This might seem like a strange thing to do, which it probably is. In my main (or how you call it in matlab) I would like to have all the information needed for the program to run. A change of variables or formulas should only happen in my main.
For example I would like to change the number of iterations and the formula of the hypothese in my main and let other function use these, instead of declaring them within the function themselves and having to edit it all over the place. The problem I face is not knowing how to do this properly for hypothese_formula and wonder if there is a better way of doing this?
function prog1()
iterations = 1;
hypothese_formula = x^2;
doSomethingWithFormulaAndIterations(hypothese_formula, iterations);
end
Practical: I would to do linear regression with a hypothesis of the formula and specific starting values of theta and don't want them to be hidden within a function. I don't know how to declare global formula's.
You can use anonymous functions.
function prog1()
iterations = 1;
hypothese_formula = #(x) x.^2
doSomethingWithFormulaAndIterations(hypothese_formula, iterations);
end
Related
I am trying to create a plot in MATLAB by iterating over values of a constant (A) with respect to a system of equations. My code is pasted below:
function F=Func1(X, A)
%X(1) is c
%X(2) is h
%X(3) is lambda
%A is technology (some constant)
%a is alpha
a=1;
F(1)=1/X(1)-X(3)
F(2)=X(3)*(X(1)-A*X(2)^a)
F(3)=-1/(24-X(2))-X(3)*A*a*X(2)^(a-1)
clear, clc
init_guess=[0,0,0]
for countA=1:0.01:10
result(countA,:)=[countA,fsolve(#Func1, init_guess)]
end
display('The Loop Has Ended')
display(result)
%plot(result(:,1),result(:,2))
The first set of code, I am defining the set of equations I am attempting to solve. In the second set of lines, I am trying to write a loop which iterates through the values of A that I want, [1,10] with an increment of 0.01. Right now, I am just trying to get my loop code to work, but I keep on getting this error:
"Failure in initial objective function evaluation. FSOLVE cannot continue."
I am not sure why this is the case. From what I understand, this is the result of my initial guess matrix not being the right size, but I believe it should be of size 3, as I am solving for three variables. Additionally, I'm fairly confident there's nothing wrong with how I'm referencing my Func1 code in my Loop code.
Any advice you all could provide would be sincerely appreciated. I've only been working on MATLAB for a few days, so if this is a rather trivial fix, pardon my ignorance. Thanks.
Couple of issues with your code:
1/X(1) in function Func1 is prone to the division by zero miscalculations. I would change it to 1/(X(1)+eps).
If countA is incrementing by 0.01, it cannot be used as an index for result. Perhaps introduce an index ind to increment.
I have included your constant A within the function to clarify what optimization variables are.
Here is the updated code. I don't know if the results are reasonable or not:
init_guess=[0,0,0];
ind = 1;
for countA=1:0.01:10
result(ind,:)=[countA, fsolve(#(X) Func1(X,countA), init_guess)];
ind = ind+1;
end
display('The Loop Has Ended')
display(result)
%plot(result(:,1),result(:,2))
function F=Func1(X,A)
%X(1) is c
%X(2) is h
%X(3) is lambda
%A is technology (some constant)
%a is alpha
a=1;
F(1)=1/(X(1)+eps)-X(3);
F(2)=X(3)*(X(1)-A*X(2)^a);
F(3)=-1/(24-X(2))-X(3)*A*a*X(2)^(a-1);
end
The following code won't work, but this is the idea I'm trying to get at.
c = #(x)constraints;
%this is where I would initialize sum as 0 but not sure how...
for i = 1:length(c)
sum = #(x)(sum(x) + (min(c(x)(i),0))^2);
end
penFunc = #(x)(funcHandle(x) + sig*sum(x));
where constraints and funcHandle are functions of x. This entire code would iterate for a sequence of sig's.
Obviously c(x)(i) isn't functional. I'm trying to write the function where the minimum of c(x) at i (c(x) is a vector) or 0 is taken and then squared.
I know I could calculate c(x) and then analyze it at each i, but I eventually want to pass penFunc as a handle to another function which calculates the minimum of penFunc, so I need to keep it as a function.
I confess I don't understand entirely what you're trying to achieve, but it appears you're trying to create a function handle of an anonymous function with a changing value sum that you precompute. MATLAB anonymous functions do allow you to do this.
It appears there might be some confusion with anonymous functions here. To start with, the line:
c = #(x)constraints;
is probably supposed to be something else, unless you really want c to be a function handle. The # at the start of the line declares a new anonymous function, when I think you just want to call the existing function constraints. It appears you really want c to be an array of constraints coming from the constraints function, in which case I think you mean to say
c = constraints(x);
Then we get to the sum, which I can't tell if you want as a vector or as a single sum. To start with, let's not name it 'sum', since that's the name of a built-in MATLAB function. Let's call it 'sumval'. If it's just a single value, then it's easy (it's easy both ways, but let's do this.) Start before the for loop with sumval=0; to initialize it, then the loop would be:
sumval = 0;
for i = 1:length(c)
sumval = sumval + (min(c(i),0))^2);
end
All four lines could be vectorized if you like to:
c(c>0) = 0; %Replace all positive values with 0
sumval = sum(c.^2); % Use .^ to do a element by element square.
The last line is obviously where you make your actual function handle, and I'm still not quite sure what is desired here. If sig is a function, then perhaps you really meant to have:
penFunc = #(x)(funcHandle(x) + sig*sumval);
But I'm not sure. If you wanted sum to be a vector, then how we specified it here wouldn't work.
Notice that it is indeed fine to have penFunc be an anonymous function with a variable within it (namely sumval), but it will continue to use the value of sumval that existed at the time of the function handle declaration.
So really the issues are A) the creation of c, which I don't think you meant to be a function handle, and B) the initialization of sum, which should probably be sumval (to not interact with MATLAB's own function), and which probably shouldn't declare a new function handle.
I'm trying to use the MATLAB function fzero properly but my program keeps returning an error message. This is my code (made up of two m-files):
friction_zero.m
function fric_zero = friction_zero(reynolds)
fric_zero = 0.25*power(log10(5.74/(power(reynolds,0.9))),-2);
flow.m
function f = flow(fric)
f = 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(power(10,4));
z = fzero(#flow,f_initial)
The goal is to return z as the root for the equation specified by f when flow.m is run.
I believe I have the correct syntax as I have spent a couple of hours online looking at examples. What happens is that it returns the following error message:
"Undefined function or variable 'fric'."
(Of course it's undefined, it's the variable I'm trying to solve!)
Can someone point out to me what I've done wrong? Thanks
EDIT
Thanks to all who helped! You have assisted me to eventually figure out my problem.
I had to add another file. Here is a full summary of the completed code with output.
friction_zero.m
function fric_zero = friction_zero(re)
fric_zero = 0.25*power(log10(5.74/(power(re,0.9))),-2); %starting value for fric
flow.m
function z = flow(fric)
re = power(10,4);
z = 1/(sqrt(fric))-1.873*log10(re*sqrt(fric))-233/((re*sqrt(fric))^0.9)-0.2361;
flow2.m
f_initial = friction_zero(re); %arbitrary starting value (Reynolds)
x = #flow;
fric_root = fzero(x,f_initial)
This returns an output of:
fric_root = 0.0235
Which seems to be the correct answer (phew!)
I realised that (1) I didn't define reynolds (which is now just re) in the right place, and (2) I was trying to do too much and thus skipped out on the line x = #flow;, for some reason when I added the extra line in, MATLAB stopped complaining. Not sure why it wouldn't have just taken #flow straight into fzero().
Once again, thanks :)
You need to make sure that f is a function in your code. This is simply an expression with reynolds being a constant when it isn't defined. As such, wrap this as an anonymous function with fric as the input variable. Also, you need to make sure the output variable from your function is z, not f. Since you're solving for fric, you don't need to specify this as the input variable into flow. Also, you need to specify f as the input into fzero, not flow. flow is the name of your main function. In addition, reynolds in flow is not defined, so I'm going to assume that it's the same as what you specified to friction_zero. With these edits, try doing this:
function z = flow()
reynolds = power(10,4);
f = #(fric) 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(reynolds);
z = fzero(#f, f_initial); %// You're solving for `f`, not flow. flow is your function name
The reason that you have a problem is because flow is called without argument I think. You should read a little more about matlab functions. By the way, reynolds is not defined either.
I am afraid I cannot help you completely since I have not been doing fluid mechanics. However, I can tell you about functions.
A matlab function definition looks something like this:
function x0 = f(xGuess)
a = 2;
fcn =#(t) a*t.^3+t; % t must not be an input to f.
disp(fcn);
a = 3;
disp(fcn);
x0 = fsolve(fcn1,xGuess); % x0 is calculated here
The function can then ne called as myX0 = f(myGuess). When you define a matlab function with arguments and return values, you must tell matlab what to do with them. Matlab cannot guess that. In this function you tell matlab to use xGuess as an initial guess to fsolve, when solving the anonymous function fcn. Notice also that matlab does not assume that an undefined variable is an independent variable. You need to tell matlab that now I want to create an anonymous function fcn which have an independent variable t.
Observation 1: I use .^. This is since the function will take an argument an evaluate it and this argument can also be a vector. In this particulat case I want pointwise evaluation. This is not really necessary when using fsolve but it is good practice if f is not a matrix equation, since "vectorization" is often used in matlab.
Observation 2: notice that even if a changes its value the function does not change. This is since matlab passes the value of a variable when defining a function and not the variable itself. A c programmer would say that a variable is passed by its value and not by a pointer. This means that fcn is really defined as fcn = #(x) 2*t.^3+t;. Using the variable a is just a conveniance (constants can may also be complicated to find, but when found they are just a value).
Armed with this knowledge, you should be able to tackle the problem in front of you. Also, the recursive call to flow in your function will eventuallt cause a crash. When you write a function that calls itself like this you must have a stopping criterium, something to tell the program when to stop. As it is now, flow will call ifself in the last row, like z = fzero(#flow,f_initial) for 500 times and then crash. Alos it is possible as well to define functions with zero inputs:
function plancksConstant = h()
plancksConstant = 6.62606957e−34;
Where the call h or h() will return Plancks constant.
Good luck!
I'm trying to plot a function on matlab that's defined such that:
Y=-100t ; 0>=t>=0.15
Y=-15 ; t>.15
I'm using the following code:
function [ Y ] = Gain( t )
for t=[0:0.01:0.15]
Y=-100*t
end
for t=0.15:0.01:2
Y=-15
end
plot (Gain)
but I'm going into an infinite loop!
Would someone please solve this problem for me.
Thank you.
Functions don't work like that in MATLAB, unfortunately. (Or at least I don't think they do). Try something like this:
function Y = Gain(t)
Y = -100*t;
Y(t >= 0.15) = -15;
end
x = 0:0.01:2;
plot(x, Gain(x))
MATLAB still uses C-esque functions, so you have to define it that way, using C-style syntax, instead of more math-like syntax, unfortunately. I'm multiplying the input values by -100, then for the ones that match up to where t is greater than 15, I replace those with -15. MATLAB is weird.
Edit: Sorry, previous code sample also used the wrong syntax.. MATLAB is weird.
I'm not certain what you're TRYING to do, but when you call the Gain function "from outside", so to speak,...
You enter the Gain function
The first for loop executes, overwriting passed-in value of t with each iteration. (Therefore, the value of t that you passed in is completely ignored, the way that you have this code written.)
The second for-loop executes, and
You call plot(Gain), which forces Gain to be called again, this time with no arguments. Back to 1.
Repeat forever.
when I am doing a function in Matlab. Sometimes I have equations and every one of these have constants. Then, I have to declare these constants inside my function. I wonder if there is a way to call the values of that constants from outside of the function, if I have their values on the workspace.
I don't want to write this values as inputs of my function in the function declaration.
In addition to the solutions provided by Iterator, which are all great, I think you have some other options.
First of all, I would like to warn you about global variables (as Iterator also did): these introduce hidden dependencies and make it much more cumbersome to reuse and debug your code. If your only concern is ease of use when calling the functions, I would suggest you pass along a struct containing those constants. That has the advantage that you can easily save those constants together. Unless you know what you're doing, do yourself a favor and stay away from global variables (and functions such as eval, evalin and assignin).
Next to global, evalin and passing structs, there is another mechanism for global state: preferences. These are to be used when it concerns a nearly immutable setting of your code. These are unfit for passing around actual raw data.
If all you want is a more or less clean syntax for calling a certain function, this can be achieved in a few different ways:
You could use a variable number of parameters. This is the best option when your constants have a default value. I will explain by means of an example, e.g. a regular sine wave y = A*sin(2*pi*t/T) (A is the amplitude, T the period). In MATLAB one would implement this as:
function y = sinewave(t,A,T)
y = A*sin(2*pi*t/T);
When calling this function, we need to provide all parameters. If we extend this function to something like the following, we can omit the A and T parameters:
function y = sinewave(t,A,T)
if nargin < 3
T = 1; % default period is 1
if nargin < 2
A = 1; % default amplitude 1
end
end
y = A*sin(2*pi*t/T);
This uses the construct nargin, if you want to know more, it is worthwhile to consult the MATLAB help for nargin, varargin, varargout and nargout. However, do note that you have to provide a value for A when you want to provide the value of T. There is a more convenient way to get even better behavior:
function y = sinewave(t,A,T)
if ~exists('T','var') || isempty(T)
T = 1; % default period is 1
end
if ~exists('A','var') || isempty(A)
A = 1; % default amplitude 1
end
y = A*sin(2*pi*t/T);
This has the benefits that it is more clear what is happening and you could omit A but still specify T (the same can be done for the previous example, but that gets complicated quite easily when you have a lot of parameters). You can do such things by calling sinewave(1:10,[],4) where A will retain it's default value. If an empty input should be valid, you should use another invalid input (e.g. NaN, inf or a negative value for a parameter that is known to be positive, ...).
Using the function above, all the following calls are equivalent:
t = rand(1,10);
y1 = sinewave(t,1,1);
y2 = sinewave(t,1);
y3 = sinewave(t);
If the parameters don't have default values, you could wrap the function into a function handle which fills in those parameters. This is something you might need to do when you are using some toolboxes that impose constraints onto the functions that are to be used. This is the case in the Optimization Toolbox.
I will consider the sinewave function again, but this time I use the first definition (i.e. without a variable number of parameters). Then you could work with a function handle:
f = #(x)(sinewave(x,1,1));
You can work with f as you would with an other function:
e.g. f(10) will evaluate sinewave(10,1,1).
That way you can write a general function (i.e. sinewave that is as general and simple as possible) but you create a function (handle) on the fly with the constants substituted. This allows you to work with that function, but also prevents global storage of data.
You can of course combine different solutions: e.g. create function handle to a function with a variable number of parameters that sets a certain global variable.
The easiest way to address this is via global variable:
http://www.mathworks.com/help/techdoc/ref/global.html
You can also get the values in other workspaces, including the base or parent workspace, but this is ill-advised, as you do not necessarily know what wraps a given function.
If you want to go that route, take a look at the evalin function:
http://www.mathworks.com/help/techdoc/ref/evalin.html
Still, the standard method is to pass all of the variables you need. You can put these into a struct, if you wish, and only pass the one struct.