I have a matrix y and a vector x, I need to find all the possible vectors resulted from the mapping of each value in x into each vector in y.
That is difficult to be understood; let's explain is with an example:
Here is an example, I have the vector x = [0.7 + 0.7i; 0.7-0.7i]; The matrix y = [1 0; 2 0; 1 2]; the resulted matrix is supposed to be like this one Z = [0.7 + 0.7i 0; 0.7-0.7i 0; 0 0.7 + 0.7i; 0 0.7-0.7i; 0.7 + 0.7i 0.7 + 0.7i; 0.7 + 0.7i 0.7-0.7i; 0.7 - 0.7i 0.7-0.7i ; 0.7 - 0.7i 0.7+0.7i]; . That is equivalent into Z = [x_1 0; x_2 0; 0 x_1; 0 x_2; x_1 x_1; x_1 x_2; x_2 x_2; x_2 x_1];. That means it map each value in x into the row of Z according to the index value in y.
Here is my try code:
clear all; clc;
y = [];
G = 2;
v = 1 : G;
for i = 1: G
x=nchoosek(v,i);
m = zeros(size(x,1),G-i);
y =[y ; x m]; % creat the matrix y
end
x = [0.7 + 0.7i; 0.7-0.7i];
Z = []; s = zeros(G,1);
for k=1:size(x,1)
for i=1:size(y,1)
n=y(i,:);
n=n(n ~= 0);
s(n)=x(k);
Z=[Z s];
s = zeros(G,1);
end
end
The problem in my code that matrix Z show the inverse, it means it takes the input x_1 from x and then map it into all possible values in y. For example the matrix Z starts with [x_1 0; 0 x_1; x_1 x_1 ….], however that should be the inverse, which means takes each values in x and map it as shown in the above example [x_1 0; x_2 0; x_3 0 …..]. The second issue, when y contains more than non-zeros values, my code cannot get all possible vectors, it can only get [x_1 x_1; x_2 x_2]; but I cannot map the other possibilities which are [x_1 x_2; x_2 x_1] and so on.
How can I solve that issue?
UPDATE
Here is the updated question with clear description. I have the vector x and matrix y, I need to fill the matrix z following the index taken from each row in matrix y. For example, if the first row in matrix y is [1 0] or [0 1]; then I will take all possible values from x and put it in z following the number taken from the row in y which is 1 in this case. Then, the same case for row 2 in matrix y which is [2 0] or [0 2]; it means that second column in z will be filled with all possible values in x.
Then, the two columns in z can be filled which is equivalent into the case [1 2] in y, so it will take the first value from x and fill it with all other possible values from x, and so on. The rows in z should not be repeated.
The matrix Z is exactly as shown with below answer of AboAmmar below, but using the loop if with longer vector x and bigger matrix y will be little bit complicated.
As you describe it, there are 4 distinct cases for each row of y and the corresponding output:
[0 1] or [1 0] => [x 0]
[0 2] or [2 0] => [0 x]
[1 2] => [x1 x1; x1 x2; x2 x2; x2 x1]
[2 1] => [x1 x1; x2 x1; x2 x2; x1 x2]
These don't seem to follow any obvious rule. So, the easiest (but not smartest) solution is to use if-else and select the suitable case from the above. We don't have all the information about the possible indices, or if rows like [1 1] and [2 2] might happen, so the following solution is by no means exhaustive; surprising errors might happen if other inputs are fed into y matrix.
y = [];
G = 2;
v = 1 : G;
for i = 1: G
x = nchoosek(v,i);
m = zeros(size(x,1),G-i);
y = [y ; x m]; % creat the matrix y
end
Z = [];
x = [0.7 + 0.7i; 0.7-0.7i]
for i = 1:size(y,1)
r = y(i,:);
if ismember(r, [1 0; 0 1], 'rows')
Z(end+1:end+2,:) = [x [0; 0]];
elseif ismember(r, [2 0; 0 2], 'rows')
Z(end+1:end+2,:) = [[0; 0] x];
elseif ismember(r, [1 2], 'rows')
Z(end+1:end+4,:) = [x(1) x(1); x(1) x(2); x(2) x(2); x(2) x(1)];
elseif ismember(r, [2 1], 'rows')
Z(end+1:end+4,:) = [x(1) x(1); x(2) x(1); x(2) x(2); x(1) x(2)];
end
end
Z =
0.7000 + 0.7000i 0.0000 + 0.0000i
0.7000 - 0.7000i 0.0000 + 0.0000i
0.0000 + 0.0000i 0.7000 + 0.7000i
0.0000 + 0.0000i 0.7000 - 0.7000i
0.7000 + 0.7000i 0.7000 + 0.7000i
0.7000 + 0.7000i 0.7000 - 0.7000i
0.7000 - 0.7000i 0.7000 - 0.7000i
0.7000 - 0.7000i 0.7000 + 0.7000i
Your code is valid if you have fix length in y, for example if each vector in y has one value and others are zeros, or two non-zeros values ...etc.
So you can do your code for each length separately and then build the matrix Z by combining all other matrices.
Related
For the below given input and output, the matrix A can be found out by pseudoinverse or mrdivision in MATLAB. Similarly, I would now like to know, how to determine A, if my output signal Y matrix contains additive zero mean, uncorrelated, Gaussian noise?
x1 = [1 1 1]';
x2 = [0 1 1]';
x3 = [0 0 1]';
x4 = [1 0 1]';
y1 = [1 2 0]';
y2 = [-1 0 3]';
y3 = [3 1 1]';
y4 = [5 3 -2]';
X = [x1 x2 x3 x4];
Y = [y1 y2 y3 y4];
A = Y/X
Also, I have modelled the unknown noisy output as below:
y1_n = y1 + sqrt(var(y1))*randn(size(y1));
y2_n = y2 + sqrt(var(y2))*randn(size(y2));
y3_n = y3 + sqrt(var(y3))*randn(size(y3));
y4_n = y4 + sqrt(var(y4))*randn(size(y4));
Y = [y1_n y2_n y3_n y4_n];
The statement A = Y/X solves the linear system of equations A*X = Y. If the system is overdetermined, as in your case, the solution given is the least squares solution. Thus, if you have additive, zero mean, uncorrelated, Gaussian noise on Y, then A = Y/X will give you the best possible, unbiased, estimate of A.
Note that the noise you add to your Y matrix is quite large, hence the estimate of A is far away from the ideal. If you add less noise, the estimate will be closer:
x1 = [1 1 1]';
x2 = [0 1 1]';
x3 = [0 0 1]';
x4 = [1 0 1]';
X = [x1 x2 x3 x4];
y1 = [1 2 0]';
y2 = [-1 0 3]';
y3 = [3 1 1]';
y4 = [5 3 -2]';
Y = [y1 y2 y3 y4];
for n = [1,0.1,0.01,0]
Y_n = Y + n*randn(size(Y));
A = Y_n/X;
fprintf('n = %f, A = \n',n)
disp(A)
end
Output:
n = 1.000000, A =
2.9728 -5.5407 2.8011
2.6563 -1.3166 0.6596
-3.3366 1.1349 1.5342
n = 0.100000, A =
2.0011 -4.0256 2.9402
1.9223 -1.0029 1.0921
-3.1383 1.9874 1.0913
n = 0.010000, A =
1.9903 -3.9912 2.9987
1.9941 -1.0001 1.0108
-3.0015 2.0001 1.0032
n = 0.000000, A =
2.0000 -4.0000 3.0000
2.0000 -1.0000 1.0000
-3.0000 2.0000 1.0000
Of course if you make X and Y larger by adding more vectors you'll get a better estimate too, and will be able to compensate more noisy data.
I want to use solve() to solve a large system of linear equations. The solve() function needs equations and variables. I use a loop to generate the equations and my variables are contained in a large array. This is a simple code of what I am trying to do:
x = sym('x',[1 3])
eqn = sym('eqn', [1,3])
eqn1 = 2*x1 + x2 + x3 - 2 == 0
eqn2 = 2*x2 -x2 -x1- x3 == 3
eqn3 = 2*x2+ x1 + 3*x3 == -10
Y = solve(eqn, x)
MATLAB does not recognize my variable x1. I have solved the same system using the following code:
syms x1 x2 x3
eqn1 = 2*x1 + x2 + x3 == 2
eqn2 = 2*x2 -x2 -x1 - x3 == 3
eqn3 = 2*x2+ x1 + 3*x3 == -10
X = solve([eqn1, eqn2, eqn3, eqn4], [x1 x2 x3])
structfun(#subs,X)
But this is useless for a very large number of equations. What am I doing wrong?
You don't need symbolic (syms) for that. This is a standard linear system of equations that can be represented as:
Ax = b where A = [2 1 1; -1 1 -1; 1 2 3], x = [x1; x2; x3] and b = [0; 3; -10]
To solve for x, you would first define
A = [2 1 1; -1 1 -1; 1 2 3]
and
b = [0; 3; -10]
and then solve using
x = A\b
PS. There are some odd things in your question, eg. in eq.2 eqn2 = 2*x2 -x2 -x1- x3 == 3 I assume you omitted simplying this to -x1 +x2 -x3 ==3.
PS2. This is pretty standard Matlab, you can find a lot of info under the standard mldivide page in the documentation along with a lot similar questions here on SO.
My question seems wierd, because I know that we can't use a matrix as input in linspace(x1,x2,n) function. But my question is more something like : I have a vector A=linspace(0,Amax,N), and I want to build a serie of vector B_k or big matrix B_k=linspace(0,A(k),N) but without making a for loop that slows down my whole calculation.
% already defined A
rGC=linspace(0,75e-7,N);
for k=1:N
r=rGC(k);
v=linspace(0,A*r,N);
y=f(r,v);
INT=trapz(v,y);
% The same for 8 more integrals
end
Maybe something using interp1, to interpolate something like:
[0 0 0 ... 0 ]
[A(1) A(2) A(3) ... A(N)]
with N rows .... For instance:
N = 5;
Amax = 15;
A = linspace(0, Amax, N);
x = [0 1];
y = zeros(2, N);
y(2, :) = A;
B = interp1(x, y, linspace(0, 1, N))
Which will give:
B =
0 0 0 0 0
0 0.9375 1.8750 2.8125 3.7500
0 1.8750 3.7500 5.6250 7.5000
0 2.8125 5.6250 8.4375 11.2500
0 3.7500 7.5000 11.2500 15.0000
Not sure it will be any faster than a for loop or that even get the point here :)
My question seems wierd, because I know that we can't use a matrix as input in linspace(x1,x2,n) function. But my question is more something like : I have a vector A=linspace(0,Amax,N), and I want to build a serie of vector B_k or big matrix B_k=linspace(0,A(k),N) but without making a for loop that slows down my whole calculation.
% already defined A
rGC=linspace(0,75e-7,N);
for k=1:N
r=rGC(k);
v=linspace(0,A*r,N);
y=f(r,v);
INT=trapz(v,y);
% The same for 8 more integrals
end
Maybe something using interp1, to interpolate something like:
[0 0 0 ... 0 ]
[A(1) A(2) A(3) ... A(N)]
with N rows .... For instance:
N = 5;
Amax = 15;
A = linspace(0, Amax, N);
x = [0 1];
y = zeros(2, N);
y(2, :) = A;
B = interp1(x, y, linspace(0, 1, N))
Which will give:
B =
0 0 0 0 0
0 0.9375 1.8750 2.8125 3.7500
0 1.8750 3.7500 5.6250 7.5000
0 2.8125 5.6250 8.4375 11.2500
0 3.7500 7.5000 11.2500 15.0000
Not sure it will be any faster than a for loop or that even get the point here :)
This is my matlab code to predict [1;1;1] given [1;0;1] :
m = 1;
alpha = .00001;
x = [1;0;1;0;0;0;0;0;0];
y = [1;1;1;0;0;0;0;0;0];
theta1 = [4.7300;3.2800;1.4600;0;0;0;4.7300;3.2800;1.4600];
theta1 = theta1 - (alpha/m .* (x .* theta1-y)' * x)'
theta1 = reshape(theta1(1:9) , 3 , 3)
sigmoid(theta1 * [1; 0; 1])
x = [1;0;1;0;0;0;0;0;0];
y = [1; 1; 1;0;0;0;0;0;0];
theta2 = [8.892;6.167;2.745;8.892;6.167;2.745;8.892;6.167;2.745];
theta2 = theta2 - (alpha/m .* (x .* theta2-y)' * x)'
theta2 = reshape(theta2(1:9) , 3 , 3)
sigmoid(theta2 * [1; 0; 1])
x = [1;0;1;0;0;0;0;0;0];
y = [1; 1; 1;0;0;0;0;0;0];
theta3 = [9.446;6.55;2.916;9.351;6.485;2.886;8.836;6.127;2.727];
theta3 = theta3 - (alpha/m .* (x .* theta3-y)' * x)'
theta3 = reshape(theta3(1:9) , 3 , 3)
sigmoid(theta3 * [1; 0; 1])
I'm computing theta1, theta2, theta3 individually but I think they
should be linked between each computation ?
Though gradient descent appears to be working as :
sigmoid(theta1 * [1; 0; 1]) =
0.9999
0.9986
0.9488
sigmoid(theta2 * [1; 0; 1]) =
1.0000
1.0000
0.9959
sigmoid(theta3 * [1; 0; 1]) =
1.0000
1.0000
0.9965
This shows for each theta value (layer in the network) the prediction is moving closer to [1;1;1]
Update : sigmoid function :
function g = sigmoid(z)
g = 1.0 ./ (1.0 + exp(-z));
end
Update2 :
After extended discussion with user davidhigh who provided key insights have made following changes :
x = [1;0;1];
y = [1;1;1];
theta1 =
4.7300 3.2800 1.4600
0 0 0
4.7300 3.2800 1.4600
theta2 =
8.8920 8.8920 8.8920
6.1670 6.1670 6.1670
2.7450 2.7450 2.7450
theta3 =
9.4460 6.5500 2.9160
9.3510 6.4850 2.8860
8.8360 6.1270 2.7270
The crux of my issue is that I don't feed output of each layer into the next layer, once I made this change I get better result :
z1 = sigmoid(theta1 * x)
z1 =
0.9980
0.5000
0.9980
z2 = sigmoid(theta2 * z1)
z2 =
1.0000
1.0000
0.9989
z3 = sigmoid(theta3 * z2)
z3 =
1.0000
1.0000
1.0000
z3 is the predicted value which correctly is [1;1;1;] whereas previously it is approx [1;1;1;]