Displaying 4 results from an estimated 4 matches for "simdat".
Did you mean:
simdata
2004 Jan 15
1
random effects with lme() -- comparison with lm()
...t
model. Could someone explain to me why lme() still gives two standard
deviation estimates? I would expect lme() to return either:
a) an error or a warning for having an unidentifiable model;
b) only one standard deviation estimate.
Thank you for your time.
Jerome Asselin
> library(nlme)
> simdat <- data.frame(A=1:4,Y=c(23,43,11,34))
> simdat
A Y
1 1 23
2 2 43
3 3 11
4 4 34
> lme(Y~1,data=simdat,random=~1|A)
<...snip...>
Random effects:
Formula: ~1 | A
(Intercept) Residual
StdDev: 12.96007 4.860027
<...snip...>
> summary(lm(Y~1,data=simdat))$sigma
[1]...
2006 Apr 18
2
Unfound objects in function
...q (0,
max.r, length = 100), correction = "trans")
diff <- sum (Ki$theo - Ki$trans)
i <- i + 0.1
if (i > max.r) stop ("no suitable kernel found within 0.2 to
max.r")
}
if (abs (olddiff) > abs (diff)) return (smo) else return (oldsmo)
}
data (simdat)
test <- kernelEst (simdat)
Error in while (diff > 0) { : missing value where TRUE/FALSE needed
> version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor...
2012 May 17
1
oldlogspline probabilities
.... I know I can just use the proportion of actual hits, but am curious to compare this to an estimate from a density estimation. Unfortunately when using poldlogspline, this probability is always=0 (simulated data example is below). How can this be, given that the density is highest at area=0?
> simdat<-c(rep(0,8),rexp(92))
> myspline<-oldlogspline(simdat,lbound=0)
> poldlogspline(fit=myspline, q=0)
[1] 0
Any help to work out the probability of an area value in my distribution = 0 would be appreciated
Terry Beutel
Agri-Science Queensland
[[alternative HTML version deleted]]
2017 Jul 06
1
Convert date to continuous variable in R
...want the axis on the top to look like, but here's what I'd be likely to do:
>
> ## note, correction to format; in your example data year comes last
> LAI_simulation$Date <- as.Date( LAI_simulation$Date, '%m/%d/%Y')
>
> plot(LAI~Date, data=LAI_simulation)
>
> simdate <- as.numeric( LAI_simulation$Date - as.Date('2009-10-7') )
> ## or
> simdate <- difftime( LAI_simulation$Date, as.Date('2009-10-7') )
>
> ## then
> axis(3, pretty(simdate) )
>
>
> Converting LAI_simulation$Date to numeric, and then applying axis.Date(...