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.
Related
I am doing a project involving scientific computing. The following are three variables and their values I got after some experiments.
There is also an equation with three unknowns, a, b and c:
x=(a+0.98)/y+(b+0.7)/z+c
How do I get values of a,b,c using the above? Is this possible in MATLAB?
This sounds like a regression problem. Assuming that the unexplained errors in measurements are Gaussian distributed, you can find the parameters via least squares. Basically, you'd have to rewrite the equation so that you get this to the form of ma + nb + oc = p and then you have 6 equations with 3 unknowns (a, b, c) and these parameters can be found through optimization by least squares. Therefore, with some algebra, we get:
za + yb + yzc = xyz - 0.98z - 0.7z
As such, m = z, n = y, o = yz, p = xyz - 0.98z - 0.7z. I'll leave that for you as an exercise to verify that my algebra is right. You can then form the matrix equation:
Ax = d
We would have 6 equations and we want to solve for x where x = [a b c]^{T}. To solve for x, you can employ what is known as the pseudoinverse to retrieve the parameters that best minimize the error between the true output and the output that is generated by these parameters if you were to use the same input data.
In other words:
x = A^{+}d
A^{+} is the pseudoinverse of the matrix A and is matrix-vector multiplied with the vector d.
To put our thoughts into code, we would define our input data, form the A matrix and d vector where each row shared between them both is one equation, and then employ the pseudoinverse to find our parameters. You can use the ldivide (\) operator to do the job:
%// Define x y and z
x = [9.98; 8.3; 8.0; 7; 1; 12.87];
y = [7.9; 7.5; 7.4; 6.09; 0.9; 11.23];
z = [7.1; 5.6; 5.9; 5.8; -1.8; 10.8];
%// Define A matrix
A = [z y y.*z];
%// Define d vector
d = x.*y.*z - 0.98*z - 0.7*z;
%// Find parameters via least-squares
params = A\d;
params stores the parameters a, b and c, and we get:
params =
-37.7383
-37.4008
19.5625
If you want to double-check how close the values are, you can simply use the above expression in your post and compare with each of the values in x:
a = params(1); b = params(2); c = params(3);
out = (a+0.98)./y+(b+0.7)./z+c;
disp([x out])
9.9800 9.7404
8.3000 8.1077
8.0000 8.3747
7.0000 7.1989
1.0000 -0.8908
12.8700 12.8910
You can see that it's not exactly close, but the parameters you got would be the best in a least-squares error sense.
Bonus - Fitting with RANSAC
You can see that some of the predicted values (right column in the output) are more off than others. This is because we used all points in your data to find the appropriate model. One technique that is used to minimize error and increase the robustness of the model estimation is to use something called RANSAC, or RANdom SAmple Consensus. The basic methodology behind RANSAC is that for a certain number of iterations, you take your data and randomly sample the least amount of points necessary to find a model. Once you find this model, you find the overall error if you were to use these parameters to describe your data. You keep randomly choosing points, finding your model, and finding the error and the iteration that produced the least amount of error would be the parameters you keep to define the overall model.
As you can see above, one error that we can define is the sum of absolute differences between the true x points and the predicted x points. There are many other measures, such as the sum of squared errors, but let's stick with something simple for now. If you take a look at the above formulation, we need a minimum of three equations in order to define a, b and c, and so for each iteration, we'd randomly select three points without replacement I might add, find our model, determine the error, and keep iterating and finding the parameters with the least amount of error.
Therefore, you could write a RANSAC algorithm like so:
%// Define cost and number of iterations
cost = Inf;
iterations = 50;
%// Set seed for reproducibility
rng(123);
%// Define x y and z
x = [9.98; 8.3; 8.0; 7; 1; 12.87];
y = [7.9; 7.5; 7.4; 6.09; 0.9; 11.23];
z = [7.1; 5.6; 5.9; 5.8; -1.8; 10.8];
for idx = 1 : iterations
%// Determine where we would need to sample
ind = randperm(numel(x), 3);
xs = x(ind); ys = y(ind); zs = z(ind); %// Sample
%// Define A matrix
A = [zs ys ys.*zs];
%// Define d vector
d = xs.*ys.*zs - 0.98*zs - 0.7*zs;
%// Find parameters via least-squares
params = A\d;
%// Determine error
a = params(1); b = params(2); c = params(3);
out = (a+0.98)./y+(b+0.7)./z+c;
err = sum(abs(x - out));
%// If error produced is less than current error
%// then save parameters
if err < cost
cost = err;
final_params = params;
end
end
When I run the above code, I get for my parameters:
final_params =
-38.1519
-39.1988
19.7472
Comparing this with our x, we get:
a = final_params(1); b = final_params(2); c = final_params(3);
out = (a+0.98)./y+(b+0.7)./z+c;
disp([x out])
9.9800 9.6196
8.3000 7.9162
8.0000 8.1988
7.0000 7.0057
1.0000 -0.1667
12.8700 12.8725
As you can see, the values are improved - especially the fourth and sixth points... and compare it to the previous version:
9.9800 9.7404
8.3000 8.1077
8.0000 8.3747
7.0000 7.1989
1.0000 -0.8908
12.8700 12.8910
You can see that the second value is worse off than the previous version, but the other numbers are much more closer to the true values.
Have fun!
I am having trouble using fminsearch: getting the error that there were not enough input arguments for my function.
f = #(x1,x2,x3) x1.^2 + 3.*x2.^2 + 4.*x3.^2 - 2.*x1.*x2 + 5.*x1 + 3.*x2 + 2.*x3;
[x, val] = fminsearch(f,0)
Is there something wrong with my function? I keep getting errors anytime I want to use it as an input function with any other command.
I am having trouble using fminsearch [...]
Stop right there and think some more about the function you're trying to minimize.
Numerical optimization (which is what fminsearch does) is unnecessary, here. Your function is a quadratic function of vector x; in other words, its value at x can be expressed as
x^T A x + b^T x
where matrix A and vector b are defined as follows (using MATLAB notation):
A = [ 1 -1 0;
-1 3 0;
0 0 4]
and
b = [5 3 2].'
Because A is positive definite, your function has one and only one minimum, which can be computed in MATLAB with
x_sol = -0.5 * A \ b;
Now, if you're curious about the cause of the error you ran into, have a look at fuesika's answer; but do without fminsearch whenever you can.
It is exactly what Matlab is telling you: your function expects three arguments. You are passing only one.
Instead of
[x, val] = fminsearch(f,0)
you should call it like
[x, val] = fminsearch(f,[0,0,0])
since you define the function f to accept a three dimensional vector as input only.
You can read more about the specification of fminsearch in the online documentation at http://mathworks.com/help/matlab/ref/fminsearch.html:
x = fminsearch(fun,x0) starts at the point x0 and returns a value x
that is a local minimizer of the function described in fun. x0 can be
a scalar, vector, or matrix. fun is a function_handle.
I have a 3x3 matrix, A. I also compute a value, g, as the maximum eigen value of A. I am trying to change the element A(3,3) = 0 for all values from zero to one in 0.10 increments and then update g for each of the values. I'd like all of the other matrix elements to remain the same.
I thought a for loop would be the way to do this, but I do not know how to update only one element in a matrix without storing this update as one increasingly larger matrix. If I call the element at A(3,3) = p (thereby creating a new matrix Atry) I am able (below) to get all of the values from 0 to 1 that I desired. I do not know how to update Atry to get all of the values of g that I desire. The state of the code now will give me the same value of g for all iterations, as expected, as I do not know how to to update Atry with the different values of p to then compute the values for g.
Any suggestions on how to do this or suggestions for jargon or phrases for me to web search would be appreciated.
A = [1 1 1; 2 2 2; 3 3 0];
g = max(eig(A));
% This below is what I attempted to achieve my solution
clear all
p(1) = 0;
Atry = [1 1 1; 2 2 2; 3 3 p];
g(1) = max(eig(Atry));
for i=1:100;
p(i+1) = p(i)+ 0.01;
% this makes a one giant matrix, not many
%Atry(:,i+1) = Atry(:,i);
g(i+1) = max(eig(Atry));
end
This will also accomplish what you want to do:
A = #(x) [1 1 1; 2 2 2; 3 3 x];
p = 0:0.01:1;
g = arrayfun(#(x) eigs(A(x),1), p);
Breakdown:
Define A as an anonymous function. This means that the command A(x) will return your matrix A with the (3,3) element equal to x.
Define all steps you want to take in vector p
Then "loop" through all elements in p by using arrayfun instead of an actual loop.
The function looped over by arrayfun is not max(eig(A)) but eigs(A,1), i.e., the 1 largest eigenvalue. The result will be the same, but the algorithm used by eigs is more suited for your type of problem -- instead of computing all eigenvalues and then only using the maximum one, you only compute the maximum one. Needless to say, this is much faster.
First, you say 0.1 increments in the text of your question, but your code suggests you are actually interested in 0.01 increments? I'm going to operate under the assumption you mean 0.01 increments.
Now, with that out of the way, let me state what I believe you are after given my interpretation of your question. You want to iterate over the matrix A, where for each iteration you increase A(3, 3) by 0.01. Given that you want all values from 0 to 1, this implies 101 iterations. For each iteration, you want to calculate the maximum eigenvalue of A, and store all these eigenvalues in some vector (which I will call gVec). If this is correct, then I believe you just want the following:
% Specify the "Current" A
CurA = [1 1 1; 2 2 2; 3 3 0];
% Pre-allocate the values we want to iterate over for element (3, 3)
A33Vec = (0:0.01:1)';
% Pre-allocate a vector to store the maximum eigenvalues
gVec = NaN * ones(length(A33Vec), 1);
% Loop over A33Vec
for i = 1:1:length(A33Vec)
% Obtain the version of A that we want for the current i
CurA(3, 3) = A33Vec(i);
% Obtain the maximum eigen value of the current A, and store in gVec
gVec(i, 1) = max(eig(CurA));
end
EDIT: Probably best to paste this code into your matlab editor. The stack-overflow automatic text highlighting hasn't done it any favors :-)
EDIT: Go with Rody's solution (+1) - it is much better!
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