I wrote a bisection method in Octave but it can't consume another function..
My bisection method code is like:
function[x,b] = bisection(f,a,b)
t = 10e-8
while abs(b-a) > t;
c = (a+b)/2;
if f(a) * f(b) <= 0
a = a;
b = c;
else
b = b;
a = c
endif
endwhile
x = (a+b)/2
endfunction
And I already have a file f1.m:
function y = f1(x)
y = x^2 - 4;
endfunction
But when I call [x,v] = bisection[f1,0,5], I get:
>> [t,v] = bisection(f1,0,5)
error: 'x' undefined near line 2 column 5
error: called from
f1 at line 2 column 3
error: evaluating argument list element number 1
what you want is to pass a pointer to f1 to your function bisection so the right call would be
[t,v] = bisection(#f1,0,5)
which outputs:
t = 1.0000e-07
a = 0.62500
a = 0.93750
a = 1.0938
a = 1.1719
a = 1.2109
a = 1.2305
a = 1.2402
a = 1.2451
a = 1.2476
a = 1.2488
a = 1.2494
a = 1.2497
a = 1.2498
a = 1.2499
a = 1.2500
a = 1.2500
a = 1.2500
a = 1.2500
a = 1.2500
a = 1.2500
a = 1.2500
a = 1.2500
a = 1.2500
a = 1.2500
x = 1.2500
t = 1.2500
v = 1.2500
Andy has given you the answer on how to fix this. I would just like to add why you get that error and what it means. Consider the following octave session:
octave:1> function Out = g1(x); Out = x+5; end
octave:2> function Out = g2(); Out = 10;end
octave:3>
octave:3> g2
ans = 10
octave:4> g1
error: 'x' undefined near line 1 column 29
error: called from
g1 at line 1 column 27
I.e., when you write g1 or g2 here, this is an actual function call. The call to g2 succeeds because g2 does not take any arguments; the syntax g2 is essentially equivalent to g2(). However, the call to g1 fails, because g1 expects an argument, and we didn't provide one.
Compare with:
octave:4> a = #g1;
octave:5> b = #g2;
octave:6> a
a = #g1
octave:7> a(1)
ans = 6
octave:8> b
b = #g2
octave:9> b()
ans = 10
where you have created handles to these functions, which you can capture into variables, and pass them as arguments into functions. These handles could then be called as a(5) or b() inside the function that received them as arguments, and it would be like calling the original g1 and g2 functions.
When you called bisection(f1,0,5), you essentially called bisection(f1(),0,5), i.e. you asked octave to evaluate the function f1 without passing any arguments, and use the result as the first input argument to the bisection function. Since function f1 is defined to take an input argument and you didn't supply any, octave complains that when it tries to evaluate y = x^2 - 4; as per the definition of f1, x was not passed as an input argument and was therefore undefined.
Therefore, to pass a "function" as an arbitrary argument that can be called inside your bisection function, you need to pass a function handle instead, which can be created using the #f1 syntax. Read up on "anonymous functions" on the octave (or matlab) documentation.
Related
I wrote my script for a class but the next step is to convert that script into a function. What do I type in?
This is my script:
% Find max volume
b = 2.75; %ft for diameter
h = 3.00; %ft for height
v = (b*h)/3; %ft^3
% use volume to find mass
p = 62.3;
m = p*v;
The following syntax is used to create a function in MATLAB:
function [y1,...,yN] = myfun(x1,...,xM)
In the example below, the maximumVolume() method takes 3 parameters (b, h, p) and returns the value of the m variable.
result = maximumVolume(2.75, 3.0, 62.3);
function m = maximumVolume(b, h, p)
v = (b * h)/3;
m = p * v;
end
References
function
Is there a way to supply arguments using varargin in MATLAB in the following manner?
Function
func myFunc(varargin)
if a not given as argument
a = 2;
if b not given as argument
b = 2;
if c not given as argument
c = a+b;
d = 2*c;
end
I want to call the above function once with b = 3 and another time while the previous one is running in the same command window with a = 3 and c = 3 and letting b take the default value in the function this time. How can it be done using varargin?
Here's the latest and greatest way to write the function (using arguments blocks from R2019b)
function out = someFcn(options)
arguments
options.A = 3;
options.B = 7;
options.C = [];
end
if isempty(options.C)
options.C = options.A + options.B;
end
out = options.A + options.B + options.C;
end
Note that this syntax does not allow you to say options.C = options.A + options.B directly in the arguments block.
In MATLAB < R2021a, you call this like so
someFcn('A', 3)
In MATLAB >= R2021a, you can use the new name=value syntax
someFcn(B = 7)
Here are two ways to do this which have been available since 2007a (i.e. a long time!). For a much newer approach, see Edric's answer.
Use nargin and ensure your inputs are always in order
Use name-value pairs and an input parser
nargin: slightly simpler but relies on consistent input order
function myFunc( a, b, c )
if nargin < 1 || isempty(a)
a = 2;
end
if nargin < 2 || isempty(b)
b = 2;
end
if nargin < 3 || isempty(c)
c = a + b;
end
end
Using the isempty check you can optionally provide just later arguments, for example myFunc( [], 4 ) would just set b=4 and use the defaults otherwise.
inputParser: more flexible but can't directly handle the c=a+b default
function myFunc( varargin )
p = inputParser;
p.addOptional( 'a', 2 );
p.addOptional( 'b', 2 );
p.addOptional( 'c', NaN ); % Can't default to a+b, default to NaN
p.parse( varargin{:} );
a = p.Results.a;
b = p.Results.b;
c = p.Results.c;
if isnan(c) % Handle the defaulted case
c = a + b;
end
end
This would get used like myFunc( 'b', 4 );. This approach is also agnostic to the input order because of the name-value pairs, so you can also do something like myFunc( 'c', 3, 'a', 1 );
For the function
function function_maker(N)
if N == 1
x = 1.0
else
x = 1
end
f(y) = x+y
end
I want the output of this to not be type-stable, but I want it to generate an f that is type-stable, i.e. uses the type of x determined by the value of N to generate a function dependent on N. Basically, I want the functions that come out of this to be performant, but the function_maker itself doesn't need to be performant because it's only used in the global scope or above a function barrier.
f = function_maker(1)
#code_warntype f(1)
Variables:
#self#::#f#9
y::Int64
Body:
begin
return ((Core.getfield)((Core.getfield)(#self#::#f#9, :x)::ANY, :contents)::ANY + y::Int64)::ANY
end::ANY
This doesn't happen by default. I tried f(y) = x::typeof(x)+y but that didn't work either. Is there a simple way to do this?
There's:
julia> function function_maker2(N)
if N == 1
let x = 1.0
return f(y) = x + y
end
else
let x = 1
return f(y) = x + y
end
end
end
function_maker2 (generic function with 1 method)
julia> f2 = function_maker2(1)
(::f) (generic function with 1 method)
julia> #code_warntype f2(1)
Variables:
#self#::#f#5{Float64}
y::Int64
Body:
begin
return (Base.add_float)((Core.getfield)(#self#::#f#5{Float64}, :x)::Float64, (Base.sitofp)(Float64, y::Int64)::Float64)::Float64
end::Float64
This version separates the x in each branch inside let blocks. Otherwise the compiler seems to get confused.
I am new to Octave, so I am trying to make some simple examples work before moving onto more complex projects.
I am trying to resolve the ODE dy/dx = a*x+b, but without success. Here is the code:
%Funzione retta y = a*x + b. Ingressi: vettore valori t; coefficienti a,b
clear all;
%Inizializza argomenti
b = 1;
a = 1;
x = ones(1,20);
function y = retta(a, x, b) %Definisce funzione
y = ones(1,20);
y = a .* x .+ b;
endfunction
%Calcola retta
x = [-10:10];
a = 2;
b = 2;
r = retta(a, x, b)
c = b;
p1 = (a/2)*x.^2+b.*x+c %Sol. analitica di dy/dx = retta %
plot(x, r, x, p1);
% Risolve eq. differenziale dy/dx = retta %
y0 = b; x0 = 0;
p2 = lsode(#retta, y0, x)
And the output is:
retta3code
r =
-18 -16 -14 -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18 20 22
p1 =
Columns 1 through 18:
82 65 50 37 26 17 10 5 2 1 2 5 10 17 26 37 50 65
Columns 19 through 21:
82 101 122
error: 'b' undefined near line 9 column 16
error: called from:
error: retta at line 9, column 4
error: lsode: evaluation of user-supplied function failed
error: lsode: inconsistent sizes for state and derivative vectors
error: /home/fabio/octave_file/retta3code.m at line 21, column 4
So, the function retta works properly the first time, but it fails when used in lsode.
Why does that happen? What needs to be changed to make the code work?
Somehow you still miss some important parts of the story. To solve an ODE y'=f(y,x) you need to define a function
function ydot = f(y,x)
where ydot has the same dimensions as y, both have to be vectors, even f they are of dimension 1. x is a scalar. For some traditional reason, lsode (a FORTRAN code used in multiple solver packages) prefers the less used order (y,x), in most text books and other solvers you find the order (x,y).
Then to get solution samples ylist over sample points xlist you call
ylist = lsode("f", y0, xlist)
where xlist(1) is the initial time.
The internals of f are independent of the sample list list and what size it has. It is a separate issue that you can use multi-evaluation to compute the exact solution with something like
yexact = solexact(xlist)
To pass parameters, use anonymous functions, like in
function ydot = f(y,x,a,b)
ydot = [ a*x+b ]
end
a_val = ...
b_val = ...
lsode(#(y,x) f(y,x,a_val, b_val), y0, xlist)
The code as modified below works, but I'd prefer to be able to define the parameters a and b out of the function and then pass them to rdot as arguments.
x = [-10,10];
a = 1;
b = 0;
c = b;
p1 = (a/2).*(x.^2)+b.*x+c %Sol. analitica di dy/dx = retta %
function ydot = rdot(ydot, x)
a = 1;
b = 0;
ydot = ones(1,21);
ydot = a.*x .+ b;
endfunction
y0 = p1(1); x0 = 0;
p2 = lsode("rdot", y0, x, x0)'
plot(x, p1, "-k", x, p2, ".r");
I've encountered this code and don't know what its performing :
yk = y == k;
Recreating :
>> a = 1
a = 1
>> b = 2
b = 2
>> c = 3
c = 3
>> a = b == c
a = 0
>>
I think it is a boolean operation. If y == k then yk = 1 else yk = 0 ?
In order to figure out what your statement means, you can refer to Octave's operator precedence. As you can see from that list, assignment (=) has the lowest precedence of any operator (including ==). As a result, the line that you have posted translates to
Perform the relational operation y == k
Assign the result to the variable yk