I think it was my mistake , I switched the sd and mean position.. On Tue, Nov 12, 2013 at 3:45 PM, Craig DeForest <[email protected]> wrote: > I think it was luck that those three were negative. grandom() gives you zero > mean and unit variance, so half the numbers that come out of it are negative. > > > On Nov 12, 2013, at 1:10 PM, mraptor <[email protected]> wrote: > >> thanks... my mistake with the byte thing.. >> What I want to do is to generate normally distributed random numbers, >> where the range is humongous (fe: 2^1000) >> I just tried : >>> $n = 1000 >>> p grandom(3) * ((2**$n)/2) + sqrt((2**$n))/2 >> [-1.9906798e+300 -3.9087665e+300 -5.6272303e+300] >> >> >> those seem like a random numbers !! but why negative ? >> Do you think I should use Math::BigInt >> >> On Mon, Nov 11, 2013 at 11:55 AM, Craig DeForest >> <[email protected]> wrote: >>> The issue is that grandom doesn't work properly on integer types, since it >>> makes a distribution with mean 0 and stddev 1. It's a bit awkward, but you >>> can use >>> >>> $vals = (grandom($n) * $stdev + $mean)->rint->byte >>> >>> instead. >>> >>> >>> On Nov 11, 2013, at 8:59 AM, mraptor <[email protected]> wrote: >>> >>>> Is there are way to generate random values based on Normal distribution. >>>> The same like rnorm() in R-language. >>>> >>>>> round( rnorm(3,mean=2^n/2,sd=sqrt(2^n)/2) ) >>>> [1] 524758 524540 524647 >>>> >>>> I see there is grandom() funtion, but it requires some sort of lib (ndtri) >>>> ? >>>> >>>> pdl> p grandom(byte,1) >>>> >>>> ndtri domain error >>>> [0] >>>> >>>> >>>> thanks >>>> >>>> _______________________________________________ >>>> Perldl mailing list >>>> [email protected] >>>> http://mailman.jach.hawaii.edu/mailman/listinfo/perldl >>>> >>> >> >> _______________________________________________ >> Perldl mailing list >> [email protected] >> http://mailman.jach.hawaii.edu/mailman/listinfo/perldl >> >
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