This is simplified but take as an example the following MATLAB function handle:
F = #(x)[x(1)-x(2);x(2)-x(3)]
The system has of course has many solutions. Is it possible to obtain a solution for a function like this one after substituting at least one variable? For example, substituting x(3)=1 the function would become:
G = #(x)[x(1)-x(2);x(2)-1]
And a solution for the other variables can be obtained. I use fsolve and it works quite well for the system of equations I have. Of course what I need to do can be done using the Symbolic Toolbox, but calling it in a big for loop makes it too slow to use for my code.
I'm trying to come up with some code that can return G given F and a set of indices in x to replace with given values.
What you're basically asking to do is have G call F with values in x reassigned based on a logical replacement index index and a set of replacement values values, which is doable but messy since anonymous functions can only have a single executable statement.
The solution is similar to what is done in this answer, but instead of using the functional form for subscripted reference we'll need to use the functional form for subscripted assignment: subsasgn. Here's what it would look like:
F = #(x)[x(1)-x(2); x(2)-x(3)];
values = [3 2 1];
index = logical([0 0 1]);
G = #(x) F(subsasgn(x, struct('type', '()', 'subs', {{index}}), values(index)));
And here's a test:
>> F([3 2 3])
ans =
1
-1
>> F([3 2 1]) % Replace the last element by 1
ans =
1
1
>> G([3 2 3]) % G handles replacing the last element by 1
ans =
1
1
Related
Lets start with an example. Define a 1-D vector in the base Workspace:
q = [1 0 2 4 2 3 4 2 1 0 2]; % Defined 1-D vector, already in Workspace.
This is in an M-file:
function vector = r(i)
vect(i) = q(3*i-2:3*i-1);
If the function is called, for example,
r(2);
Matlab throws an error: "Undefined function or method 'q' for input arguments of type 'double'".
My question is: is there a way to make Matlab take the vector q "for granted" when calling the function r – q is already defined so technically it could collected from the base Workspace to use it.
Of course the obvious solution is to add q to the argument list:
function vector = r(i,q)
...
and then call it with
r(2,q)
Eventually I will, but I just want to know if there is something I don't know since this could be a pretty useful feature.
You can use global variables, but you shouldn't. They're inefficient too.
If you for some reason don't just want to pass q in as a second argument, here are two better alternatives.
1. Closure via anonymous function
If your function is simple and can be written on one line you can create an anonymous function after defining your q vector:
q = [1 0 2 4 2 3 4 2 1 0 2];
r = #(i)q(3*i-2:3*i-1);
In this case, the function r is a closure in that it captures the predefined q (note that r in this case is slightly different from that in your question). You can call this now with:
out = r(2)
2. Shared variables via sub-function
A more general option is to use a sub-function inside your main M-file function:
function mainFun
q = [1 0 2 4 2 3 4 2 1 0 2];
r(i);
function vect = r(i)
vect(i) = q(3*i-2:3*i-1);
end
end
The vector q is shared between the outer mainFun and the sub-function r. The Matlab Editor should make this clear to you by coloring q differently. More details and examples here.
You can use global variables MATLAB:
1) in the workspace, declare q as global, than set its value
global q;
q = [1 0 2 4 2 3 4 2 1 0 2];
2) in your function, declare q as global and use it:
function vector = r(i)
global q;
vect(i) = q(3*i-2:3*i-1);
I've been searching the net for a couple of mornings and found nothing, hope you can help.
I have an anonymous function like this
f = #(x,y) [sin(2*pi*x).*cos(2*pi*y), cos(2*pi*x).*sin(2*pi*y)];
that needs to be evaluated on an array of points, something like
x = 0:0.1:1;
y = 0:0.1:1;
w = f(x',y');
Now, in the above example everything works fine, the result w is a 11x2 matrix with in each row the correct value f(x(i), y(i)).
The problem comes when I change my function to have constant values:
f = #(x,y) [0, 1];
Now, even with array inputs like before, I only get out a 1x2 array like w = [0,1];
while of course I want to have the same structure as before, i.e. a 11x2 matrix.
I have no idea why Matlab is doing this...
EDIT 1
Sorry, I thought it was pretty clear from what I wrote in the original question, but I see some of you asking, so here is a clarification: what I want is to have again a 11x2 matrix, since I am feeding the function with arrays with 11 elements.
This means I expect to have an output exactly like in the first example, just with changed values in it: a matrix with 11 rows and 2 columns, with only values 0 in the first column and only values 1 in the second, since for all x(i) and y(i) the answer should be the vector [0,1].
It means I expect to have:
w = [0 1
0 1
0 1
...
0 1]
seems pretty natural to me...
You are defining a function f = #(x,y) [0, 1]; which has the input parameters x,y and the output [0,1]. What else do you expect to happen?
Update:
This should match your description:
g=#(x,y)[zeros(size(x)),ones(size(y))]
g(x',y')
Defining an anonymous function f as
f = #(x,y) [0,1];
naturally returns [0,1] for any inputs x and y regardless of the length of those vectors.
This behavior puzzled me also until I realized that I expected f(a,b) to loop over a and b as if I had written
for inc = 1:length(a)
f(a(inc), b(inc))
end
However, f(a,b) does not loop over the length of its inputs, so it merely returns [0,1] regardless of the length of a and b.
The desired behavior can be obtained by defining f as
g=#(x,y)[zeros(size(x)),ones(size(y))]
as Daniel stated in his answer.
I'm trying to use MatLab code as a way to learn math as a programmer.
So reading I'm this post about subspaces and trying to build some simple matlab functions that do it for me.
Here is how far I got:
function performSubspaceTest(subset, numArgs)
% Just a quick and dirty function to perform subspace test on a vector(subset)
%
% INPUT
% subset is the anonymous function that defines the vector
% numArgs is the the number of argument that subset takes
% Author: Lasse Nørfeldt (Norfeldt)
% Date: 2012-05-30
% License: http://creativecommons.org/licenses/by-sa/3.0/
if numArgs == 1
subspaceTest = #(subset) single(rref(subset(rand)+subset(rand))) ...
== single(rref(rand*subset(rand)));
elseif numArgs == 2
subspaceTest = #(subset) single(rref(subset(rand,rand)+subset(rand,rand))) ...
== single(rref(rand*subset(rand,rand)));
end
% rand just gives a random number. Converting to single avoids round off
% errors.
% Know that the code can crash if numArgs isn't given or bigger than 2.
outcome = subspaceTest(subset);
if outcome == true
display(['subset IS a subspace of R^' num2str(size(outcome,2))])
else
display(['subset is NOT a subspace of R^' num2str(size(outcome,2))])
end
And these are the subset that I'm testing
%% Checking for subspaces
V = #(x) [x, 3*x]
performSubspaceTest(V, 1)
A = #(x) [x, 3*x+1]
performSubspaceTest(A, 1)
B = #(x) [x, x^2, x^3]
performSubspaceTest(B, 1)
C = #(x1, x3) [x1, 0, x3, -5*x1]
performSubspaceTest(C, 2)
running the code gives me this
V =
#(x)[x,3*x]
subset IS a subspace of R^2
A =
#(x)[x,3*x+1]
subset is NOT a subspace of R^2
B =
#(x)[x,x^2,x^3]
subset is NOT a subspace of R^3
C =
#(x1,x3)[x1,0,x3,-5*x1]
subset is NOT a subspace of R^4
The C is not working (only works if it only accepts one arg).
I know that my solution for numArgs is not optimal - but it was what I could come up with at the current moment..
Are there any way to optimize this code so C will work properly and perhaps avoid the elseif statments for more than 2 args..?
PS: I couldn't seem to find a build-in matlab function that does the hole thing for me..
Here's one approach. It tests if a given function represents a linear subspace or not. Technically it is only a probabilistic test, but the chance of it failing is vanishingly small.
First, we define a nice abstraction. This higher order function takes a function as its first argument, and applies the function to every row of the matrix x. This allows us to test many arguments to func at the same time.
function y = apply(func,x)
for k = 1:size(x,1)
y(k,:) = func(x(k,:));
end
Now we write the core function. Here func is a function of one argument (presumed to be a vector in R^m) which returns a vector in R^n. We apply func to 100 randomly selected vectors in R^m to get an output matrix. If func represents a linear subspace, then the rank of the output will be less than or equal to m.
function result = isSubspace(func,m)
inputs = rand(100,m);
outputs = apply(func,inputs);
result = rank(outputs) <= m;
Here it is in action. Note that the functions take only a single argument - where you wrote c(x1,x2)=[x1,0,x2] I write c(x) = [x(1),0,x(2)], which is slightly more verbose, but has the advantage that we don't have to mess around with if statements to decide how many arguments our function has - and this works for functions that take input in R^m for any m, not just 1 or 2.
>> v = #(x) [x,3*x]
>> isSubspace(v,1)
ans =
1
>> a = #(x) [x(1),3*x(1)+1]
>> isSubspace(a,1)
ans =
0
>> c = #(x) [x(1),0,x(2),-5*x(1)]
>> isSubspace(c,2)
ans =
1
The solution of not being optimal barely scratches the surface of the problem.
I think you're doing too much at once: rref should not be used and is complicating everything. especially for numArgs greater then 1.
Think it through: [1 0 3 -5] and [3 0 3 -5] are both members of C, but their sum [4 0 6 -10] (which belongs in C) is not linear product of the multiplication of one of the previous vectors (e.g. [2 0 6 -10] ). So all the rref in the world can't fix your problem.
So what can you do instead?
you need to check if
(randn*subset(randn,randn)+randn*subset(randn,randn)))
is a member of C, which, unless I'm mistaken is a difficult problem: Conceptually you need to iterate through every element of the vector and make sure it matches the predetermined condition. Alternatively, you can try to find a set such that C(x1,x2) gives you the right answer. In this case, you can use fminsearch to solve this problem numerically and verify the returned value is within a defined tolerance:
[s,error] = fminsearch(#(x) norm(C(x(1),x(2)) - [2 0 6 -10]),[1 1])
s =
1.999996976386119 6.000035034493023
error =
3.827680714104862e-05
Edit: you need to make sure you can use negative numbers in your multiplication, so don't use rand, but use something else. I changed it to randn.
The problem:
I want to index into the result of a function call that returns a variable number of output arguments without storing the result in a temporary.
getel = #(x,i) x(i); #% simple anonymous function to index into a vector
x = zeros(2,2,2);
row = getel(ind2sub(size(x), 8), 1) #% desired: 2 (row 2)
#% actual: 8 (linear index)-because ind2sub is returning 1 value only
[row col dep]=ind2sub(size(x),8) #% row=2, ind2sub returning 3 values
Example usage:
x(1).val1 = [1 2 3];
x(1).val2 = [2 1 2];
x(2).val1 = [2 1 2];
x(2).val2 = [1 0 0];
#% The normal way I would do this, with a temporary variable
[~,ind] = min(x(1).val2); #% ind=2
v(1) = x(1).val1(ind);
[~,ind] = min(x(2).val2); #% ind=2
v(2) = x(2).val1(ind);
#% I'd like to be able to do this with arrayfun:
v = arrayfun(#(s) s.val1(min(s.val2), x);
-------^ returns value of minimum, not index
The above arrayfun doesn't work - the form of min that is called returns one output: the minimum value. To make it work right, one option would be the following hypothetical function call:
v = arrayfun(#(s) s.val1(getoutputnum(2, 2, #min, s.val2)), x);
hypothetical function -----------^ ^ ^ ^-func ^--func args
which form (nargout) of func ---| |- which arg to return
I realize that for the above scenario, I could use
s.val1(find(s.val2==min(s.val2),1,'first'))
or other tricks, but that isn't possible in all cases.
In the case of ind2sub, I may want to know the index into a particular dimension (columns, say) - but the 1-output form of the function returns only a linear index value - the n-dimensional form needs to be called, even if the value of dimension 1 is what I care about.
Note: I realize that writing a function file would make this trivial: use ~ and the [out] = func(in) form. However, when writing scripts or just on the command line, it would be nice to be able to do this all within anonymous functions. I also realize that there are undoubtedly other ways to get around the problem; I would just like to know if it is possible to specify which form of a function to call, and perhaps which output number to be returned, without using the out=func(in) syntax, thus allowing functions to be nested much more nicely.
Could you do something like this?
In its own file:
function idx=mymin(x)
[~,idx] = min(x);
In your code:
v = arrayfun(#(s) s.val1(mymin(s.val2), x);
Might have syntax errors; I don't have MATLAB on the computer I'm writing this on. The idea is there though: just wrap MATLAB's min and capture the second argument, which is the logical indexing for the position of the minimum value in x.
I can get ind2sub() to return the variable number of args like this:
x = zeros(2,2,2);
c = cell(ndims(x),1);
[c{:}] = ind2sub(size(x), 8);
The c cell array will now have the 3D indices c = {2;2;2}.
[c{:}] = ind2sub(size(x), 2);
would produce c = {2;1;1}.
Is this what you were looking for?
I'm a little surprised that MATLAB doesn't have a Map function, so I hacked one together myself since it's something I can't live without. Is there a better version out there? Is there a somewhat-standard functional programming library for MATLAB out there that I'm missing?
function results = map(f,list)
% why doesn't MATLAB have a Map function?
results = zeros(1,length(list));
for k = 1:length(list)
results(1,k) = f(list(k));
end
end
usage would be e.g.
map( #(x)x^2,1:10)
The short answer: the built-in function arrayfun does exactly what your map function does for numeric arrays:
>> y = arrayfun(#(x) x^2, 1:10)
y =
1 4 9 16 25 36 49 64 81 100
There are two other built-in functions that behave similarly: cellfun (which operates on elements of cell arrays) and structfun (which operates on each field of a structure).
However, these functions are often not necessary if you take advantage of vectorization, specifically using element-wise arithmetic operators. For the example you gave, a vectorized solution would be:
>> x = 1:10;
>> y = x.^2
y =
1 4 9 16 25 36 49 64 81 100
Some operations will automatically operate across elements (like adding a scalar value to a vector) while others operators have a special syntax for element-wise operation (denoted by a . before the operator). Many built-in functions in MATLAB are designed to operate on vector and matrix arguments using element-wise operations (often applied to a given dimension, such as sum and mean for example), and thus don't require map functions.
To summarize, here are some different ways to square each element in an array:
x = 1:10; % Sample array
f = #(x) x.^2; % Anonymous function that squares each element of its input
% Option #1:
y = x.^2; % Use the element-wise power operator
% Option #2:
y = f(x); % Pass a vector to f
% Option #3:
y = arrayfun(f, x); % Pass each element to f separately
Of course, for such a simple operation, option #1 is the most sensible (and efficient) choice.
In addition to vector and element-wise operations, there's also cellfun for mapping functions over cell arrays. For example:
cellfun(#upper, {'a', 'b', 'c'}, 'UniformOutput',false)
ans =
'A' 'B' 'C'
If 'UniformOutput' is true (or not provided), it will attempt to concatenate the results according to the dimensions of the cell array, so
cellfun(#upper, {'a', 'b', 'c'})
ans =
ABC
A rather simple solution, using Matlab's vectorization would be:
a = [ 10 20 30 40 50 ]; % the array with the original values
b = [ 10 8 6 4 2 ]; % the mapping array
c = zeros( 1, 10 ); % your target array
Now, typing
c( b ) = a
returns
c = 0 50 0 40 0 30 0 20 0 10
c( b ) is a reference to a vector of size 5 with the elements of c at the indices given by b. Now if you assing values to this reference vector, the original values in c are overwritten, since c( b ) contains references to the values in c and no copies.
It seems that the built-in arrayfun doesn't work if the result needed is an array of function:
eg:
map(#(x)[x x^2 x^3],1:10)
slight mods below make this work better:
function results = map(f,list)
% why doesn't MATLAB have a Map function?
for k = 1:length(list)
if (k==1)
r1=f(list(k));
results = zeros(length(r1),length(list));
results(:,k)=r1;
else
results(:,k) = f(list(k));
end;
end;
end
If matlab does not have a built in map function, it could be because of efficiency considerations. In your implementation you are using a loop to iterate over the elements of the list, which is generally frowned upon in the matlab world. Most built-in matlab functions are "vectorized", i. e. it is more efficient to call a function on an entire array, than to iterate over it yourself and call the function for each element.
In other words, this
a = 1:10;
a.^2
is much faster than this
a = 1:10;
map(#(x)x^2, a)
assuming your definition of map.
You don't need map since a scalar-function that is applied to a list of values is applied to each of the values and hence works similar to map. Just try
l = 1:10
f = #(x) x + 1
f(l)
In your particular case, you could even write
l.^2
Vectorizing the solution as described in the previous answers is the probably the best solution for speed. Vectorizing is also very Matlaby and feels good.
With that said Matlab does now have a Map container class.
See http://www.mathworks.com/help/matlab/map-containers.html