how to make random vector with a certain profile

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 Comments

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.
the data is random, it is just an example.
the profile is what matter. do you still need the data? i will send it

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Answers (2)

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 Comments

just edit with c calculation
that's good, but how can i get a different output every time im running the code?
the reason is that im using different fv everytime, to build a library of results using this random fv.
If you want change for different pdf fv then change it, it in the line #1 & 2 of my code example.
actualy r is given and constant. that i can not change.
for fv (#2 line), if i play with numbers, wont it mess up the profile?

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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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Release

R2019b

Asked:

on 21 Jul 2020

Commented:

on 21 Jul 2020

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