How do I Loop a code every 6 rows until the end?

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sruizpp 2015-4-22

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评论： Star Strider 2015-5-26

采纳的回答： Star Strider

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I need to calculate daily n-factors for a complete year. I use this code;

trapz(x(1:6,1),min(y(1:6,3),0)/trapz(x(1:6,1),min(y(1:6,2),0)

1:6 is 1st day, 7:12 is 2nd and so on... I have 1800 measurements and I need to perform this code every 6 rows from matrix Y so I have another matrix corresponding to the DAILY calculations of the n-factor.

How do I loop this code every 6 rows?

-

Regards!

Sebastián.-

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Answer 1

Star Strider 2015-4-23

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I would use the reshape function to redefine x(:,1), y(:,2) and y(:,3) as:

X1 = reshape(x(:,1), 6, []);
Y2 = reshape(y(:,2), 6, []);
Y3 = reshape(y(:,3), 6, []);

then:

for k1 = 1:size(X1,2)
    W(k1) = trapz(X1(:,k1),min(Y3(:,k1),0)/trapz(X1(:,k1),min(Y3(:,k1),0))
end

There is something wrong with the code you posted, since it doesn’t make sense, so I didn’t run this code. (I assume it produces a scalar.) I will leave you to sort that out.

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sruizpp 2015-5-26

Greetings, If X1 is reshaped, it is no longer recognized as vector and throws an error while running the code. And yes, my result needs to be a scalar. I need that value each 6 terms.

Regards!

Star Strider 2015-5-26

Did you run the for loop? It treats every column of ‘X1’ (and ‘Y3’) as a separate vector, so there should be no problem.

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Answer 2

Geoff Hayes 2015-4-23

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Try setting the step size in your for loop to be 6. Something like the following may work where we assume that both X and Y have 1800 rows.

 numRows  = size(Y,1);
 stepSize = 6;
 for k=1:stepSize:numRows
     value = trapz(X(k:k+stepSize-1,1),min(Y(k:k+stepSize-1,3),0)/...
             trapz(X(k:k+stepSize-1,1),min(Y(k:k+stepSize-1,2),0);
     % save value to other matrix
 end

Given that stepSize is six, then our k becomes 1, 7, 13, 19, ..., 1795 and we get the correct groupings of six elements of data (since k + stepSize - 1 becomes 6, 12, 18, ..., 1800).

Try the above and see what happens!