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R help

3 messages · Shutnik, Spencer Graves, Patrick Burns

#
> Sd <- sqrt(diag(Var))
 > Var/outer(Sd, Sd)
      [,1] [,2]
[1,] 1.00 0.25
[2,] 0.25 1.00

hope this helps.  spencer graves
Shutnik wrote:
#
If the variance matrix is large, then

t(Var / Sd) / Sd

is probably more efficient.

Patrick Burns

Burns Statistics
patrick at burns-stat.com
+44 (0) 208 525 0696
http://www.burns-stat.com/  
(home of S Poetry and "A Guide for the Unwilling S User")
Spencer Graves wrote: