How to concatenate or save iteration from a loop when the answers are vectors?

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Priti Dhiman 2014-12-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/164886-how-to-concatenate-or-save-iteration-from-a-loop-when-the-answers-are-vectors

编辑： Mohammad Abouali 2014-12-1

I'm very new to MatLab and need some help with saving iterations from a loop. I know that when the answers from a loop are scalar, you can make an array of answers. However, is there any way to do this when the answers from your loop are a vector? Here is my code:

Curve = input('What is your data: ' );
plot (Curve(1,1:end))
n = 0;
while n < size(Curve,1)
    n = n+1;
    TS(n) = max(Curve(n,:));
end
% Yield Point
k = 0;
while k < size(Curve,1)
        k = k+1;
        [obs,YP_1] = findpeaks(Curve(k,:));%in MatLab 2013b
        c = (length(YP_1)-1);
        Y_P(k) = YP_1(c)+1;
end
% Elasticity
b = 0;
a = 0;
while b < size(Curve,1)
    b = b+1;
    YP(b) = Curve(b,Y_P)
end

Since Y_P is a vector, I am getting my answer for YP as a vector which is what I want. But each vector gets overwritten with the next iteration and I need all the data saved. I would really appreciate the help! Thank you!

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Image Analyst 2014-12-1

Can you give an example for data that is supposed to be inputted for "Curve"?

Priti Dhiman 2014-12-1

I'm currently working with a 18x3509 matrix which is actually a concatenation of two matrices of different dimensions.

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Answer 1

Mohammad Abouali 2014-12-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/164886-how-to-concatenate-or-save-iteration-from-a-loop-when-the-answers-are-vectors#answer_160757

编辑：Mohammad Abouali 2014-12-1

在 MATLAB Online 中打开

add an extra dimension

So instead of

Answer(i)= ....

write

Answer(i,j)= ....

where j is the loop index.

If the size is changing between each loop iteration you can also use cell arrays.

NOTE: Your code is all jammed up I can't really see what is what. Try to use "{} Code" when posting so your code is readable.

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Mohammad Abouali 2014-12-1

编辑：Mohammad Abouali 2014-12-1

在 MATLAB Online 中打开

Ok couple of things here:

(1) This part of your code:

n = 0;
while n < size(Curve,1)
    n = n+1;
    TS(n) = max(Curve(n,:));
end

Gives you the max value of each row. You can simplify the whole thing in one line

TS=max(Curve,[],2);

(2) Although this part of your code is perfectly correct:

k = 0;
while k < size(Curve,1)
        k = k+1;
        [obs,YP_1] = findpeaks(Curve(k,:));%in MatLab 2013b
        c = (length(YP_1)-1);
        Y_P(k) = YP_1(c)+1;
end

you can change it to:

for k=1: (size(Curve,1)-1) 
        [obs,YP_1] = findpeaks(Curve(k,:));%in MatLab 2013b
        Y_P(k) = YP_1(end-1)+1;
end

NOTE: Why k<size(Curve,1)? This ignores the last row of data in Curve. so if you have 5 rows of data in Curve the last row is not processed. If this is a bug and you want all rows then change "for k=1: (size(Curve,1)-1)" to "for k=1:size(Curve,1)"

(3) What is the porpose of this part of the code?

b = 0;
a = 0;
while b < size(Curve,1)
    b = b+1;
    YP(b) = Curve(b,Y_P)
end

what is "a" doing here? it is not used.

You can change this to

    for b=1: (size(Curve,1)-1)
        YP(b,:) = Curve(b,Y_P); 
    end

This should work. as long as size(Y_P) is 1*N or N*1.

NOTE: again you are not processing the last row. If this is a bug, change the for-loop as mentioned above.

Priti Dhiman 2014-12-1

Thank you, thank you, thank you! Amazing answer, I truly appreciate your help!

Mohammad Abouali 2014-12-1

编辑：Mohammad Abouali 2014-12-1

you are welcome.

One note: actually k < size(Curve,1) was not ignoring the last row. That was correct (my mistake). So make sure that you use for k=1:size(Curve,1) every where to process all rows.

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