Help solving an error
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Here is the code:
load PollutionData.txt;
array = PollutionData(:,1); %#ok<NODEF>
timearray = (0:.5:250);
timearray = reshape(timearray,[],1);
constant = reaeration(array);
array(3,:) = array(3,:) .* 0.001076; %conversion
array(6,:) = array(6,:) .* 0.0002778; %conversion
resultarray = deOxygen(array , timearray, constant);
plot(timearray(:,1) , resultarray(:,1))
return
function constant = reaeration(array)
constant = ((array(5,:) * array(3,:)) ^ 0.5) / (array(4,:) .^ 1.5);
return
function resultarray = deOxygen(array , timearray, constant)
resultarray = zeros(1,1000);
for a = (0:.5:250)
resultarray(a) = (((array(6,:) * array(2,:)) / (constant - array(6,:))) * (exp(-array(6,:) * timearray(a) * 3600) - exp(-constant * timearray(a) * 3600))) + (array(1,:) * exp(-constant * timearray(a) * 3600));
end
return
I'm getting in error in the deOxygen function at the "resultarray(a) = (((..." portion. Also, the two following errors also appear:
Attempted to access timearray(0); index must be a positive integer or logical.
Error in myHW7 (line 15) resultarray = deOxygen(array , timearray, constant);
And I have no idea why. Any ideas?
Thanks
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Accepted Answer
Star Strider
on 8 Dec 2014
In ‘deOxygen’, define the ‘a’ vector first:
a = (0:.5:250);
then set the loop up as:
for k1 = 1:length(a)
and replace the ‘(a)’ subscripts with ‘(k1)’ subscripts (or whatever loop counter variable you choose). MATLAB does not allow subscripts to be negative, zero, or non-integers.
Also, I’m not sure what ‘a’ is doing. You don’t use it anywhere in ‘deOxygen’ that I can find.
6 Comments
Star Strider
on 9 Dec 2014
Did you implement the array element subscript idea I suggested in my initial Answer? (You need to change all the ‘(a)’ subscripts to ‘k1’ or some variable of your choice. Keeping them as they are now simply will not work.) If so, did it at least solve part of the problem?
There are other potential problems in the way you are calculating ‘resultarray’ even with the subscripts corrected, but until you fix the subscripts it’s impossible to say what other problems may exist.
You’re currently doing matrix (as distinguished from element-wise) operations to create ‘resultarray’. This may or may not be appropriate, depending on what your ‘array’ variables are and what you want to calculate from them.
More Answers (1)
Image Analyst
on 8 Dec 2014
Because indexes start with one. Why don't you do this:
index = 1;
for a = (0:.5:250)
resultarray(index) =..........
index = index + 1;
% more code......
end
Or else this:
aArray = 0:.5:250;
for index = 1 : length(aArray)
a = aArray(index); % Extract the value of a that we need.
resultarray(index) =..........
% more code......
end
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