Matrix processing problem; creating new matrix

Question

Laura Steel 2023-1-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1894905-matrix-processing-problem-creating-new-matrix

评论： Kevin Holly 2023-2-3

Hi,

I have the following experimental set up. An object is able to move between three areas (A, B and C). It can only access C via B.

I am interested in knowing at what point the object moved from C to B, and then whether it moved onto A or back to C.

i.e. if the data matrix looked like the following:

0 0 0

0 0 1

0 1 0

1 0 0

0 1 0

0 0 1

0 1 0

0 0 1

0 0 0

I would want to output a new matrix:

0

1 (marking C to B transition)

0

2 (marking B to A transition, following a C to B transition)

0

1 (marking C to B transition)

3 (marking B to C transition, following a C to B transition)

0

I already have the following code to mark transitions from C to B:

matrix =[0 0; 0 1; 1 0; 0 1; 0 1; 0 1; 1 0]; % column 1 = B, column 2 = C

if matrix(1,1)

output1 = nan;

else

output1 = 0;

end

output2end = matrix(2:end,1) & matrix(1:end-1,2);

output_Cage1 = [output1; output2end];

output_Cage1 =

0

1

0

1

But I am not sure how to extend this to work for 3 columns. Any help would be much appreciated. Many thanks.

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Answer 1

Kevin Holly 2023-1-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1894905-matrix-processing-problem-creating-new-matrix#answer_1149655

编辑：Kevin Holly 2023-1-16

在 MATLAB Online 中打开

matrix = [0 0 0 
0 0 1
0 1 0
0 1 0
1 0 0
0 1 0
0 1 0
0 0 1
0 1 0
0 0 1
0 0 1
0 0 0];
if matrix(1,1)
    output1 = nan;
else
    output1 = 0;
end
output2end = matrix(2:end,2) & matrix(1:end-1,3);
output_Cage1 = [output1; output2end] % C to B
output_Cage1 = 12×1
     0
     0
     1
     0
     0
     0
     0
     0
     1
     0
    % Identify all B to A
    output2end = matrix(2:end,1) & matrix(1:end-1,2);
    output_Cage2 = output_Cage1+[output1; 2*output2end] % B to A
output_Cage2 = 12×1
     0
     0
     1
     0
     2
     0
     0
     0
     1
     0
    % Identify all B to C
    output2end = matrix(2:end,3) & matrix(1:end-1,2);
    output_Cage3 = output_Cage2+[output1; 3*output2end] % B to C
output_Cage3 = 12×1
     0
     0
     1
     0
     2
     0
     0
     3
     1
     3
% Remove B to C or B to A that does not follow C to B
for ii = 1:length(output_Cage3) % iteratively go through array
    if output_Cage3(ii) == 1 % if C to B happens, then set count to 1
        count = 1;
    elseif output_Cage3(ii) ~= 0 % if the value is not 0 or 1
        if count == 0 % and count is 0 (no previous C to B detected)
            output_Cage3(ii) = 0; % then remove the B to C or B to A present
        end
        count = 0; % set count back to 0 indicating that there is no previous C to B.
    end
end
output_Cage3
output_Cage3 = 12×1
     0
     0
     1
     0
     2
     0
     0
     0
     1
     3

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Kevin Holly 2023-1-16

在 MATLAB Online 中打开

Why would you consider the 3 row as going from B to A (2)? How do you want to handle the data if all 3 sensors are active? Below I ignore rows when more than one channel was active.

matrix = [0 0 1
0 1 1
1 1 1
1 1 0
0 1 0
0 1 1
0 0 1
0 1 1
0 1 1 
0 0 1
0 0 0];
if matrix(1,1)
    output1 = nan;
else
    output1 = 0;
end
output2end = matrix(2:end,2) & matrix(1:end-1,3);
output_Cage1 = [output1; output2end] % C to B
output_Cage1 = 11×1
     0
     1
     1
     1
     0
     0
     0
     1
     1
     0
    % Identify all B to A
    output2end = matrix(2:end,1) & matrix(1:end-1,2);
    output_Cage2 = output_Cage1+[output1; 2*output2end]; % B to A
    output_Cage2(output_Cage2>2)=0
output_Cage2 = 11×1
     0
     1
     0
     0
     0
     0
     0
     1
     1
     0
    % Identify all B to C
    output2end = matrix(2:end,3) & matrix(1:end-1,2);
    output_Cage3 = output_Cage2+[output1; 3*output2end]; % B to C
    output_Cage3(output_Cage3>3)=0
output_Cage3 = 11×1
     0
     1
     3
     0
     0
     3
     3
     1
     0
     3
% Remove B to C or B to A that does not follow C to B
for ii = 1:length(output_Cage3) % iteratively go through array
    if output_Cage3(ii) == 1 % if C to B happens, then set count to 1
        count = 1;
    elseif output_Cage3(ii) ~= 0 % if the value is not 0 or 1
        if count == 0 % and count is 0 (no previous C to B detected)
            output_Cage3(ii) = 0; % then remove the B to C or B to A present
        end
        count = 0; % set count back to 0 indicating that there is no previous C to B.
    end
end
output_Cage3
output_Cage3 = 11×1
     0
     1
     3
     0
     0
     0
     0
     1
     0
     3

Laura Steel 2023-1-19

Hi Kevin,

Unfortunately selecting the highest value sensor didn't work as planned. However, I know what I need to do:

I would like to scan through my data matrix rows and create a new matrix based on the following rules:

If all columns are 0, assign three 0s to the new matrix.
If all columns are 1, assign a 1 to column 2 of the new matrix and a 0 to column 1 and 3.
If 2 columns =1, move up rows until you hit a 0 in at least one of those rows and assign a 1 to this column and a 0 to the other column.
(If you hit a 0 at the same time, then assign a 1 to both columns) and continue.

My plan is to run this loop twice so that even if point 4. occurs, once I run it twice, this should not be a problem.

e.g. original matrix = [1 0 0; 1 1 0; 1 0 0; 1 0 0; 1 1 1; 0 1 1; 0 1 1; 1 1 1; 1 1 0; 1 1 0; 1 0 0; 1 1 0; 1 0 0]

Should turn into, after one round of data processing:

[1 0 0; 0 1 0; 1 0 0; 1 0 0; 0 1 0; 0 1 1; 0 1 1; 0 1 0; 1 0 0; 1 0 0; 1 0 0; 0 1 0; 1 0 0]

and after two rounds:

[1 0 0; 0 1 0; 1 0 0; 1 0 0; 0 1 0; 0 0 1; 0 0 1; 0 1 0; 1 0 0; 1 0 0; 1 0 0; 0 1 0; 1 0 0]

If there is a way to put this all into one function, that would be brilliant. If you were able to direct me to some resources on if/else loops that would be much appreciated as I know what I want to achieve, but am not sure how to assign new values to new matrices based on an original matrix.

Many thanks again,

Laura

Kevin Holly 2023-1-20

在 MATLAB Online 中打开

information: https://www.mathworks.com/help/matlab/ref/if.html

x2 =[1 0 0; 0 1 0; 1 0 0; 1 0 0; 0 1 0; 0 1 1; 0 1 1; 0 1 0; 1 0 0; 1 0 0; 1 0 0; 0 1 0; 1 0 0];
x3= [1 0 0; 0 1 0; 1 0 0; 1 0 0; 0 1 0; 0 0 1; 0 0 1; 0 1 0; 1 0 0; 1 0 0; 1 0 0; 0 1 0; 1 0 0];
x = [1 0 0; 1 1 0; 1 0 0; 1 0 0; 1 1 1; 0 1 1; 0 1 1; 1 1 1; 1 1 0; 1 1 0; 1 0 0; 1 1 0; 1 0 0]
x = 13×3
     1     0     0
     1     1     0
     1     0     0
     1     0     0
     1     1     1
     0     1     1
     0     1     1
     1     1     1
     1     1     0
     1     1     0
newx = insertfunctionname(x)
newx = 13×3
     1     0     0
     0     1     0
     1     0     0
     1     0     0
     0     1     0
     0     0     1
     0     0     1
     0     1     0
     0     1     0
     0     1     0
newx2 = insertfunctionname(newx)
newx2 = 13×3
     1     0     0
     0     1     0
     1     0     0
     1     0     0
     0     1     0
     0     0     1
     0     0     1
     0     1     0
     0     1     0
     0     1     0
function [newx] = insertfunctionname(x)
newx = zeros(size(x)); % If all columns are 0, assign three 0s to the new matrix. 
for ii = 1:size(x,1)
    if sum(x(ii,:)) == 1
        newx(ii,:) = x(ii,:);
    elseif sum(x(ii,:)) == 3
        newx(ii,2) = 1; %If all columns are 1, assign a 1 to column 2 of the new matrix and a 0 to column 1 and 3. 
    elseif sum(x(ii,:)) == 2 %If 2 columns =1, 
        index = find(x(ii,:)==1);
        for iii = ii+1:size(x,1) %move up rows 
            if sum(x(iii,index))<2 %until you hit a 0 in at least one of those rows
                newx(ii,index) = x(iii,index)==0; %and assign a 1 to this column and a 0 to the other column.
                break
            end
        end
    end
end
end

Laura Steel 2023-2-3

Ah, I see, thanks. I do not have a function called count, but I hadn't created a variable called "count" prior to the loop. When I create count before hand, it then runs properly. I am assuming I should set count to 0 when I define it? Many thanks!

Kevin Holly 2023-2-3

Yes, that is right.

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Matrix processing problem; creating new matrix

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