I faced this problem for the first time and I can't figure it out.
Let's say we have an array and a foreach loop.
Something like this:
var v = array(10,20,50);
var i = 0
write(foo(v, v[i++]));
function foo(ref int[] v, name int y){
foreach(int j in v){
write(y);
}
return y;
}
Am I wrong or something is not gonna work here? I mean, every time I will loop thru the foreach I will evaluate the y (by name) so, being v[i++] I will increase the value of my i variable by one.
So first step y = v[0] so write(10) then i++ (i=1).
Second step y = v[1] so write(20) then I increase i by one (i=2).
Third and last step y = v[2] so write(50) and i++ again, which now equals to 3.
Now, what value should it return?! If I evaluate again y, I can't do y = v[3] cause I go out of bounds. Am I doing anything wrong? Should I just evaluate y once, before the foreach loop? There must be something with the foreach I'm not taking in account when calling parameters by name.
Dunno, I'm a bit confused.
Thanks in advance!
Related
I have a scilab function that looks something like this (very simplified code just to get the concept of how it works):
function [A, S, Q]=myfunc(a)
A = a^2;
S = a+a+a;
if S > A then
Q = "Bigger";
else
Q = "Lower";
end
endfunction
And I get the expected result if I run:
--> [A,S,Q]=myfunc(2)
Q =
Bigger
S =
6.
A =
4.
But if I put matrices into the function I expect to get equivalent matrices back as an answer with a result but instead I got this:
--> [A,S,Q]=myfunc([2 4 6 8])
Q =
Lower
S =
6. 12. 18. 24.
A =
4. 16. 36. 64.
Why isn't Q returning matrices of values like S and A? And how do I achieve that it will return "Bigger. Lower. Lower. Lower." as an answer? That is, I want to perform the operation on each element of the matrix.
Because in your program you wrote Q = "Bigger" and Q = "Lower". That means that Q will only have one value. If you want to store the comparisons for every value in A and S, you have to make Scilab do that.
You can achieve such behavior by using loops. This is how you can do it by using two for loops:
function [A, S, Q]=myfunc(a)
A = a^2;
S = a+a+a;
//Get the size of input a
[nrows, ncols] = size(a)
//Traverse all rows of the input
for i = 1 : nrows
//Traverse all columns of the input
for j = 1 : ncols
//Compare each element
if S(i,j) > A(i,j) then
//Store each result
Q(i,j) = "Bigger"
else
Q(i,j) = "Lower"
end
end
end
endfunction
Beware of A = a^2. It can break your function. It has different behaviors if input a is a vector (1-by-n or n-by-1 matrix), rectangle matrix (m-by-n matrix, m ≠ n ), or square matrix (n-by-n matrix):
Vector: it works like .^, i.e. it raises each element individually (see Scilab help).
Rectangle: it won't work because it has to follow the rule of matrix multiplication.
Square: it works and follows the rule of matrix multiplication.
I will add that in Scilab, the fewer the number of loop, the better : so #luispauloml answer may rewrite to
function [A, S, Q]=myfunc(a)
A = a.^2; // used element wise power, see luispauloml advice
S = a+a+a;
Q(S > A) = "Bigger"
Q(S <= A) = "Lower"
Q = matrix(Q,size(a,1),size(a,2)) // a-like shape
endfunction
Say that myFunction has been invoked from somewhere like this
...
myFunction (b);
...
Now, in the definition of myFunction, can I obtain the name of the variable in the call?
function myFunction (a)
...
inputVble = whoscalling; % this will result in "b"
...
May be it can be done if b is a global variable?
Note: this is not like in a couple other questions, where you want to know the name of the input argument, i.e. a.
function ret = doit (a, b)
inputname (1)
inputname (2)
ret = a + b;
endfunction
x = 4;
y = 5;
doit (x, y)
returns
ans = x
ans = y
ans = 9
But I want to mention that in my opinion, making a function somewhat dependent on the function names is bad style and shouldn't be done even if it's possible.
EDIT: I think the main reasonable use for inputname is inside a display routine for #classes. For example http://hg.savannah.gnu.org/hgweb/octave/file/51a1d1164449/examples/code/%40polynomial/display.m
function display (p)
...
fprintf ("%s =", inputname (1));
Or inside celldisp:
octave:2> a = {"huhu", pi}
a =
{
[1,1] = huhu
[1,2] = 3.1416
}
octave:3> celldisp (a)
a{1} =
huhu
a{2} =
3.1416
So I have a for-loop in MATLAB, where either a vector x will be put through one function, say, cos(x).^2, or a different choice, say, sin(x).^2 + 9.*x. The user will select which of those functions he wants to use before the for-loop.
My question is, I dont want the loop to check what the user selected on every iteration. Is there a way to use a pointer to a function, (user defined, or otherwise), that every iteration will use automatically?
This is inside a script by the way, not a function.
Thanks
You can use function_handles. For your example (to run on all available functions using a loop):
x = 1:10; % list of input values
functionList = {#(x) cos(x).^2, #(x) sin(x).^2 + 9*x}; % function handle cell-array
for i=1:length(functionList)
functionOut{i} = functionList{i}(x); % output of each function to x
end
You can try something like the following:
userChoice = 2;
switch userChoice
case 1
myFun = #(x) sin(x).^2 + 9.*x;
case 2
myFun = #(x) cos(x).^2;
end
for k = 1:10
x(k,:) = myFun(rand(1,10));
end
I am learning MatLab on my own, and I have this assignment in my book which I don't quite understand. Basically I am writing a function that will calculate sine through the use of Taylor series. My code is as follows so far:
function y = sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
The book is now asking me to include an optional output err which will calculate the difference between sin(x) and y. The book hints that I may use nargout to accomplish this, but there are no examples in the book on how to use this, and reading the MatLab help on the subject did not make my any wiser.
If anyone can please help me understand this, I would really appreciate it!
The call to nargout checks for the number of output arguments a function is called with. Depending on the size of nargout you can assign entries to the output argument varargout. For your code this would look like:
function [y varargout]= sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
if nargout ==2
varargout{1} = sin(x)'-y;
end
Compare the output of
[y] = sine_series(rand(1,10),3)
and
[y err] = sine_series(rand(1,10),3)
to see the difference.
I have a problem with estimation.
I have a function, which is dependent on the values of an unknown vector V = [v1, …, v4].
I also have a vector of reference data YREF = [yref1, …, yrefn].
I would like to write a function, which returns the vector Y (in order to compare it later, say using lsqnonlin). I am aware of the “arrayfun”, but it seems not to work.
I have a subfunction, which returns a concrete value from the range [-100, 100],
%--------------------------------------------------------------------------
function y = SubFunction(Y, V)
y = fzero(#(x) v(1).*sinh(x./v(2)) + v(3).*x - Y, [-100 100]);
end
%--------------------------------------------------------------------------
then I make some operations on the results:
%--------------------------------------------------------------------------
function y = SomeFunction(Y,V)
temp = SubFunction (Y,V);
y = temp + v(4).*Y;
end
%--------------------------------------------------------------------------
These functions work well for a single value of Y, but not for the whole vector. How to store the results into a matrix for future comparison?
Thanks in advance
Chris
If Y is a vector, then the anonymous function defined as an argument to fzero returns a vector, not a scalar.
You can solve it by using a loop (notice the Y(k) inside the anonymous function definition):
function y = SubFunction(Y, v)
y = zeros (size(Y));
for k = 1 : length (Y)
y(k) = fzero(#(x) v(1).*sinh(x./v(2)) + v(3).*x - Y(k), [-100 100]);
end
end