Predict function

Predict function

Usage

# S3 method for class 'LMMsolve'
predict(
  object,
  newdata,
  type = c("response", "link"),
  se.fit = FALSE,
  deriv = NULL,
  ...
)

Arguments

object

an object of class LMMsolve.

newdata

A data.frame containing new points for which the smooth trend should be computed. Column names should include the names used when fitting the spline model.

type

When this has the value "link" the linear predictor fitted values or predictions (possibly with associated standard errors) are returned. When type = "response" (default) fitted values or predictions on the scale of the response are returned (possibly with associated standard errors).

se.fit

calculate standard errors, default FALSE.

deriv

Character string of variable for which to calculate the first derivative; default NULL.

...

other arguments. Not yet implemented.

Value

A data.frame with predictions for the smooth trend on the specified grid. The standard errors are saved if `se.fit=TRUE`.

Examples

## simulate some data
f <- function(x) { 0.3 + 0.4*x + 0.2*sin(20*x) }
set.seed(12)
n <- 150
x <- seq(0, 1, length = n)
sigma2e <- 0.04
y <- f(x) + rnorm(n, sd = sqrt(sigma2e))
dat <- data.frame(x, y)

## fit the model
obj <- LMMsolve(fixed = y ~ 1,
         spline = ~spl1D(x, nseg = 50), data = dat)

## make predictions
newdat <- data.frame(x = seq(0, 1, length = 5))
pred <- predict(obj, newdata = newdat, se.fit = TRUE)
pred
#>      x     ypred         se
#> 1 0.00 0.2051355 0.09713998
#> 2 0.25 0.1842741 0.05188443
#> 3 0.50 0.3714877 0.05202181
#> 4 0.75 0.7024734 0.05188443
#> 5 1.00 0.6767744 0.09713998

## make predictions for derivative of x:
pred2 <- predict(obj, newdata = newdat, se.fit = TRUE, deriv = "x")
pred2
#>      x      ypred       se
#> 1 0.00  3.5771184 3.395408
#> 2 0.25  0.5155952 1.549117
#> 3 0.50 -3.5814981 1.474648
#> 4 0.75 -3.1879787 1.549117
#> 5 1.00 -1.0076817 3.395408