how to make random vector with a certain profile
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Well, im looking for a way to produce a random vector that behave like the profile in the picture.
It needs to start from let's say around 2, then climbs up to a max value not higher than 20, and then drops to zero.
That is the general idea, the distribution doesnt have to be exactly like in the graph.
does anyone have a good idea how to make that happened?
2 comentarios
Walter Roberson
el 21 de Jul. de 2020
Do you happen to have the data for that plot available? Rather than us having to read it off the graph and enter the values by hand.
samuel
el 21 de Jul. de 2020
Respuestas (2)
Bruno Luong
el 21 de Jul. de 2020
Editada: Bruno Luong
el 21 de Jul. de 2020
r=linspace(0,1.5);
fv=(1.5-r).*(2+10*exp(-((r-1)./0.3).^2)); % whatever unnormaized pdf
% NOTE: This method assumes r is an equidistance vector
% otherwise you need to multiply fv by dr before cumulative sum
c = cumsum(fv);
c=(c-c(1))/(c(end)-c(1));
[cu,loc] = unique(c);
rs=r(loc);
x=interp1(cu, rs, rand(1,1e6)); % your random vector
% Check
subplot(2,1,1);
plot(r,fv);
subplot(2,1,2);
hist(x,100)

5 comentarios
Walter Roberson
el 21 de Jul. de 2020
What is c?
Bruno Luong
el 21 de Jul. de 2020
just edit with c calculation
samuel
el 21 de Jul. de 2020
Bruno Luong
el 21 de Jul. de 2020
If you want change for different pdf fv then change it, it in the line #1 & 2 of my code example.
samuel
el 21 de Jul. de 2020
Bruno Luong
el 21 de Jul. de 2020
Editada: Bruno Luong
el 21 de Jul. de 2020
Modified code in case r is not equidistance (but monotonic)
rmid = 0.5*(r(1:end-1) + r(2:end));
dr = r(2:end)-r(1:end-1);
fvfun = @(r)(1.5-r).^3.*(2+50*exp(-((r-1)./0.3).^2)); % whatever
fv = fvfun(rmid);
c = cumsum(fv.*dr);
c = [0, c]/c(end);
[cu, loc] = unique(c);
x = interp1(cu, r(loc), rand(1,1e6));
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