User defined function to output matrix

2017 9 23
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更新时间：2017 9 23
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I am in need of some help with a user defined function. I have it so it will output a column array of zeros however I need to function to fill the array dependant on a number of conditions and I have no idea how to get from where I am to where I need to be.
function [Q] = O_R_B(H)
%Operating rule B for varying dam heights
% Determining the usage rates for the dam for different dam heights
H_max = 30.9; %dam height in m (scaled 1:300)
Q_out = 138; %normal output rate
Q = zeros(length(H),1);
for k = 1:length(H)
if H >= 1*H_max;
Q = string('issue warning');
elseif H >= 0.9*H_max;
Q(k) = 4 * Q_out;
elseif H < 0.9*H_max & H>= 0.85*H_max;
Q(k) = 2* Q_out;
elseif H < 0.85*H_max & H>= 0.4*H_max;
Q(k) = Q_out;
elseif H < 0.4*H_max & H>= 0.2*H_max;
Q(k) = 0.5*Q_out;
else H< 0.2*H_max;
Q(k) = 0;
end
end
end

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James Tursa
James Tursa 2017-9-23
编辑：James Tursa 2017-9-23

0 个投票

You need to index H also in your code. E.g.,
if H(k) >= 1*H_max;
error('H too large');
elseif H(k) >= 0.9*H_max;
Q(k) = 4 * Q_out;
:
etc

6 个评论
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Rhys Webb
Rhys Webb 2017-9-23
thank you this has gotten me closer to the answer however it is only outputting values in the first and last location (for testing purposes the range I am using is 10:10:30) and the centre value is being produced as a 0. Thank you for your assistance with this.
ans =
69
0
552
James Tursa
James Tursa 2017-9-23
编辑：James Tursa 2017-9-23
Please post your current code. Did you change both of the H's into H(k)'s in the middle tests?
Also, since these are scalar tests you are doing, it would be more appropriate to use the && operator instead of the & operator.
Rhys Webb
Rhys Webb 2017-9-23
sorry should have made sure to post my code in my previous comment.
function [Q] = O_R_B(H)
%Operating rule B for varying dam heights
% Determining the usage rates for the dam for different dam heights
H_max = 30.9; %dam height in m (scaled 1:300)
Q_out = 138; %normal output rate
Q = zeros(length(H),1);
for k = 1:length(H)
if H(k) >= 1*H_max;
Q = string('issue warning');
elseif H(k) >= 0.9*H_max;
Q(k) = 4 * Q_out;
elseif H(k) < 0.9*H_max & H>= 0.85*H_max;
Q(k) = 2* Q_out;
elseif H(k) < 0.85*H_max & H>= 0.4*H_max;
Q(k) = Q_out;
elseif H(k) < 0.4*H_max & H>= 0.2*H_max;
Q(k) = 0.5*Q_out;
else H(k) < 0.2*H_max;
Q(k) = 0;
end
end
end
This is where I am at currently. The other thing I also need to achieve is that if a value greater than H_max occurs i need an output of issue warning in the matrix as well and I have no clue on how to achieve this.
James Tursa
James Tursa 2017-9-23
编辑：James Tursa 2017-9-23
You didn't change both of the H's to H(k)'s. E.g., this line
elseif H(k) < 0.9*H_max & H>= 0.85*H_max;
should be this instead:
elseif H(k) < 0.9*H_max & H(k) >= 0.85*H_max
And, to simplify things, since your tests cover the entire range of 0 to H_max, you don't even need the < part of the tests. You know you will pass that part of the test guaranteed because you just failed the previous test. So you only need the >= part of the test. E.g.,
if H(k) >= 1*H_max;
Q = string('issue warning'); % <-- YOU NEED TO CHANGE THIS!
elseif H(k) >= 0.9*H_max;
Q(k) = 4 * Q_out;
elseif H(k) >= 0.85*H_max;
Q(k) = 2* Q_out;
elseif H(k) >= 0.4*H_max;
Q(k) = Q_out;
elseif H(k) >= 0.2*H_max;
Q(k) = 0.5*Q_out;
else
Q(k) = 0;
end
You need to change your warning line. You can't mix a string in with your double data. I would either put a NaN there or maybe just error() the function at that point.
Rhys Webb
Rhys Webb 2017-9-23
You sir are a life saver I have been looking at this for 2 days trying to figure it out.
James Tursa
James Tursa 2017-9-23
编辑：James Tursa 2017-9-23
You could issue a warning and fill the element with NaN. E.g.,
if H(k) >= 1*H_max;
Q(k) = NaN;
warning(sprintf('H(%d) element too large',k));

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