Executing a "for" loop, but getting "array indices must be positive integers or logical values" error

Hi,

I am trying to carry out a calculation for array "y" for each value in the array, starting from the second value. I get the "array indices must be positive integers or logical values" error, and not sure why. I think it may be getting hung up on dividing the previous "y" element in to the current "y" element. I am currently specifying "y(i - 1)" when really I want it to be "y(i) / y(previous element to the current i'th value)"

Code is as follows:

clear
clc

y = [2.903E-04, 7.538E-05, 2.525E-05, 2.136E-03, 2.051E-02];
x = [0, 6.25, 12.5, 18.75, 25];

for i = y(2):y(end)
    dBa(i) = 10*log10(y(i)/(y(i-1)))
end

Thanks.

Answer 1

[Steven Lord]

What values does your loop variable take on when you use the code as you've written it?

y = [2.903E-04, 7.538E-05, 2.525E-05, 2.136E-03, 2.051E-02];
for k = y(2):y(end)
    disp(k)
end

There's no such thing as element 7.538e-5 of an array in MATLAB. You could loop from 2 to numel(y) but that would leave dBa(1) as 0. [I'm not sure whether that's intended or not based on your code.]

for k = 2:numel(y)
    fprintf('y(%d) is %g.\n', k, y(k))
end

Or you could vectorize the computation of dBa using element-wise division (./) instead of matrix division (/). For scalar values, ./ and / behave the same. For non-scalars they don't.

k = 2:numel(y);
dBa(k) = 10*log10(y(k)./(y(k-1)))

5./10
5/10

[Will Crowson]

Thanks for the feedback.

Why would MATLAB not recognise the element 7.538E-05?

The element-wise division gives the same error also. N.B. I were to carry out this "loop" in Excel, I would carry out the following in the neighbouring cell (for example):

10*LOG(B12/B11)

10*LOG(B13/B12)

10*LOG(B14/B13)

10*LOG(B15/B14)

etc..

Hopefully that clears up what I am trying to achieve with the loop - the number of times the loop goes round may vary depending on length of the column vector.

Code is currently as follows, but I gather the error is down to the "y" values?:

clear
clc

y = [2.903E-04, 7.538E-05, 2.525E-05, 2.136E-03, 2.051E-02];

for k = y(2):numel(y)
    dBa(k) = 10*log10(y(k)./(y(k-1)))
end

Thanks in advance.