Boris Steipe writes:
> Sorry - two typos coorected:
>
> If you need x[,"a"] + x[,"b"] equal to 1, then replace any non-zero initial
> value of x[,b] with 1-x[,a].
> But if you really need "less than" h, you'll need to specify what your
> desired distribution of h - (x[,"a"] + x[,"b"]) should loo
Sorry - two typos coorected:
If you need x[,"a"] + x[,"b"] equal to 1, then replace any non-zero initial
value of x[,b] with 1-x[,a].
But if you really need "less than" h, you'll need to specify what your desired
distribution of h - (x[,"a"] + x[,"b"]) should look like.
On Oct 15, 2015, at 9:5
If you need h equal to 1, then replace any non-zero initial value of x[,b] with
1-x[,a].
But if you really need "less than", you'll need to specify what your desired
distribution of h - x[,"a"] + x[,"b"] should look like.
No?
B.
On Oct 15, 2015, at 9:27 AM, Rainer M Krug wrote:
> Boris St
Boris Steipe writes:
> I don't think the problem is well defined. Otherwise you could just
> pick very small numbers from a range that is guaranteed to keep the
> sum < h.
What further information is missing? That the variables should be
covering the whole range from 0 to 1?
OK - forgotten to s
I don't think the problem is well defined. Otherwise you could just pick very
small numbers from a range that is guaranteed to keep the sum < h.
B.
On Oct 15, 2015, at 8:48 AM, Rainer M Krug wrote:
> Hi
>
> I need a Latin Hypercube with the following conditions:
>
> 0 < x[,"a"] < 1
> 0 < x
Hi
I need a Latin Hypercube with the following conditions:
0 < x[,"a"] < 1
0 < x[,"b"] < 1
0 < x[,"c"] < 1
but also
x[,"a"] + x[,"b"] < h
The first three are easy:
--8<---cut here---start->8---
n <- 1000
lhc <- lhs::randomLHS(n=n, k=3
colnames(lhc) <- c("a
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