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?
댓글 수: 2
Walter Roberson
2020년 7월 21일
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.
답변 (2개)
Bruno Luong
2020년 7월 21일
편집: Bruno Luong
2020년 7월 21일
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)

댓글 수: 5
Bruno Luong
2020년 7월 21일
If you want change for different pdf fv then change it, it in the line #1 & 2 of my code example.
Bruno Luong
2020년 7월 21일
편집: Bruno Luong
2020년 7월 21일
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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