... and here is a a simple 2-liner without a sort that I think is linear in
time and space (but please correct if this is wrong):
x <- cumsum(runif(10))
x/x[10] * runif(1, 0, 55) + seq.int(0, 45,5)
Question: Does this give the same distribution as Peter's method using the
order statistics? I bel
Richard:
"The "use an upper bound of 100 - (n+1)*5" and then "add back
cumsum(rep(5,n)) at the end" (or equivalent) trick ensures the gaps
are right but does nothing about the distribution.."
If I understand you correctly, I think the above is wrong. Here is a
one-line version of Peter's code for
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