In GNU Octave, would like to calculate an n-day exponential moving average of a vector without using a for-loop.
I am able to do this with a for loop but it is inefficient. I would like to use the filter function, however I am unsure how to get this to work correctly.
After piecing together the bits from this thread
http://octave.1599824.n4.nabble.com/vectorized-moving-average-td2132090.html
I built this function using Octave's filter function.
function meanV = movingEMean(V, window)
simpleAvg = mean(V(1:window));
alpha = 1/window;
X = V(window:end);
X(1) = simpleAvg;
meanV = filter(alpha, [1 alpha-1], X, simpleAvg*(1-alpha));
end
It starts with the simple moving average as the basis. V is the column vector of numbers to calculate the exponential moving average. window is an integer as a number of days. I used 12.
Here is a mathematical explanation of this function.
http://en.wikipedia.org/wiki/Moving_average#Exponential_moving_average
Note that the page uses 2/(n+1) (where n is window or the number of days) as alpha, but I use 1/n because that value of alpha fits my needs. Adjust alpha as needed.
Alternatively, I sometimes need my input and output vector's dimensions to match. I fill invalid values with NaN by adding meanV = [NaN(window-1,1); meanV]; as the last line in the movingEMean function. You could also fill it with simpleAvg if you want a rough estimate.
Reinventing the wheel on octave exponential moving average for a vector is silly. Just copy and paste movavg.m function defined in the octave financial package here: https://octave.sourceforge.io/financial:
function [varargout] = movavg(asset, lead, lag, alpha = 0)
if nargin < 3 || nargin > 4
print_usage ();
endif
if lead > lag
error ("lead must be <= lag")
elseif ischar (alpha)
if ! strcmpi (alpha, "e")
error ("alpha must be 'e' if it is a char");
endif
elseif ! isnumeric (alpha)
error ("alpha must be numeric or 'e'")
endif
## Compute the weights
if ischar (alpha)
lead = exp(1:lead);
lag = exp(1:lag);
else
lead = (1:lead).^alpha;
lag = (1:lag).^alpha;
endif
## Adjust the weights to equal 1
lead = lead / sum (lead);
lag = lag / sum (lag);
short = asset;
long = asset;
for i = 1:length (asset)
if i < length (lead)
## Compute the run-in period
r = length (lead) - i + 1:length(lead);
short(i) = dot (asset(1:i), lead(r))./sum (lead(r));
else
short(i) = dot (asset(i - length(lead) + 1:i), lead);
endif
if i < length (lag)
r = length (lag) - i + 1:length(lag);
long(i) = dot (asset(1:i), lag(r))./sum (lag(r));
else
long(i) = dot (asset(i - length(lag) + 1:i), lag);
endif
endfor
if nargout > 0
varargout{1} = short;
else
plot((1:length(asset))', [asset(:), long(:), short(:)]);
endif
if nargout > 1
varargout{2} = long;
endif
endfunction
And invoke thustly:
foo = [NaN; 1;4;8;10;-3;3;4;0;0;3;4;5;6;7;8;9];
lead = 7
lag = 7
alpha = 'e'
movavg(foo, lead, lag, 'e')
Which prints:
NaN
NaN
NaN
NaN
NaN
NaN
NaN
3.39851
1.24966
0.45742
2.06175
3.28350
4.37315
5.40325
6.41432
7.42128
8.42441
Related
%z = ratio of damping co-efficients , z<1
%wn = natural frequency in rad/sec
%wd = frequency of damped osciallations
%x_0 = amp
%phi = initial phase
%t = time
%%
z = 0.6943;
wn = 50;
wd = sqrt(1-(z^2))*wn;
x_0 = 42;
phi = pi/12;
t = linspace(0,100,1000);
x = x_0.*exp(-z*wn*t).*sin(phi+(wd*t));
plot(t,x);
error: operator *: nonconformant arguments (op1 is 1x1000, op2 is 1x1000)
error: called from
/home/koustubhjain/Documents/Damped_Oscialltion_(z<1).m at line 14 column 3
I am completely new to Octave/MATLAB, I just want to plot my equations and get a graph for them. Did I do something wrong with the multiplication ? Please someone help
Also the curve I am trying to plot should look something like a sinusoidal with decreasing amplitude, that's what my teacher told. But If I replace the multiplication signs with .*, all I get is a sort of a straight line.
The curve tends to 0 and the range of t is so wide to see something. Try to plot for t from 0 to 0.5 (instead of from 0 to 100) and you will see your curve.
I'm making a graph out of calculations by putting energy over the overall distance traveled. I used the equation E/D = F (Energy/Distance = Force) to try and got values in order to create a 2D line graph from them. However, I'm getting errors such as "nonconformant arguments", one of my variables being randomly turned to 0 and that the vector lengths aren't matching, here's the code:
% Declaring all the variables for the drag equation
p = 1.23;
v = 0:30;
C = 0.32;
A = 3.61;
D = 100000;
% This next line of code uses the variables above in order to get the force.
Fd = (p*(v.^2)*C*A)/2
% This force is then used to calculate the energy used to overcome the drag force
E = Fd*D
kWh = (E/3.6e+6);
Dist = (D/1000);
x = 0:Dist
y = 0:kWh
plot(x,y)
xlabel('x, Distance( km )')
ylabel('y, Energy Used Per Hour ( kWh )')
The outputs:
Hi there I need some help!
I am new to Gnuplot and have difficulties with the scripts.
Actually my equation is a lot more complicated. It is a parametric equation consisting of 7 parts each defined into a specific interval, with a bunch of parameters.
I just need a lead. So let me simplify the problem.
Suppose I have a function defined as follows f(x)= a*x+cos(x) : for 0 <= x <= 3;
and f(x)= b*1/cos(x) : for 3 < x <=10
my question is how do I instruct GNUPLOT:
1-) to consider "a" and "b" as parameters
2-) to plot "my user-defined" equation into the intervals of definition of f(x)
So far I have used the "set parametric" command but the problem is always at the end at the "PLOT f(x)" command for which I really don't know how to deal with the intervals.
I am using Windows 7 with the latest gnuplot.
Please help
You can define your f(x) as a (conditional) piece-wise function:
f(x) = 0 <= x && x <= 3 ? a*x+cos(x) : 3 < x && x <= 10 ? b/cos(x) : 1/0
The 1/0 above makes sure the function is not defined outside of the given intervals. The parameters a and b are already implicitly treated as parameters by gnuplot. When you change their values, f(x) is updated automatically. Example:
set xrange [-2:12]
a = 1.; b = 1.
plot f(x)
If you want more flexibility, you can take a and b as variables and do the following:
f(x,a,b) = 0 <= x && x <= 3 ? a*x+cos(x) : 3 < x && x <= 10 ? b/cos(x) : 1/0
set xrange [-2:12]
plot f(x,1,1), f(x,2,3)
32-bit Octave has a limit on the maximum number of elements in an array. I have recompiled from source (following the script at https://github.com/calaba/octave-3.8.2-enable-64-ubuntu-14.04 ), and now have 64-bit indexing.
Nevertheless, when I attempt to perform elementwise multiplication using a broadcast function, I get error: out of memory or dimension too large for Octave's index type
Is this a bug, or an undocumented feature? If it's a bug, does anyone have a reasonably efficient workaround?
Minimal code to reproduce the problem:
function indexerror();
% both of these are formed without error
% a = zeros (2^32, 1, 'int8');
% b = zeros (1024*1024*1024*3, 1, 'int8');
% sizemax % returns 9223372036854775806
nnz = 1000 % number of non-zero elements
rowmax = 250000
colmax = 100000
irow = zeros(1,nnz);
icol = zeros(1,nnz);
for ind =1:nnz
irow(ind) = round(rowmax/nnz*ind);
icol(ind) = round(colmax/nnz*ind);
end
sparseMat = sparse(irow,icol,1,rowmax,colmax);
% column vector to be broadcast
broad = 1:rowmax;
broad = broad(:);
% this gives "dimension too large" error
toobig = bsxfun(#times,sparseMat,broad);
% so does this
toobig2 = sparse(repmat(broad,1,size(sparseMat,2)));
mult = sparse( sparseMat .* toobig2 ); % never made it this far
end
EDIT:
Well, I have an inefficient workaround. It's slower than using bsxfun by a factor of 3 or so (depending on the details), but it's better than having to sort through the error in the libraries. Hope someone finds this useful some day.
% loop over rows, instead of using bsxfun
mult_loop = sparse([],[],[],rowmax,colmax);
for ind =1:length(broad);
mult_loop(ind,:) = broad(ind) * sparseMat(ind,:);
end
The unfortunate answer is that yes, this is a bug. Apparently #bsxfun and repmat are returning full matrices rather than sparse. Bug has been filed here:
http://savannah.gnu.org/bugs/index.php?47175
I am a new user of MATLAB. I want to find the value that makes f(x) = 0, using the Newton-Raphson method. I have tried to write a code, but it seems that it's difficult to implement Newton-Raphson method. This is what I have so far:
function x = newton(x0, tolerance)
tolerance = 1.e-10;
format short e;
Params = load('saved_data.mat');
theta = pi/2;
zeta = cos(theta);
I = eye(Params.n,Params.n);
Q = zeta*I-Params.p*Params.p';
% T is a matrix(5,5)
Mroot = Params.M.^(1/2); %optimization
T = Mroot*Q*Mroot;
% Find the eigenvalues
E = real(eig(T));
% Find the negative eigenvalues
% Find the smallest negative eigenvalue
gamma = min(E);
% Now solve for lambda
M_inv = inv(Params.M); %optimization
zm = Params.zm;
x = x0;
err = (x - xPrev)/x;
while abs(err) > tolerance
xPrev = x;
x = xPrev - f(xPrev)./dfdx(xPrev);
% stop criterion: (f(x) - 0) < tolerance
err = f(x);
end
% stop criterion: change of x < tolerance % err = x - xPrev;
end
The above function is used like so:
% Calculate the functions
Winv = inv(M_inv+x.*Q);
f = #(x)( zm'*M_inv*Winv*M_inv*zm);
dfdx = #(x)(-zm'*M_inv*Winv*Q*M_inv*zm);
x0 = (-1/gamma)/2;
xRoot = newton(x0,1e-10);
The question isn't particularly clear. However, do you need to implement the root finding yourself? If not then just use Matlab's built in function fzero (not based on Newton-Raphson).
If you do need your own implementation of the Newton-Raphson method then I suggest using one of the answers to Newton Raphsons method in Matlab? as your starting point.
Edit: The following isn't answering your question, but is just a note on coding style.
It is useful to split your program up into reusable chunks. In this case your root finding should be separated from your function construction. I recommend writing your Newton-Raphson method in a separate file and call this from the script where you define your function and its derivative. Your source would then look some thing like:
% Define the function (and its derivative) to perform root finding on:
Params = load('saved_data.mat');
theta = pi/2;
zeta = cos(theta);
I = eye(Params.n,Params.n);
Q = zeta*I-Params.p*Params.p';
Mroot = Params.M.^(1/2);
T = Mroot*Q*Mroot; %T is a matrix(5,5)
E = real(eig(T)); % Find the eigen-values
gamma = min(E); % Find the smallest negative eigen value
% Now solve for lambda (what is lambda?)
M_inv = inv(Params.M);
zm = Params.zm;
Winv = inv(M_inv+x.*Q);
f = #(x)( zm'*M_inv*Winv*M_inv*zm);
dfdx = #(x)(-zm'*M_inv*Winv*Q*M_inv*zm);
x0 = (-1./gamma)/2.;
xRoot = newton(f, dfdx, x0, 1e-10);
In newton.m you would have your implementation of the Newton-Raphson method, which takes as arguments the function handles you define (f and dfdx). Using your code given in the question, this would look something like
function root = newton(f, df, x0, tol)
root = x0; % Initial guess for the root
MAXIT = 20; % Maximum number of iterations
for j = 1:MAXIT;
dx = f(root) / df(root);
root = root - dx
% Stop criterion:
if abs(dx) < tolerance
return
end
end
% Raise error if maximum number of iterations reached.
error('newton: maximum number of allowed iterations exceeded.')
end
Notice that I avoided using an infinite loop.