cessary corrections, i'd be very very greatful.
>
> Thanks heaps guys for taking the time to look at this
>
>> Date: Mon, 18 May 2009 15:06:47 +0200
>> From: waclaw.marcin.kusnierc...@idi.ntnu.no
>> To: konk2...@hotmail.com
>> CC: mike.lawre...@dal.ca; r-help@r-
Mike Lawrence wrote:
> why not simply
>
> vars=list()
> for (i in 1:1000) vars[[i]] = var(z[[i]])
>
>
... or, much simpler,
vars = sapply(z, var)
vQ
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PLEASE do rea
why not simply
vars=list()
for (i in 1:1000) vars[[i]] = var(z[[i]])
On Mon, May 18, 2009 at 6:51 AM, Kon Knafelman wrote:
>
> Hi,
>
>> g=list()
>> for(i in 1:1000){z[[i]]=rnorm(15,0,1)}
>
> I've attempted a similar problem based on the above method. Now, if i want to
> find the sample varianc
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