    end
end

Life is Wonderful 2020-11-19

Hi Adam,

Can you please help me wtih below issue ?

https://www.mathworks.com/matlabcentral/answers/653278-detect-overflow-for-full-range-and-resolution

Thank you!

Life is Wonderful 2020-12-5

Hi Adam,

Can you please help with me below issue

https://www.mathworks.com/matlabcentral/answers/680158-3d-x-y-z-audio-signal

Thank you!

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Answer 1

Image Analyst 2020-10-26

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https://ww2.mathworks.cn/matlabcentral/answers/626663-find-sorrunding-elements-and-element-from-an-array#answer_524973

在 MATLAB Online 中打开

Try this:

fprintf('Beginning to run %s.m ...\n', mfilename);
y = [
	0   0   0   0   0   0   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   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   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]
yMax = movmax(y, 5)
props = regionprops(yMax > 0, 'PixelIdxList') % Requires the Image Processing Toolbox.
for k = 1 : length(props)
	fprintf('\nFor block #%d, indexes = ', k);
	indexes{k} = props(k).PixelIdxList;
	fprintf('%d ', indexes{k});
end
fprintf('\nDone running %s.m ...\n', mfilename);

You'll see:

For block #1, indexes = 5 6 7 8 9

For block #2, indexes = 39 40 41 42 43

For block #3, indexes = 73 74 75 76 77

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Life is Wonderful 2020-10-26

编辑：Life is Wonderful 2020-10-26

在 MATLAB Online 中打开

==================================================================

Updated comment [Image Analyst]

Your code will NOT work in case first elemet is set to 1. The block size will be reduced to 3 instead of 5 and this does not fit anymore to a predefined block size ( as you took 5 in movmax fcn)

For block #1, indexes = 1 2 3 ==> Bug "doesn't fit in a block"

For block #2, indexes = 5 6 7 8 9

For block #3, indexes = 39 40 41 42 43

For block #4, indexes = 73 74 75 76 77

Done running .m ...

Probably You have to change properties of movmax fcn

==================================================================

Looks good! Since I don't have a image processing toolbox to work on. Could you please help in imroving the below code?

siz_y = size(y,2);
[Val,idx] = find(y>0);
blockBits  = zeros(1,5,size(idx,2));
blockindcs = zeros(1,5,size(idx,2));
for i = 1:size(idx,2)
    a = idx(i);
    if a>1 && a< siz_y
         blockBits(:,:,i)  = y(a-2:1:a+2);   
         blockindcs(:,:,i) =   a-2:1:a+2;
    elseif a==1  
         blockBits(:,:,i)  = y(a:1:a+4);   
         blockindcs(:,:,i) =   (a:1:a+4);
    elseif a== siz_y       
         blockBits(:,:,i)  = y(a-4:1:a);   
         blockindcs(:,:,i) =  (a-4:1:a);   
    end
end

Image Analyst 2020-10-27

编辑：Image Analyst 2020-10-27

在 MATLAB Online 中打开

So did you end up using my code? Or did you change it to use findgroups()? If so, attach that code.

If you're able to use the Image Processing Toolbox, you can use bwareaopen() to get rid of regions less than 5 in length:

fprintf('Beginning to run %s.m ...\n', mfilename);
y = [
	0   0   0   0   0   0   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   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   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]
yMax = movmax(y, 5) > 0; % A logical vector.
yMax = bwareaopen(yMax, 5); % Keep runs of 5 or longer ONLY.
props = regionprops(yMax, 'PixelIdxList') % Requires the Image Processing Toolbox.
for k = 1 : length(props)
	fprintf('\nFor block #%d, indexes = ', k);
	blockIndexes{k} = props(k).PixelIdxList;
	fprintf('%d ', blockIndexes{k});
end
fprintf('\nDone running %s.m ...\n', mfilename);

You'll see

For block #1, indexes = 5 6 7 8 9

For block #2, indexes = 39 40 41 42 43

For block #3, indexes = 73 74 75 76 77

Life is Wonderful 2020-10-27

编辑：Life is Wonderful 2020-10-27

在 MATLAB Online 中打开

Looks pretty good, But still improvement is required. Following Issues are observed & listed below

For block #1, indexes = 5 6 7 8 9 
For block #2, indexes = 39 40 41 42 43 
For block #3, indexes = 73 74 75 76 77 

Issue -1:

Starting index is expected to be error bit i.e.I am expecting like below

For block #1, indexes = 7 8 9  6 5
For block #2, indexes = 41 42 43 40 39
For block #3, indexes = 75 76 77  74 73

Issue -2:

There is also a case where we have higher than block size is achieved. This mean higher polynomial order. This will NOT fit into pre-allocated buffer

[For block #1, indexes = 1 2 3 4 5 6 7 8 9 ]

Issue -3:

In case , if I modify y = errors positions(set to 1), I don't get the required output, For example, You can give a try by setting 1 and 101 locations ,

For block #1, indexes = 5 6 7 8 9 
For block #2, indexes = 39 40 41 42 43 
For block #3, indexes = 73 74 75 76 77 
Done running .m ...
>> find(y>0)
ans =
     1     7    41    75   101

Also please keep "bwareaopen" function c code translation, coder support is needed for c/c++ code generation, otherwise it is pretty difficult to implement on a 32 bit processor.

Thank you!

Life is Wonderful 2020-10-28

在 MATLAB Online 中打开

What is the case where you had indexes 1-9?

In case - data altered for testing & verification

y = [
	0   1   0   0   0   0   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   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   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   1]
props = 
  3×1 struct array with fields:
    PixelIdxList
For block #1, indexes = 1 2 3 4 5 6 7 8 9 

What do you expect to happen if you have 1's that are closer than 5 elements?

It's OK and they may be , after all 1 are the error bits. The expectation is , we should have constant block size and we don't need to process the already processed error bit in a block . Block must start with a error bit ( leading coefficient of a the polynomial must be 1). And block size should not change. In case block does

not accomadete all error bit, then process the error bit in next block . We should not repeat the already processed error bit.

Is the solution to just add 2 to all the indexes?

No. If I understood your question correctly , you say adding/pad 2 additional bits. then my answer is you can do but finally we are not altering the output

Area you saying that bwareaopen() is a function that is explicitly not included in the MATLAB Coder Toolbox?

No. I am saying bwareaopen(), I have to write the c code that might be difficult for me if i don't understand the details of bwareaopen()

Image Analyst 2020-10-29

在 MATLAB Online 中打开

You said "Also please keep "bwareaopen" function c code translation, coder support is needed for c/c++ code generation, otherwise it is pretty difficult to implement on a 32 bit processor." so I was assuming that you needed the whole program to be converted to C code with the Coder Toolbox because you were going to embed this algorithm in a custom chip on some device. If you don't need Coder support and are not going to generate C code, then I don't know why you said that. Please elaboarate. Also, I don't know that Coder can generate code for 32 bit processors. I know MATLAB stopped supporting 32 bit processors with version 2016b.

If you want

For block #1, indexes = 7 8 9 6 5

For block #2, indexes = 41 42 43 40 39

For block #3, indexes = 75 76 77 74 73

then you can just tack on the first two indexes to the end, like this:

fprintf('Beginning to run %s.m ...\n', mfilename);
y = [
	0   0   0   0   0   0   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   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   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]
windowWidth = 5;
halfWindowWidth = floor(windowWidth/2);
yMax = movmax(y, windowWidth) > 0; % A logical vector.
yMax = bwareaopen(yMax, windowWidth); % Keep runs of 5 or longer ONLY.
props = regionprops(yMax, 'PixelIdxList') % Requires the Image Processing Toolbox.
for k = 1 : length(props)
	fprintf('\nFor block #%d, indexes = ', k);
	theseIndexes = props(k).PixelIdxList';
	blockIndexes{k} = [theseIndexes(end-halfWindowWidth : end), fliplr(theseIndexes(1:halfWindowWidth))];
	fprintf('%d ', blockIndexes{k});
end
fprintf('\nDone running %s.m ...\n', mfilename);

It prints out:

For block #1, indexes = 7 8 9 6 5

For block #2, indexes = 41 42 43 40 39

For block #3, indexes = 75 76 77 74 73

Life is Wonderful 2020-11-8

Thank you! I'II update further with my comments on your queries with additional information. For the time being this is sufficient

Life is Wonderful 2021-1-28

Can you please help me with below query?

https://www.mathworks.com/matlabcentral/answers/729113-colon-expression-to-use-integer-operands/?s_tid=ans_lp_feed_leaf

Thank you!

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Find sorrunding elements and element from an array

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Find sorrunding elements and element from an array

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