Why this simple for loop doesn't work?

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Fatemeh Sharafi
Fatemeh Sharafi 2021-7-12
编辑： Stephen23 2021-7-12
I have a for loop and I want it to calculate the mean of sum of square of elements of an array, in a fixed-length period;
The array is b (1 by 30). the segments that I want to calculate the mean(sum(square(b))) are 3 by 3 steps: for example first calculate it for the first 3 elements. then calculate for the second three elements, and so forth.
Now the problem is this loop doesn't work for all i and j values. it only calculates for the last i an last j.
Please guide me with this. thanks in advance
here's the code:
b=[31,12,14,8,32,38,45,29,39,4,44,6,21,29,19,26,2,36,42,47,3,47,6,26,44,26,5,43,22,36];
R1405_bar=zeros(1,10);
for i=1:1:10
for j=3:3:30
R1405_bar(1,i)=(sum(b(j-2:j).^2));
end
end
  2 个评论
Fatemeh Sharafi
Fatemeh Sharafi 2021-7-12
Thanks @Aakash Deep Chhonkar. you're right. only i needs to be in "for iteration". thanks a lot
Stephen23
Stephen23 2021-7-12
编辑：Stephen23 2021-7-12
Ran in:
b = [31,12,14,8,32,38,45,29,39,4,44,6,21,29,19,26,2,36,42,47,3,47,6,26,44,26,5,43,22,36];
The simple MATLAB approach:
R1405_bar = sum(reshape(b,3,[]).^2)
R1405_bar = 1×10
1301 2532 4387 1988 1643 1976 3982 2921 2637 3629
Vs. the complex approach:
R1405_bar = zeros(1,10);
j = 1;
for i=1:3:30
R1405_bar(j)=(sum(b(i:i+2).^2));
j = j + 1;
end
R1405_bar
R1405_bar = 1×10
1301 2532 4387 1988 1643 1976 3982 2921 2637 3629

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采纳的回答

Aakash Deep Chhonkar
The issue is in your nested for loop. For every iteration of i, you are computing the same set of commands hence, the output is same everything. Your nested for loop index j should depend on the outer for loop index i.
As far as I understand the problem statement, there is no need of the nested loop. You can compute the mean of sum of squares of elements of array using single loop. Refer below code,
b=[31,12,14,8,32,38,45,29,39,4,44,6,21,29,19,26,2,36,42,47,3,47,6,26,44,26,5,43,22,36];
R1405_bar=zeros(1,10);
j = 1;
for i=1:3:30
R1405_bar(j)=(sum(b(i:i+2).^2));
j = j + 1;
end

更多回答（3 个）

Yongjian Feng
Yongjian Feng 2021-7-12
Use a debugger to check why.
Simon Chan
Simon Chan 2021-7-12
For each i, the value on R1405_bar is keep overwrite as j runs from 3 to 30.
On the other hand, you just calculate sum of square of element only, missing the mean.
Jogesh Kumar Mukala
Hi,
Assuming you want to find the mean of squares of consecutive three elements of 'b' array and store in R1405_bar, the issue with your code is the 'j' loop is running completely for every 'i' and it is storing the same result( i.e end result of 'j' loop) in every element of R1405_bar. You may find the below code which may solve the issue.
You may also refer the sumsqr function that gives sum of squared elements.
b=[31,12,14,8,32,38,45,29,39,4,44,6,21,29,19,26,2,36,42,47,3,47,6,26,44,26,5,43,22,36];
R1405_bar=zeros(1,10);
j=[3:3:30]
for i=1:1:10
R1405_bar(1,i)=(sumsqr(b(j(i)-2:j(i))))/3;
end

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