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agridat (version 1.23)

harris.wateruse: Water use by horticultural trees

Description

Water use by horticultural trees

Arguments

Format

A data frame with 1040 observations on the following 6 variables.

species

species factor, 2 levels

age

age factor, 2 levels

tree

tree factor, 40 (non-consecutive) levels

day

day, numeric

water

water use, numeric

Details

Ten trees in each of four groups (two species, by two ages) were assessed for water usage, approximately every five days.

Missing values are included for the benefit of asreml, which needs a 'balanced' data set due to the kronecker-like syntax of the R matrix.

Used with permission of Roger Harris at Virginia Polytechnic.

Examples

Run this code
if (FALSE) {

  library(agridat)
  data(harris.wateruse)
  dat <- harris.wateruse

  # Compare to Schabenberger & Pierce, fig 7.23
  libs(latticeExtra)
  useOuterStrips(xyplot(water ~ day|species*age,dat, as.table=TRUE,
                        group=tree, type=c('p','smooth'),
                        main="harris.wateruse 2 species, 2 ages (10 trees each)"))


  # Note that measurements on day 268 are all below the trend line and
  # thus considered outliers.  Delete them.
  dat <- subset(dat, day!=268)


  # Schabenberger figure 7.24
  xyplot(water ~ day|tree,dat, subset=age=="A2" & species=="S2",
         as.table=TRUE, type=c('p','smooth'),
         ylab="Water use profiles of individual trees",
         main="harris.wateruse (Age 2, Species 2)")
  
  # Rescale day for nicer output, and convergence issues, add quadratic term
  dat <- transform(dat, ti=day/100)
  dat <- transform(dat, ti2=ti*ti)

  # Start with a subgroup: age 2, species 2
  d22 <- droplevels(subset(dat, age=="A2" & species=="S2"))

  # ----- Model 1, for subgroup A2,S2

  # First, a fixed quadratic that is common to all trees, plus
  # a random quadratic deviation for each tree.
  
  ## Schabenberger, Output 7.26
  ## proc mixed;
  ##   class tree;
  ##   model water = ti ti*ti / s;
  ##   random intercept ti ti*ti/subject=tree;

  libs(nlme,lucid)
  ## We use pdDiag() to get uncorrelated random effects
  m1n <- lme(water ~ 1 + ti + ti2, data=d22, na.action=na.omit,
             random = list(tree=pdDiag(~1+ti+ti2)))
  # lucid::vc(m1n)
  ##       effect variance    stddev
  ##  (Intercept)   0.2691 0.5188
  ##           ti   0      0.0000144
  ##          ti2   0      0.0000039
  ##     Residual   0.1472 0.3837


  # Various other models with lme4 & asreml
    
  libs(lme4, lucid)
  m1l <- lmer(water ~ 1 + ti + ti2 + (1|tree) +
                (0+ti|tree) + (0+ti2|tree), data=d22)
  # lucid::vc(m1l)
  ##      grp        var1 var2   vcov  sdcor
  ##     tree (Intercept)  0.2691 0.5188
  ##   tree.1          ti  0      0
  ##   tree.2         ti2  0      0
  ## Residual          0.1472 0.3837


  # Once the overall quadratic trend has been removed, there is not
  # too much evidence for consecutive observations being correlated
  ## d22r <- subset(d22, !is.na(water))
  ## d22r$res <- resid(m1n)
  ## xyplot(res ~ day|tree,d22r,
  ##        as.table=TRUE, type=c('p','smooth'),
  ##        ylab="residual",
  ##        main="harris.wateruse - Residuals of individual trees")
  ## op <- par(mfrow=c(4,3))
  ## tapply(d22r$res, d22r$tree, acf)
  ## par(op)
  
  # ----- Model 2, add correlation of consecutive measurements
  
  ## Schabenberger (page 516) adds correlation.
  ## Note how the fixed quadratic model is on the "ti = day/100" scale
  ## and the correlated observations are on the "day" scale.  The
  ## only impact this has on the fitted model is to increase the
  ## correlation parameter by a factor of 100, which was likely
  ## done to get better convergence.
  
  ## proc mixed data=age2sp2;
  ##   class tree;
  ##   model water = ti ti*ti / s ;
  ##   random intercept /subject=tree s;
  ##   repeated /subject=tree type=sp(exp)(day);

  ## Same as SAS, use ti for quadratic, day for correlation
  m2l <- lme(water ~ 1 + ti + ti2, data=d22,
             random = ~ 1|tree,
             cor = corExp(form=~ day|tree),
             na.action=na.omit)
  m2l # Match output 7.27.  Same fixef, ranef, variances, exp corr

  # lucid::vc(m2l)
  ##       effect variance stddev
  ##  (Intercept)   0.2656 0.5154
  ##     Residual   0.1541 0.3926
  
  # ---
  
  ## Now use asreml.  When I tried rcov=~tree:exp(ti),
  ## the estimated parameter value was on the 'boundary', i.e. 0.
  ## Changing rcov to the 'day' scale produced a sensible estimate
  ## that matched SAS.
  ## Note: SAS and asreml use different parameterizations for the correlation
  ## SAS uses exp(-d/phi) and asreml uses phi^d.
  ## SAS reports 3.79, asreml reports 0.77, and exp(-1/3.7945) = 0.7683274
  ## Note: normally a quadratic would be included as 'pol(day,2)'

  if(require("asreml", quietly=TRUE)){
    libs(asreml)
    d22 <- d22[order(d22$tree, d22$day),]
    m2a <- asreml(water ~ 1 + ti + ti2,
                  data=d22,
                  random = ~ tree,
                  residual=~tree:exp(day))
    lucid::vc(m2a)
    ##         effect component std.error z.ratio constr
    ##  tree!tree.var    0.2656   0.1301      2      pos
    ##     R!variance    0.1541   0.01611     9.6    pos
    ##      R!day.pow    0.7683   0.04191    18    uncon
  }
      
  # ----- Model 3. Full model for all species/ages.  Schabenberger p. 518
  
  ## /* Continuous AR(1) autocorrelations included      */
  ## proc mixed data=wateruse;
  ##   class age species tree;
  ##   model water = age*species age*species*ti age*species*ti*ti / noint s;
  ##   random intercept ti / subject=age*species*tree s;
  ##   repeated / subject=age*species*tree type=sp(exp)(day);
  
  
  m3l <- lme(water ~ 0 + age:species + age:species:ti + age:species:ti2,
             data=dat, na.action=na.omit,
             random = list(tree=pdDiag(~1+ti)),
             cor = corExp(form=~ day|tree) )
  
  m3l # Match Schabenberger output 7.27.  Same fixef, ranef, variances, exp corr

  # lucid::vc(m3l)
  ##       effect variance stddev
  ##  (Intercept)  0.1549  0.3936
  ##           ti  0.02785 0.1669
  ##     Residual  0.16    0.4

  # --- asreml
    
  if(require("asreml", quietly=TRUE)){
    dat <- dat[order(dat$tree,dat$day),]
    m3a <- asreml(water ~ 0 + age:species + age:species:ti + age:species:ti2,
                  data=dat,
                  random = ~ age:species:tree + age:species:tree:ti,
                  residual = ~ tree:exp(day) )
    
    # lucid::vc(m3a) # Note: day.pow = .8091 = exp(-1/4.7217)
    ##                       effect component std.error z.ratio constr
    ##     age:species:tree!age.var   0.1549   0.07192      2.2    pos
    ##  age:species:tree:ti!age.var   0.02785  0.01343      2.1    pos
    ##                   R!variance   0.16     0.008917    18      pos
    ##                    R!day.pow   0.8091   0.01581     51    uncon
  }
  
}

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