Lets say I want to simulate a particle state, which can be normal (0) or excited (1) in given frame. The particle is in excited state f % of time. If the particle is in excited state, it lasts for ~L frames (with poisson distribution). I want to simulate that state for N time points. So the input is for example:
N = 1000;
f = 0.3;
L = 5;
and the result will be something like
state(1:N) = [0 0 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 ... and so on]
with sum(state)/N close to 0.3
How to do that?
Thanks!
%% parameters
f = 0.3; % probability of state 1
L1 = 5; % average time in state 1
N = 1e4;
s0 = 1; % init. state
%% run simulation
L0 = L1 * (1 / f - 1); % average time state 0 lasts
p01 = 1 / L0; % probability to switch from 0 to 1
p10 = 1 / L1; % probability to switch from 1 to 0
p00 = 1 - p01;
p11 = 1 - p10;
sm = [p00, p01; p10, p11]; % build stochastic matrix (state machine)
bins = [0, 1]; % possible states
states = zeros(N, 1);
assert(all(sum(sm, 2) == 1), 'not a stochastic matrix');
smc = cumsum(sm, 2); % cummulative matrix
xi = find(bins == s0);
for k = 1 : N
yi = find(smc(xi, :) > rand, 1, 'first');
states(k) = bins(yi);
xi = yi;
end
%% check result
ds = [states(1); diff(states)];
idx_begin = find(ds == 1 & states == 1);
idx_end = find(ds == -1 & states == 0);
if idx_end(end) < idx_begin(end)
idx_end = [idx_end; N + 1];
end
df = idx_end - idx_begin;
fprintf('prob(state = 1) = %g; avg. time(state = 1) = %g\n', sum(states) / N, mean(df));
The average length of the excited state is 5. The average length of the normal state, should thus be around 12 to obtain.
The strategy can be something like this.
Start in state 0
Draw a random number a from a Poisson distribution with mean L*(1-f)/f
Fill the state array with a zeroes
Draw a random number b from a Poission distribution with mean L
Fill the state array witb b ones.
Repeat
Another option would be to think in terms of switching probabilities, where the 0->1 and 1->0 probabilities are unequal.
Related
I created random connectivity information matrix. From that I have plotted x,y and z axis points in 3D graph .Now I just want to apply the zaxis values in the corresponding connectivity such that where ever 1 is present in connectivity it should be multiplied by corresponding zaxis values (eg: if in conn matrix (1,3)place =1 then it should take particular zaxis values and multiply). But I getting the same values for all the places. Suggestions.
%Conncectivity Matrix
success = 0;
n = input('Enter the No. of Nodes'); %size of matrix
k = input('Enter the max connectivity'); %maximal number of 1s
p = 0.5;
Result_Matrix = zeros(n,n);
while (success == 0)
Result_Matrix = (rand(n,n) < p);
Result_Matrix(logical(eye(n))) = 0;
Result_Matrix = max(Result_Matrix, Result_Matrix');
s = sum(Result_Matrix,1);
success = 1;
if min(s) == 0
success = 0; p = p*2; % too few 1s, increase p
end
if max(s) > k
success = 0; p = p/2; % too many 1s, decrease p
end
end
m=Result_Matrix;
conn_mat=m;
disp('connection matrix');
disp(m);
[r,c] = find(m);
A = [r,c]
%3D-GRAPH
PlotSizex=100;
PlotSizey=100;
PlotSizez=-100;
x=PlotSizex*rand(1,n)
y=PlotSizey*rand(1,n)
z=PlotSizez*rand(1,n)
plot3(x(A).', y(A).',z(A).', 'O-')
%Zaxis values multiply with Connectivity
d=zeros(n,n);
z % values of zaxis
for i=1:n
for j=i+1:n
d(i,j)= z(i);
d(j,i)=d(i,j);
end
end
New matrix= d.*m %d is zaxis values and m is connectivity matrix.
I do obtain different values in new_matrix:
new_matrix =
0 -63.4303 -63.4303 0 0
-63.4303 0 0 -23.9408 0
-63.4303 0 0 -24.5725 0
0 -23.9408 -24.5725 0 -76.5436
0 0 0 -76.5436 0
My connection matrix is:
connection matrix
0 1 1 0 0
1 0 0 1 0
1 0 0 1 0
0 1 1 0 1
0 0 0 1 0
and z values are:
z =
-63.4303 -23.9408 -24.5725 -76.5436 -86.3677
I find it strange to multiply the elements in your connection matrix with a single z value, because each element in the connection matrix is related to two points in space (and thus two z values). So, it would make more sense to use the following:
for i=1:n
for j=i:n
d(i,j)= z(i)*z(j); % or another combination of z(i) and z(j)
d(j,i)=d(i,j);
end
end
I have indices
I = [nGrid x 9] matrix % mesh on fine grid (9 point rectangle)
J = [nGrid x 4] matrix % mesh on coarse grid (4 point rectangle)
Here, nGrid is large number depending on the mesh (e.g. 1.e05)
Then I want to do like
R_ref = [4 x 9] matrix % reference restriction matrix from fine to coarse
P_ref = [9 x 4] matrix % reference prolongation matrix from coarse to fine
R = sparse(size) % n_Coarse x n_Fine
P = sparse(size) % n_Fine x n_Coarse
for k = 1 : nGrid % number of elements on coarse grid
R(I(k,:),J(k,:)) = R_ref;
P(J(k,:),I(k,:)) = P_ref;
end
size is predetermined number.
Note that even if there is same index in (I,J), I do not want to accumulate. I just want to put stencils Rref and Pref at each indices respectively.
I know that this is very slow due to the data structure of sparse.
Usually, I use
sparse(row,col,entry,n_row,n_col)
I can use this by manipulate I, J, R_ref, P_ref by bsxfun and repmat.
However, this cannot be done because of the accumulation of sparse function
(if there exists (i,j) such that [row(i),col(i)]==[row(j),col(j)], then R(row(i),row(j)) = entry(i)+entry(j))
Is there any suggestion for this kind of assembly procedure?
Sample code
%% INPUTS
% N and M could be much larger
N = 2^5+1; % number of fine grid in x direction
M = 2^5+1; % number of fine grid in y direction
% [nOx * nOy] == nGrid
nOx = floor((M)/2)+1; % number of coarse grid on x direction
nOy = floor((N)/2)+1; % number of coarse grid on y direction
Rref = [4 4 -1 4 -2 0 -1 0 0
-1 -1 -2 4 4 4 -1 4 4
-1 -1 4 -2 4 -2 4 4 -1
0 4 4 0 0 4 -1 -2 -1]/8;
Pref = [2 1 0 1 0 0 0 0 0
0 0 0 1 1 1 0 1 2
0 0 1 0 1 0 2 1 0
0 2 1 0 0 1 0 0 0]'/2;
%% INDEX GENERATION
tri_ref = reshape(bsxfun(#plus,[0,1,2]',[0,N,2*N]),[],1)';
TRI_ref = reshape(bsxfun(#plus,[0,1]',[0,nOy]),[],1)';
I = reshape(bsxfun(#plus,(1:2:N-2)',0:2*N:(M-2)*N),[],1);
J = reshape(bsxfun(#plus,(1:nOy-1)',0:nOy:(nOx-2)*nOy),[],1);
I = bsxfun(#plus,I,tri_ref);
J = bsxfun(#plus,J,TRI_ref);
%% THIS PART IS WHAT I WANT TO CHANGE
R = sparse(nOx*nOy,N*M);
P = R';
for k = 1 : size(I,1)
R(J(k,:),I(k,:)) = Rref;
P(I(k,:),J(k,:)) = Pref;
end
I have a large column vector y containing integer values from 1 to 10. I wanted to convert it to a matrix where each row is full of 0s except for a 1 at the index given by the value at the respective row of y.
This example should make it clearer:
y = [3; 4; 1; 10; 9; 9; 4; 2; ...]
% gets converted to:
Y = [
0 0 1 0 0 0 0 0 0 0;
0 0 0 1 0 0 0 0 0 0;
1 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 1;
0 0 0 0 0 0 0 0 1 0;
0 0 0 0 0 0 0 0 1 0;
0 0 0 1 0 0 0 0 0 0;
0 1 0 0 0 0 0 0 0 0;
...
]
I have written the following code for this (it works):
m = length(y);
Y = zeros(m, 10);
for i = 1:m
Y(i, y(i)) = 1;
end
I know there are ways I could remove the for loop in this code (vectorizing). This post contains a few, including something like:
Y = full(sparse(1:length(y), y, ones(length(y),1)));
But I had to convert y to doubles to be able to use this, and the result is actually about 3x slower than my "for" approach, using 10.000.000 as the length of y.
Is it likely that doing this kind of vectorization will lead to better performance for a very large y? I've read many times that vectorizing calculations leads to better performance (not only in MATLAB), but this kind of solution seems to result in more calculations.
Is there a way to actually improve performance over the for approach in this example? Maybe the problem here is simply that acting on doubles instead of ints isn't the best thing for comparison, but I couldn't find a way to use sparse otherwise.
Here is a test to comapre:
function [t,v] = testIndicatorMatrix()
y = randi([1 10], [1e6 1], 'double');
funcs = {
#() func1(y);
#() func2(y);
#() func3(y);
#() func4(y);
};
t = cellfun(#timeit, funcs, 'Uniform',true);
v = cellfun(#feval, funcs, 'Uniform',false);
assert(isequal(v{:}))
end
function Y = func1(y)
m = numel(y);
Y = zeros(m, 10);
for i = 1:m
Y(i, y(i)) = 1;
end
end
function Y = func2(y)
m = numel(y);
Y = full(sparse(1:m, y, 1, m, 10, m));
end
function Y = func3(y)
m = numel(y);
Y = zeros(m,10);
Y(sub2ind([m,10], (1:m).', y)) = 1;
end
function Y = func4(y)
m = numel(y);
Y = zeros(m,10);
Y((y-1).*m + (1:m).') = 1;
end
I get:
>> testIndicatorMatrix
ans =
0.0388
0.1712
0.0490
0.0430
Such a simple for-loop can be dynamically JIT-compiled at runtime, and would run really fast (even slightly faster than vectorized code)!
It seems you are looking for that full numeric matrix Y as the output. So, you can try this approach -
m = numel(y);
Y1(m,10) = 0; %// Faster way to pre-allocate zeros than using function call `zeros`
%// Source - http://undocumentedmatlab.com/blog/preallocation-performance
linear_idx = (y-1)*m+(1:m)'; %//'# since y is mentioned as a column vector,
%// so directly y can be used instead of y(:)
Y1(linear_idx)=1; %// Y1 would be the desired output
Benchmarking
Using Amro's benchmark post and increasing the datasize a bit -
y = randi([1 10], [1.5e6 1], 'double');
And finally doing the faster pre-allocation scheme mentioned earlier of using Y(m,10)=0; instead of Y = zeros(m,10);, I got these results on my system -
>> testIndicatorMatrix
ans =
0.1798
0.4651
0.1693
0.1457
That is the vectorized approach mentioned here (the last one in the benchmark suite) is giving you more than 15% performance improvement over your for-loop code (the first one in the benchmark suite). So, if you are using large datasizes and intend to get full versions of sparse matrices, this approach would make sense (in my personal opinion).
Does something like this not work for you?
tic;
N = 1e6;
y = randperm( N );
Y = spalloc( N, N, N );
inds = sub2ind( size(Y), y(:), (1:N)' );
Y = sparse( 1:N, y, 1, N, N, N );
toc
The above outputs
Elapsed time is 0.144683 seconds.
I am new to MATLAB and I want to formulate the following lease square expression in Matlab. I have some codes that I am typing here. But the optimization problem solution seems not to be correct. Does anyone has an idea why?
First, I want to solve the heat equation
$$T_t(x,t) = - L_x . T(x,t) + F(x,t)$$
where L_x is Laplacian matrix of the graph.
then find y from the following least square.
$$ \min_y \sum_{j} \sum_{i} (\hat{T}_j(t_i) - T_j(t_i, y))^2$$
Thanks in advance!!
Here is my code:
%++++++++++++++++ main ++++++++++++++++++++
% incidence matrix for original graph
C_hat = [ 1 -1 0 0 0 0;...
0 1 -1 0 0 -1;...
0 0 0 0 -1 1;...
0 0 0 1 1 0;...
-1 0 1 -1 0 0];
% initial temperature for each vertex in original graph
T_hat_0 = [0 7 1 9 4];
[M_bar,n,m_bar,T_hat_heat,T_hat_temp] = simulate_temp(T_hat_0,C_hat);
C = [ 1 1 -1 -1 0 0 0 0 0 0;...
0 -1 0 0 1 -1 1 0 0 0;...
0 0 1 0 0 1 0 -1 -1 0;...
0 0 0 1 0 0 -1 0 1 -1;...
-1 0 0 0 -1 0 0 1 0 1];
%
% initial temperature for each vertex in original graph
T_0 = [0 7 1 9 4];
%
% initial temperature simulation
[l,n,m,T_heat,T_temp] = simulate_temp(T_0,C);
%
% bounds for variables
lb = zeros(m,1);
ub = ones(m,1);
%
% initial edge weights
w0 = ones(m,1);
% optimization problem
% w = fmincon(#fun, w0, [], [], [], [], lb, ub);
%++++++++++++++++++++ function++++++++++++++++++++++++++++
function [i,n,m,T_heat,T_temp] = simulate_temp(T,C)
%
% initial conditions
delta_t = 0.1;
M = 20; %% number of time steps
t = 1;
[n,m] = size(C);
I = eye(n);
L_w = C * C';
T_ini = T';
Temp = zeros(n,1);
% Computing Temperature
%
for i=1:M
K = 2*I + L_w * delta_t;
H = 2*I - L_w * delta_t;
%
if i == 1
T_heat = (K \ H) * T_ini;
%
t = t + delta_t;
else
T_heat = (K \ H) * Temp;
%
t = t + delta_t;
end
% replacing column of T_final with each node temperature in each
% iteration. It adds one column to the matrix in each step
T_temp(:,i) = T_heat;
%
Temp = T_heat;
end
end
%++++++++++++++++++ function+++++++++++++++++++++++++++++++++++++++++
function w_i = fun(w);
%
for r=1:n
for s=1:M_bar
w_i = (T_hat_temp(r,s) - T_temp(r,s)).^2;
end
end
To give a more clear answer, I need more information about what form you have the functions F_j and E_j in.
I've assumed that you feed each F_j a value, x_i, and get back a number. I've also assumed that you feed E_j a value x_i, and another value (or vector) y, and get back a value.
I've also assumed that by 'i' and 'j' you mean the indices of the columns and rows respectively, and that they're finite.
All I can suggest without knowing more info is to do this:
Pre-calculate the values of the functions F_j for each x_i, to give a matrix F - where element F(i,j) gives you the value F_j(x_i).
Do the same thing for E_j, giving a matrix E - where E(i,j) corresponds to E_j(x_i,y).
Perform (F-E).^2 to subtract each element of F and E, then square them element-wise.
Take sum( (F-E).^2**, 2)**. sum(M,2) will sum across index i of matrix M, returning a column vector.
Finally, take sum( sum( (F-E).^2, 2), 1) to sum across index j, the columns, this will finally give you a scalar.
I have some code below, and I cant seem to get the matrices formatted correctly. I have been trying to get the matrices to look more professional (close together) with \t and fprintf, but cant seem to do so. I am also having some trouble putting titles for each columns of the matrix. Any help would be much appreciated!
clear all
clc
format('bank')
% input file values %
A = [4 6 5 1 0 0 0 0 0; 7 8 4 0 1 0 0 0 0; 6 5 9 0 0 1 0 0 0; 1 0 0 0 0 0 -1 0 0; 0 1 0 0 0 0 0 -1 0; 0 0 1 0 0 0 0 0 -1];
b = [480; 600; 480; 24; 20; 25];
c = [3000 4000 4000 0 0 0 0 0 0];
% Starting xb %
xb = [1 2 3 4 5 6]
% Starting xn %
xn = [7 8 9]
cb = c(xb)
cn = c(xn)
% Get B from A %
B = A(:,xb)
% Get N from A %
N = A(:,xn)
% Calculate z %
z = ((cb*(inv(B))*A)-c)
% Calculate B^(-1) %
Binv = inv(B)
% Calculate RHS of row 0 %
RHS0 = cb*Binv*b
% Calculates A %
A = Binv*A
%STARTING Tableau%
ST = [z RHS0;A b]
for j=1:A
fprintf(1,'\tz%d',j)
end
q = 0
while q == 0
m = input('what is the index value of the ENTERING variable? ')
n = input('what is the index value of the LEAVING variable? ')
xn(xn==m)= n
xb(xb==n) = m
cb = c(xb)
cn = c(xn)
B = A(:,xb)
N = A(:,xn)
Tableuz = (c-(cb*(B^(-1))*A))
RHS0 = (cb*(B^(-1))*b)
TableuA = ((B^(-1))*A)
Tableub = ((B^(-1))*b)
CT = [Tableuz RHS0; TableuA Tableub];
disp(CT)
q = input('Is the tableau optimal? Y-1, N-0')
end
I didn't dig into what you are doing really deeply, but a few pointers:
* Put semicolons at the end of lines you don't want printing to the screen--it makes it easier to see what is happening elsewhere.
* Your for j=1:A loop only prints j. I think what you want is more like this:
for row = 1:size(A,1)
for column = 1:size(A,2)
fprintf('%10.2f', A(row,column));
end
fprintf('\n');
end
If you haven't used the Matlab debugger yet, give it a try; it makes a lot of these problems easier to spot. All you have to do to start it is to add a breakpoint to the file by clicking on the dash(-) next to the line numbers and starting the script. Quick web searches can turn up the solution very quickly too--someone else has usually already had any problem you're going to run into.
Good luck.
Try using num2str with a format argument of your desired precision. It's meant for converting matrices to strings. (note: this is different than mat2str which serializes matrices so they can be deserialized with eval)