Automatic curve fitting in R

Question

Is there any package that automatically fits a curve using many simple models?
By simple models I mean:

• ax+b
• ax^2+bx+c
• a*log(x) + b

Answer 1

You could also look at packages providing functions to evaluate fractional polynomials. So far, these appear to be mboost (with the function FP) and mfp (with the function mfp). Although I haven't tried the packages, the theory behind them fits what you're after.

The mfp package was described in R-News in 2005.

Two references that might be of interest are

Royston P, Altman D (1994) Regression using fractional polynomials of continuous covariates. Appl Stat. 3: 429–467.

Sauerbrei W, Royston P (1999) Building multivariable prognostic and diagnostic models: transformation of the predictors by using fractional polynomials. Journal of the Royal Statistical Society (Series A) 162: 71–94.

Answer 2

Try this. rhs is a character vector of right sides and x and y are the data. It constructs the formula fo for each and then extracts the parameters and sets each to 1 for the starting value. Finally it runs nls and returns the SSEs sorted so that the result is a vector of SSE's named via the right hand sides. If verbose=TRUE (which it is by default) then it also displays the output from each fit.

sse <- function(rhs, x, y) sort(sapply(rhs, function(rhs, x, y, verbose = TRUE) {
    fo <- as.formula(paste("y", rhs, sep = "~"))
    nms <- setdiff(all.vars(fo), c("x", "y"))
    start <- as.list(setNames(rep(1, length(nms)), nms))
    fm <- nls(fo, data.frame(x, y), start = start)
    if (verbose) { print(fm); cat("---\n") }
    deviance(fm)
}, x = x, y = y))

## test

set.seed(123)
x <- 1:10
y <- rnorm(10, x)

# modify to suit
rhs <- c("a*x+b", "a*x*x+b*x+c")

sse(rhs, x, y)

Answer 3

You can fit a Regression Splines and find a good fit by manually adjusting the degrees of freedom a few times. Try the following function:

spline.fit <- function(x, y, df=5) {
  ## INPUT: x, y are two vectors (predictor and response);
  ##        df is the number of spline basis.  Increase "df" to fit more adaptively to the data.
  require(splines) # available as default R Package.
  bx <- bs(x, df)  # B-spline basis matrix as New Predictors (dimension is "length(x)" by "df")
  f <- lm(y ~ bx)  # Linear Regression on Spline Basis (that is, "df" number of new predictors)
  fy <- fitted(f)  # Fitted Response
  plot(x, y); lines(x, fy, col="blue", lwd=2) # Make a plot to show the fit.
  invisible(list(x=bx, y=fy, f=f))    # Return the Basis (new predictors), Fitted Y, Regression
}

if (F) {                                # Unit Test
  spline.fit(1:100, rnorm(100))
  spline.fit(1:100, rnorm(100), df=20)
}