how are you?.
I need to simulate a disturbance in a control system using scilab, that is, the csim function is used to simulate the response of a system by using a step, impulse, ramp or any other input, but, I need to input a disturbance for example in t = 0.5s to see the system behavior.
That drags another problem to me because I don't know how to make csim or syslin to acknowledge two different inputs, or its as simple as defining two systems, one with the referent input and other with the disturbance entrance and sum both?.
Thanks in advance for the help.
Suppose you have the following linear time invariant system (A,B,C)
x'=A*x+B1*v+B2*d
y=C*x
with B=[B1,B2], where v is the control/input and d the disturbance. If you want e.g. simulate a step response and a disturbance you have to define your own overall input [v;d] and decide when to apply the disturbance. Here is an example:
function ud = step(t)
ud = [1;0];
endfunction
function ud = input(t)
ud = zeros(t);
ud(1,:) = 1;
s = abs(t-0.5);
// 0.2 is the half-width of disturbance
ud(2,:) = -0.1*(1-s/0.2).*(s<0.2)
endfunction
A = [-2 1;
1 -2];
B1 = [1;0];
B2 = [0;4];
C = [0 1];
sl = syslin('c',A,[B1 B2],C);
t = linspace(0,5,1000);
x = csim(step,t,sl)
xd = csim(input,t,sl)
clf
plot(t,x,t,xd,t,input(t)(2,:))
legend('step','step and disturbance','disturbance',2)
I made here two csim calls, one for the usual step response and the second one for the perturbed step response. However, I warn you about the ode solver used by csim: discontinuous inputs can be easily missed, that's why I applied here a hat-shaped disturbance. The code of the perturbed input is designed to allow vector time input, in order to easily plot the perturbation.
Related
I have the following problem: given a 3D irregular geometry A with
(i,j,k)-coordinates, which are the centroids of connected voxels, create a table with the (i_out,j_out,k_out)-coordinates of the cells that represent the complementary set B of the bounding box of A, which we may call C. That is to say, I need the voxel coordinates of the set B = C - A.
To get this done, I am using the Matlab code below, but it is taking too much time to complete when C is fairly large. Then, I would like to speed up the code. To make it clear: cvc is the matrix of voxel coordinates of A; allcvc should produce C and B results from outcvc after setdiff.
Someone has a clue regarding the code performance, or even to improve my strategy?
Problem: the for-loop seems to be the villain.
My attempts: I have tried to follow some hints of Yair Altman's book by doing some tic,toc analyses, using pre-allocation and int8 since I do not need double values. deal yet gave me a slight improvement with min,max. I have also checked this discussion here, but, parallelism, for instance, is a limitation that I have for now.
% A bounding box limits
m = min(cvc,[],1);
M = max(cvc,[],1);
[im,jm,km,iM,jM,kM] = deal(m(1),m(2),m(3),M(1),M(2),M(3));
% (i,j,k) indices of regular grid
I = im:iM;
J = jm:jM;
K = km:kM;
% (i,j,k) table
m = length(I);
n = length(J);
p = length(K);
num = m*n*p;
allcvc = zeros(num,3,'int8');
for N = 1:num
for i = 1:m
for j = 1:n
for k = 1:p
aux = [I(i),J(j),K(k)];
allcvc(N,:) = aux;
end
end
end
end
% operation of exclusion: out = all - in
[outcvc,~] = setdiff(allcvc,cvc,'rows');
To avoid all for-loops in the present code you can use ndgrid or meshgrid functions. For example
[I,J,K] = ndgrid(im:iM, jm:jM, km:kM);
allcvc = [I(:),J(:),K(:)];
instead of your code between % (i,j,k) indices of regular grid and % operation of exclusion: out =.
I was persuaded some time ago to drop my comfortable matlab programming and start programming in Julia. I have been working for a long with neural networks and I thought that, now with Julia, I could get things done faster by parallelising the calculation of the gradient.
The gradient need not be calculated on the entire dataset in one go; instead one can split the calculation. For instance, by splitting the dataset in parts, we can calculate a partial gradient on each part. The total gradient is then calculated by adding up the partial gradients.
Though, the principle is simple, when I parallelise with Julia I get a performance degradation, i.e. one process is faster then two processes! I am obviously doing something wrong... I have consulted other questions asked in the forum but I could still not piece together an answer. I think my problem lies in that there is a lot of unnecessary data moving going on, but I can't fix it properly.
In order to avoid posting messy neural network code, I am posting below a simpler example that replicates my problem in the setting of linear regression.
The code-block below creates some data for a linear regression problem. The code explains the constants, but X is the matrix containing the data inputs. We randomly create a weight vector w which when multiplied with X creates some targets Y.
######################################
## CREATE LINEAR REGRESSION PROBLEM ##
######################################
# This code implements a simple linear regression problem
MAXITER = 100 # number of iterations for simple gradient descent
N = 10000 # number of data items
D = 50 # dimension of data items
X = randn(N, D) # create random matrix of data, data items appear row-wise
Wtrue = randn(D,1) # create arbitrary weight matrix to generate targets
Y = X*Wtrue # generate targets
The next code-block below defines functions for measuring the fitness of our regression (i.e. the negative log-likelihood) and the gradient of the weight vector w:
####################################
## DEFINE FUNCTIONS ##
####################################
#everywhere begin
#-------------------------------------------------------------------
function negative_loglikelihood(Y,X,W)
#-------------------------------------------------------------------
# number of data items
N = size(X,1)
# accumulate here log-likelihood
ll = 0
for nn=1:N
ll = ll - 0.5*sum((Y[nn,:] - X[nn,:]*W).^2)
end
return ll
end
#-------------------------------------------------------------------
function negative_loglikelihood_grad(Y,X,W, first_index,last_index)
#-------------------------------------------------------------------
# number of data items
N = size(X,1)
# accumulate here gradient contributions by each data item
grad = zeros(similar(W))
for nn=first_index:last_index
grad = grad + X[nn,:]' * (Y[nn,:] - X[nn,:]*W)
end
return grad
end
end
Note that the above functions are on purpose not vectorised! I choose not to vectorise, as the final code (the neural network case) will also not admit any vectorisation (let us not get into more details regarding this).
Finally, the code-block below shows a very simple gradient descent that tries to recover the parameter weight vector w from the given data Y and X:
####################################
## SOLVE LINEAR REGRESSION ##
####################################
# start from random initial solution
W = randn(D,1)
# learning rate, set here to some arbitrary small constant
eta = 0.000001
# the following for-loop implements simple gradient descent
for iter=1:MAXITER
# get gradient
ref_array = Array(RemoteRef, nworkers())
# let each worker process part of matrix X
for index=1:length(workers())
# first index of subset of X that worker should work on
first_index = (index-1)*int(ceil(N/nworkers())) + 1
# last index of subset of X that worker should work on
last_index = min((index)*(int(ceil(N/nworkers()))), N)
ref_array[index] = #spawn negative_loglikelihood_grad(Y,X,W, first_index,last_index)
end
# gather the gradients calculated on parts of matrix X
grad = zeros(similar(W))
for index=1:length(workers())
grad = grad + fetch(ref_array[index])
end
# now that we have the gradient we can update parameters W
W = W + eta*grad;
# report progress, monitor optimisation
#printf("Iter %d neg_loglikel=%.4f\n",iter, negative_loglikelihood(Y,X,W))
end
As is hopefully visible, I tried to parallelise the calculation of the gradient in the easiest possible way here. My strategy is to break the calculation of the gradient in as many parts as available workers. Each worker is required to work only on part of matrix X, which part is specified by first_index and last_index. Hence, each worker should work with X[first_index:last_index,:]. For instance, for 4 workers and N = 10000, the work should be divided as follows:
worker 1 => first_index = 1, last_index = 2500
worker 2 => first_index = 2501, last_index = 5000
worker 3 => first_index = 5001, last_index = 7500
worker 4 => first_index = 7501, last_index = 10000
Unfortunately, this entire code works faster if I have only one worker. If add more workers via addprocs(), the code runs slower. One can aggravate this issue by create more data items, for instance use instead N=20000.
With more data items, the degradation is even more pronounced.
In my particular computing environment with N=20000 and one core, the code runs in ~9 secs. With N=20000 and 4 cores it takes ~18 secs!
I tried many many different things inspired by the questions and answers in this forum but unfortunately to no avail. I realise that the parallelisation is naive and that data movement must be the problem, but I have no idea how to do it properly. It seems that the documentation is also a bit scarce on this issue (as is the nice book by Ivo Balbaert).
I would appreciate your help as I have been stuck for quite some while with this and I really need it for my work. For anyone wanting to run the code, to save you the trouble of copying-pasting you can get the code here.
Thanks for taking the time to read this very lengthy question! Help me turn this into a model answer that anyone new in Julia can then consult!
I would say that GD is not a good candidate for parallelizing it using any of the proposed methods: either SharedArray or DistributedArray, or own implementation of distribution of chunks of data.
The problem does not lay in Julia, but in the GD algorithm.
Consider the code:
Main process:
for iter = 1:iterations #iterations: "the more the better"
δ = _gradient_descent_shared(X, y, θ)
θ = θ - α * (δ/N)
end
The problem is in the above for-loop which is a must. No matter how good _gradient_descent_shared is, the total number of iterations kills the noble concept of the parallelization.
After reading the question and the above suggestion I've started implementing GD using SharedArray. Please note, I'm not an expert in the field of SharedArrays.
The main process parts (simple implementation without regularization):
run_gradient_descent(X::SharedArray, y::SharedArray, θ::SharedArray, α, iterations) = begin
N = length(y)
for iter = 1:iterations
δ = _gradient_descent_shared(X, y, θ)
θ = θ - α * (δ/N)
end
θ
end
_gradient_descent_shared(X::SharedArray, y::SharedArray, θ::SharedArray, op=(+)) = begin
if size(X,1) <= length(procs(X))
return _gradient_descent_serial(X, y, θ)
else
rrefs = map(p -> (#spawnat p _gradient_descent_serial(X, y, θ)), procs(X))
return mapreduce(r -> fetch(r), op, rrefs)
end
end
The code common to all workers:
#= Returns the range of indices of a chunk for every worker on which it can work.
The function splits data examples (N rows into chunks),
not the parts of the particular example (features dimensionality remains intact).=#
#everywhere function _worker_range(S::SharedArray)
idx = indexpids(S)
if idx == 0
return 1:size(S,1), 1:size(S,2)
end
nchunks = length(procs(S))
splits = [round(Int, s) for s in linspace(0,size(S,1),nchunks+1)]
splits[idx]+1:splits[idx+1], 1:size(S,2)
end
#Computations on the chunk of the all data.
#everywhere _gradient_descent_serial(X::SharedArray, y::SharedArray, θ::SharedArray) = begin
prange = _worker_range(X)
pX = sdata(X[prange[1], prange[2]])
py = sdata(y[prange[1],:])
tempδ = pX' * (pX * sdata(θ) .- py)
end
The data loading and training. Let me assume that we have:
features in X::Array of the size (N,D), where N - number of examples, D-dimensionality of the features
labels in y::Array of the size (N,1)
The main code might look like this:
X=[ones(size(X,1)) X] #adding the artificial coordinate
N, D = size(X)
MAXITER = 500
α = 0.01
initialθ = SharedArray(Float64, (D,1))
sX = convert(SharedArray, X)
sy = convert(SharedArray, y)
X = nothing
y = nothing
gc()
finalθ = run_gradient_descent(sX, sy, initialθ, α, MAXITER);
After implementing this and run (on 8-cores of my Intell Clore i7) I got a very slight acceleration over serial GD (1-core) on my training multiclass (19 classes) training data (715 sec for serial GD / 665 sec for shared GD).
If my implementation is correct (please check this out - I'm counting on that) then parallelization of the GD algorithm is not worth of that. Definitely you might get better acceleration using stochastic GD on 1-core.
If you want to reduce the amount of data movement, you should strongly consider using SharedArrays. You could preallocate just one output vector, and pass it as an argument to each worker. Each worker sets a chunk of it, just as you suggested.
I am trying to verify my RK4 code and have a state space model to solve the same system. I have a 14 state system with initial conditions, but the conditions change with time (each iteration). I am trying to formulate A,B,C,D matrices and use sys and lsim in order to compile the results for all of my states for the entire time span. I am trying to do it similar to this:
for t=1:1:5401
y1b=whatever
.
.
y14b = whatever
y_0 = vector of ICs
A = (will change with time)
B = (1,14) with mostly zeros and 3 ones
C = ones(14,1)
D = 0
Q = eye(14)
R = eye(1)
k = lqr(A,B,C,D)
A_bar = A - B*k
sys = ss(A_bar,B,C,D)
u = zeros(14,1)
sto(t,14) = lsim(sys,u,t,y_0)
then solve for new y1b-y14b from outside function
end
In other words I am trying to use sto(t,14) to store each iteration of lsim and end up with a matrix of all of my states for each time step from 1 to 5401. I keep getting this error message:
Error using DynamicSystem/lsim (line 85)
In time response commands, the time vector must be real, finite, and must contain
monotonically increasing and evenly spaced time samples.
and
Error using DynamicSystem/lsim (line 85)
When simulating the response to a specific input signal, the input data U must be a
matrix with as many rows as samples in the time vector T, and as many columns as
input channels.
Any helpful input is greatly appreciated. Thank you
For lsim to work, t has to contain at least 2 points.
Also, the sizes of B and C are flipped. You have 1 input and 1 output so u should be length of t in lsim by 1.
Lastly, it looks like you try to put all initials conditions at once in lsim with y_0 where you just want the part relevant to this iteration.
s = [t-1 t];
u = [0; 0];
if t==1
y0 = y_0;
else
y0 = sto(t-1,1:14);
end
y = lsim(sys, u, s, y0);
sto(t,1:14) = y(end,:);
I'm not sure I understood correctly your question but I hope it helps.
I am trying to save some calculation time. I am doing some Image processing with the well known Lucas Kanade algorithm. Starting point was this paper by Baker / Simon.
I am doing this Matlab and I also use a background substractor. I want the substractor to set all background to 0 or have a logical mask with 1 as foreground and 0 as background.
What I want to have is to exclude all matrix elements which are background from the calculation. My goal is to save time for the calculation. I am aware that I can use syntax like
A(A>0) = ...
But that doesn't work in a way like
B(A>0) = A.*C.*D
because I am getting an error:
In an assignment A(I) = B, the number of elements in B and I must be the same.
This is probably because A,B and C all together have more elements than only matrix A.
In c-code I would just loop the matrix and check if the pixel has the value 0 and the continue. In this case a save a whole bunch of calculations.
In matlab however it's not very fast to loop through the matrix. So is there a fast way to solve my Problem? I couldn't find a sufficient answere to my problem here.
I case anybody is interested: I am trying to use robust error function instead of quadratic ones.
Update:
I tried the following approach to test the speed as suggested by #Acorbe:
function MatrixTest()
n = 100;
A = rand(n,n);
B = rand(n,n);
C = rand(n,n);
D = rand(n,n);
profile clear, profile on;
for i=1:10000
tests(A,B,C,D);
end
profile off, profile report;
function result = tests(A,B,C,D)
idx = (B>0);
t = A(idx).*B(idx).*C(idx).*D(idx);
LGS1a(idx) = t;
LGS1b = A.*B.*C.*D;
And i got the folloing results with the profiler of matlab:
t = A(idx).*B(idx).*C(idx).*D(idx); 1.520 seconds
LGS1a(idx) = t; 0.513 seconds
idx = (B>0); 0.264 seconds
LGS1b = A.*B.*C.*D; 0.155 seconds
As you can see, the overhead of accessing the matrix by index hast far more costs than just
What about the following?
mask = A>0;
B = zeros(size(A)); % # some initialization
t = A.*C.*D;
B( mask ) = t( mask );
in this way you select just the needed elements of t. Maybe there is some overhead in the calculation, although likely negligible with respect to for loops slowness.
EDIT:
If you want more speed, you can try a more selective approach which uses the mask everywhere.
t = A(mask).*C(mask).*D(mask);
B( mask ) = t;
How can I convert a uniform distribution (as most random number generators produce, e.g. between 0.0 and 1.0) into a normal distribution? What if I want a mean and standard deviation of my choosing?
There are plenty of methods:
Do not use Box Muller. Especially if you draw many gaussian numbers. Box Muller yields a result which is clamped between -6 and 6 (assuming double precision. Things worsen with floats.). And it is really less efficient than other available methods.
Ziggurat is fine, but needs a table lookup (and some platform-specific tweaking due to cache size issues)
Ratio-of-uniforms is my favorite, only a few addition/multiplications and a log 1/50th of the time (eg. look there).
Inverting the CDF is efficient (and overlooked, why ?), you have fast implementations of it available if you search google. It is mandatory for Quasi-Random numbers.
The Ziggurat algorithm is pretty efficient for this, although the Box-Muller transform is easier to implement from scratch (and not crazy slow).
Changing the distribution of any function to another involves using the inverse of the function you want.
In other words, if you aim for a specific probability function p(x) you get the distribution by integrating over it -> d(x) = integral(p(x)) and use its inverse: Inv(d(x)). Now use the random probability function (which have uniform distribution) and cast the result value through the function Inv(d(x)). You should get random values cast with distribution according to the function you chose.
This is the generic math approach - by using it you can now choose any probability or distribution function you have as long as it have inverse or good inverse approximation.
Hope this helped and thanks for the small remark about using the distribution and not the probability itself.
Here is a javascript implementation using the polar form of the Box-Muller transformation.
/*
* Returns member of set with a given mean and standard deviation
* mean: mean
* standard deviation: std_dev
*/
function createMemberInNormalDistribution(mean,std_dev){
return mean + (gaussRandom()*std_dev);
}
/*
* Returns random number in normal distribution centering on 0.
* ~95% of numbers returned should fall between -2 and 2
* ie within two standard deviations
*/
function gaussRandom() {
var u = 2*Math.random()-1;
var v = 2*Math.random()-1;
var r = u*u + v*v;
/*if outside interval [0,1] start over*/
if(r == 0 || r >= 1) return gaussRandom();
var c = Math.sqrt(-2*Math.log(r)/r);
return u*c;
/* todo: optimize this algorithm by caching (v*c)
* and returning next time gaussRandom() is called.
* left out for simplicity */
}
Where R1, R2 are random uniform numbers:
NORMAL DISTRIBUTION, with SD of 1:
sqrt(-2*log(R1))*cos(2*pi*R2)
This is exact... no need to do all those slow loops!
Reference: dspguide.com/ch2/6.htm
Use the central limit theorem wikipedia entry mathworld entry to your advantage.
Generate n of the uniformly distributed numbers, sum them, subtract n*0.5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to (1/12) * (1/sqrt(N)) (see wikipedia on uniform distributions for that last one)
n=10 gives you something half decent fast. If you want something more than half decent go for tylers solution (as noted in the wikipedia entry on normal distributions)
I would use Box-Muller. Two things about this:
You end up with two values per iteration
Typically, you cache one value and return the other. On the next call for a sample, you return the cached value.
Box-Muller gives a Z-score
You have to then scale the Z-score by the standard deviation and add the mean to get the full value in the normal distribution.
It seems incredible that I could add something to this after eight years, but for the case of Java I would like to point readers to the Random.nextGaussian() method, which generates a Gaussian distribution with mean 0.0 and standard deviation 1.0 for you.
A simple addition and/or multiplication will change the mean and standard deviation to your needs.
The standard Python library module random has what you want:
normalvariate(mu, sigma)
Normal distribution. mu is the mean, and sigma is the standard deviation.
For the algorithm itself, take a look at the function in random.py in the Python library.
The manual entry is here
This is a Matlab implementation using the polar form of the Box-Muller transformation:
Function randn_box_muller.m:
function [values] = randn_box_muller(n, mean, std_dev)
if nargin == 1
mean = 0;
std_dev = 1;
end
r = gaussRandomN(n);
values = r.*std_dev - mean;
end
function [values] = gaussRandomN(n)
[u, v, r] = gaussRandomNValid(n);
c = sqrt(-2*log(r)./r);
values = u.*c;
end
function [u, v, r] = gaussRandomNValid(n)
r = zeros(n, 1);
u = zeros(n, 1);
v = zeros(n, 1);
filter = r==0 | r>=1;
% if outside interval [0,1] start over
while n ~= 0
u(filter) = 2*rand(n, 1)-1;
v(filter) = 2*rand(n, 1)-1;
r(filter) = u(filter).*u(filter) + v(filter).*v(filter);
filter = r==0 | r>=1;
n = size(r(filter),1);
end
end
And invoking histfit(randn_box_muller(10000000),100); this is the result:
Obviously it is really inefficient compared with the Matlab built-in randn.
This is my JavaScript implementation of Algorithm P (Polar method for normal deviates) from Section 3.4.1 of Donald Knuth's book The Art of Computer Programming:
function normal_random(mean,stddev)
{
var V1
var V2
var S
do{
var U1 = Math.random() // return uniform distributed in [0,1[
var U2 = Math.random()
V1 = 2*U1-1
V2 = 2*U2-1
S = V1*V1+V2*V2
}while(S >= 1)
if(S===0) return 0
return mean+stddev*(V1*Math.sqrt(-2*Math.log(S)/S))
}
I thing you should try this in EXCEL: =norminv(rand();0;1). This will product the random numbers which should be normally distributed with the zero mean and unite variance. "0" can be supplied with any value, so that the numbers will be of desired mean, and by changing "1", you will get the variance equal to the square of your input.
For example: =norminv(rand();50;3) will yield to the normally distributed numbers with MEAN = 50 VARIANCE = 9.
Q How can I convert a uniform distribution (as most random number generators produce, e.g. between 0.0 and 1.0) into a normal distribution?
For software implementation I know couple random generator names which give you a pseudo uniform random sequence in [0,1] (Mersenne Twister, Linear Congruate Generator). Let's call it U(x)
It is exist mathematical area which called probibility theory.
First thing: If you want to model r.v. with integral distribution F then you can try just to evaluate F^-1(U(x)). In pr.theory it was proved that such r.v. will have integral distribution F.
Step 2 can be appliable to generate r.v.~F without usage of any counting methods when F^-1 can be derived analytically without problems. (e.g. exp.distribution)
To model normal distribution you can cacculate y1*cos(y2), where y1~is uniform in[0,2pi]. and y2 is the relei distribution.
Q: What if I want a mean and standard deviation of my choosing?
You can calculate sigma*N(0,1)+m.
It can be shown that such shifting and scaling lead to N(m,sigma)
I have the following code which maybe could help:
set.seed(123)
n <- 1000
u <- runif(n) #creates U
x <- -log(u)
y <- runif(n, max=u*sqrt((2*exp(1))/pi)) #create Y
z <- ifelse (y < dnorm(x)/2, -x, NA)
z <- ifelse ((y > dnorm(x)/2) & (y < dnorm(x)), x, z)
z <- z[!is.na(z)]
It is also easier to use the implemented function rnorm() since it is faster than writing a random number generator for the normal distribution. See the following code as prove
n <- length(z)
t0 <- Sys.time()
z <- rnorm(n)
t1 <- Sys.time()
t1-t0
function distRandom(){
do{
x=random(DISTRIBUTION_DOMAIN);
}while(random(DISTRIBUTION_RANGE)>=distributionFunction(x));
return x;
}