I have a simple model where I want to minimize the RMSE between my dependent variable y and my model values. The model is: y = alpha + beta'*x.
For minimization, I am using Matlab's fmincon function and am struggling with multiplying my parameter p(2) by x.
MWE:
% data
y = [5.072, 7.1588, 7.263, 4.255, 6.282, 6.9118, 4.044, 7.2595, 6.898, 4.8744, 6.5179, 7.3434, 5.4316, 3.38, 5.464, 5.90, 6.80, 6.193, 6.070, 5.737]
x = [18.3447, 79.86538, 85.09788, 10.5211, 44.4556, 69.567, 8.960, 86.197, 66.857, 16.875, 52.2697, 93.971, 24.35, 5.118, 25.126, 34.037, 61.4445, 42.704, 39.531, 29.988]
% initial values
p_initial = [0, 0];
% function: SEE BELOW
objective = #(p) sqrt(mean((y - y_mod(p)).^2));
% optimization
[param_opt, fval] = fmincon(objective, p_initial)
If I specify my function as follows then it works.
y_mod = #(p) p(1) + p(2).*x
However, it does not work if I use the following code. How can I multiply p(2) with x? Where x is not optimized, because the values are given.
function f = y_mod(p)
f = p(1) + p(2).*x
end
Here is the output from a script that has the function declaration:
>> modelFitExample2a
RMS Error=0.374, intercept=4.208, slope=0.0388
And here is code for the above. It has many commented lines because it includes alternate ways to fit the data: an inline declaration of y_mod(), or a multi-line declaration of y_mod(), or no y_mod() at all. This version uses the multi-line declaration of y_mod().
%modelFitExample2a.m WCR 2021-01-19
%Reply to stack exchange question on parameter fitting
clear;
global x %need this if define y_mod() separately, and in that case y_mod() must declare x global
% data
y = [5.0720, 7.1588, 7.2630, 4.2550, 6.2820, 6.9118, 4.0440, 7.2595, 6.8980, 4.8744...
6.5179, 7.3434, 5.4316, 3.3800, 5.4640, 5.9000, 6.8000, 6.1930, 6.0700, 5.7370];
x = [18.3447,79.8654,85.0979,10.5211,44.4556,69.5670, 8.9600,86.1970,66.8570,16.8750,...
52.2697,93.9710,24.3500, 5.1180,25.1260,34.0370,61.4445,42.7040,39.5310,29.9880];
% initial values
p_initial = [0, 0];
%predictive model with parameter p
%y_mod = #(p) p(1) + p(2)*x;
% objective function
%If you use y_mod(), then you must define it somewhere
objective = #(p) sqrt(mean((y - y_mod(p)).^2));
%objective = #(p) sqrt(mean((y-p(1)-p(2)*x).^2));
% optimization
options = optimset('Display','Notify');
[param_opt, fval] = fmincon(objective,p_initial,[],[],[],[],[],[],[],options);
% display results
fprintf('RMS Error=%.3f, intercept=%.3f, slope=%.4f\n',...
fval,param_opt(1),param_opt(2));
%function declaration: predictive model
%This is an alternative to the inline definition of y_mod() above.
function f = y_mod(p)
global x
f = p(1) + p(2)*x;
end
carl,
The second method, in which you declare y_mod() explicitly (at the end of your script, or in a separate file y_mod.m), does not work because y_mod() does not know what x is. Fix it by declaring x global in the main program at the top, and declare x global in y_mod().
%function declaration
function f = y_mod(p)
global x
f = p(1) + p(2)*x;
end
Of course you don't need y_mod() at all. The code also works if you use the following, and in this case, no global x is needed:
% objective function
objective = #(p) sqrt(mean((y-p(1)-p(2)*x).^2));
By the way, you don't need to multiply with .* in y_mod. You may use *, because you are multiplying a scalar by a vector.
Goal: Plot the graph using a non-linear function.
Function and graph
This is my first time working at Octave. To plot the graph, I need to calculate a function in the range Fx (0.1 ... 10).
I tried to implement this by looping the function through the for loop, writing the results to an array (x-axis - Fn, y-axis - function value), then loading the arrays into the plot() function.
Fn = 1
Ln = 5
Q = 0.5
function retval = test (Fn, Ln, Q)
# Fn squared (for common used)
Fn = Fn^2
# Node A + Node B
nodeA = Fn * (Ln - 1)
nodeB = (Ln * Fn - 1)^2 + Fn * (Fn - 1)^2 * (Ln - 1)^2 * Q^2
nodeB = sqrt(nodeB)
# Result
result = nodeA / nodeB
retval = result
return;
endfunction
frequencyArray = {}
gainArray = {}
fCount = 1
gCount = 1
for i = 0:0.5:5
# F
Fn = i
frequencyArray{fCount} = Fn
fCount = fCount + 1
# G
gainArray{gCount} = test(Fn, Ln, Q)
gCount = gCount + 1
end
plot(frequencyArray, gainArray);
As a result, I get an error about the format of the arrays.
>> plot(frequencyArray, gainArray);
error: invalid value for array property "xdata"
error: __go_line__: unable to create graphics handle
error: called from
__plt__>__plt2vv__ at line 495 column 10
__plt__>__plt2__ at line 242 column 14
__plt__ at line 107 column 18
plot at line 223 column 10
In addition to the error, I believe that these tasks are solved in more correct ways, but I did not quite understand what to look for.
Questions:
Did I choose the right way to solve the problem? Are there any more elegant ways?
How can I fix this error?
Thank you!
If I have correctly interpreted what you are trying to do, the following should work. Firstly, you need to use the term-by-term versions of all arithmetic operators that act on Fn. These are the same as the normal operators except preceded by a dot. Next, you need to put Fn equal to a vector containing the x-values of all the points you wish to plot and put Q equal to a vector containing the values of Q for which you want to draw curves. Use a for-loop to loop through the values of Q and plot a single curve in each iteration of the loop. You don't need a loop to plot each curve because Octave will apply your "test" function to the whole Fn vector and return the result as a vector of the same size. To plot the curves on a log axis, use the function "semilogx(x, y)" insetad of "plot(x, y)". To make the plots appear on the same figure, rather than separate ones put "hold on" before the loop and "hold off" afterwards. You used cell arrays instead of vectors in your for-loop, which the plotting functions don't accept. Also, you don't need an explicit return statement in an Octave function.
The following code produces a set of curves that look like the ones in the figure you pasted in your question:
Ln = 5
function result = test (Fn, Ln, Q)
# Fn squared (for common used)
Fn = Fn.^2;
# Node A + Node B
nodeA = Fn .* (Ln - 1);
nodeB = (Ln .* Fn .- 1).^2 + Fn .* (Fn .- 1).^2 .* (Ln - 1)^2 * Q^2;
nodeB = sqrt(nodeB);
# Result
result = nodeA ./ nodeB;
endfunction
Fn = linspace(0.1, 10, 500);
Q = [0.1 0.2 0.5 0.8 1 2 5 8 10];
hold on
for q = Q
K = test(Fn, Ln, q);
semilogx(Fn, K);
endfor
hold off
I am trying to make a function in Octave where you give octave a function f(x,y) as a string, a change in X, a change in Y, a starting point, and the size of a matrix, the function will create a matrix populated with the values of f(x,y) at each point in the matrix.
This is for an application that displays a 3d graph, using the matrix to map each value to a block
# funcStr: The function whose Z values are being calculated
# dx: the change in x that each block in the x direction represents
# dy: the change in y that each block in the y direction represents
# startPt: the point (in an array of x, y) that center block represents
# res: the side length (in blocks) of the plane
pkg load symbolic
syms x y
function[zValues] = calculateZValues(funcStr, dx, dy, startPt, res)
zValues = zeros(res);
eqn = #(x, y) inline(funcStr);
startX = startPt{1};
startY = startPt{2};
for yOffset = 1:res
for xOffset = 1:res
xCoord = startX + dx * xOffset;
yCoord = startY + dy * yOffset;
zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
endfor
endfor
endfunction
The error I am getting is:
>> calculateZValues("x*y", 1, 1, {0,0}, 10)
parse error near line 20 of file /home/rahul/Documents/3dGraph/graph/calculateZValues.m
anonymous function bodies must be single expressions
>>> zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
I have no idea what the issue is. I have replaced the #(x,y) part with {x,y} in the line referenced by the error but it says nothing or it raises an error about the function subs not being declared. I have also tried moving the pkg and syms lines above the function header
I would like to implement a function duration = timer(n, f, arguments_of_f) that would measure how much time does a method f with arguments arguments_of_f need to run n times. My attempt was the following:
function duration = timer(n, f, arguments_of_f)
duration = 0;
for i=1:n
t0 = cputime;
f(arguments_of_f);
t1 = cputime;
duration += t1 - t0;
end
In another file, I have
function y = f(x)
y = x + 1;
end
The call d1 = timer(100, #f, 3); works as expected.
In another file, I have
function y = g(x1, x2)
y = x1 + x2;
end
but the call d2 = timer(100, #g, 1, 2); gives an error about undefined
argument x2, which is, when I look back, somehow expected, since I pass only
1 to g and 2 is never used.
So, how to implement the function timer in Octave, so that the call like
timer(4, #g, x1, ... , xK) would work? How can one pack the xs together?
So, I am looking for the analogue of Pythons *args trick:
def use_f(f, *args):
f(*args)
works if we define def f(x, y): return x + y and call use_f(f, 3, 4).
You don't need to pack all the arguments together, you just need to tell Octave that there is more than one argument coming and that they are all necessary. This is very easy to do using variadic arguments.
Your original implementation is nearly spot on: the necessary change is minimal. You need to change the variable arguments_to_f to the special name varargin, which is a magical cell array containing all your arbitrary undeclared arguments, and pass it with expansion instead of directly:
function duration = timer(n, f, varargin)
duration = 0;
for i=1:n
t0 = cputime;
f(varargin{:});
t1 = cputime;
duration += t1 - t0;
end
That's it. None of the other functions need to change.
I have followed the tutorial on http://www.mit.edu/people/abbe/matlab/ode.html and prepared a function as follows:
function dxy = diffxy(xy)
%
%split xy into variables in our equations
%
x = xy(1);
xdot = xy(2);
y = xy(3);
%
% define the derivatives of these variables from equations
%
xdot = xdot;
ydot = 3*x + 2*y + 5;
xdoubledot = 3 - ydot + 2*xdot;
%
%return the derivatives in dxy in the right order
%
dxy = [xdot; xdoubledot; ydot]
end
When I call it using
[T, XY] = ode45('diffxy',0,10,[0 1 0])
I get an error
??? Error using ==> diffxy
Too many input arguments.
I also tried
XY= ode45(#diffxy,[0 10],[0;1;0])
Anybody have any idea?
haven't read the whole tutorial but aren't you supposed to define your function as
function dxy = diffxy(t, xy)
where t is time vector