# [R] finding interpolated values along an empirical parametric curve

John Fox jfox at mcmaster.ca
Mon Dec 5 15:50:06 CET 2011

```Hi Michael,

I can get what appears to be a good interpolation with a regression spline
in a multivariate LM, playing around with the tuning parameter to leave 1
residual df. Try this:

library(splines)
mod <- lm(cbind(log.det, norm.beta) ~ bs(lambda, df=4), data=pd)
summary(mod)

x <- data.frame(lambda=seq(min(pd\$lambda), max(pd\$lambda), length=100))
fit <- predict(mod, newdata=x)
points(fit[, "norm.beta"], fit[, "log.det"], pch=16, cex=0.5)

x.2 <- data.frame(lambda=c(lambda.HKB, lambda.LW))
fit.2 <- predict(mod, x.2)
points(fit.2[, "norm.beta"], fit.2[, "log.det"], pch=15, col="green")

That doesn't solve the problem of calculating the normal, however.

I hope this helps,
John

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Michael Friendly
> Sent: December-05-11 9:16 AM
> To: R-help
> Subject: [R] finding interpolated values along an empirical parametric
> curve
>
> Given the following data, I am plotting log.det ~ norm.beta, where the
> points depend on a parameter, lambda (but there is no functional form).
> I want to find the (x,y) positions along this curve corresponding to
> two special values of lambda
>
> lambda.HKB <- 0.004275357
> lambda.LW <- 0.03229531
>
> and draw reference lines at ~ -45 degrees (or normal to the curve) thru
> these points.
> How can I do this?  A complete example is below
>
>  > pd
>        lambda   log.det norm.beta
> 0.000  0.000 -12.92710  3.806801
> 0.005  0.005 -14.41144  2.819460
> 0.010  0.010 -15.41069  2.423197
> 0.020  0.020 -16.82581  2.010924
> 0.040  0.040 -18.69819  1.611304
> 0.080  0.080 -21.05065  1.283928
>  >
>
> pd <-
> structure(list(lambda = c(0, 0.005, 0.01, 0.02, 0.04, 0.08),
>      log.det = c(-12.9270978142337, -14.411442487768, -
> 15.4106886674014,
>      -16.8258120792945, -18.6981870228698, -21.050646106925),
>      norm.beta = c(3.8068008759562, 2.81945995964196, 2.42319655878575,
>      2.01092421747594, 1.6113040561427, 1.28392804825009)), .Names =
> c("lambda", "log.det", "norm.beta"), class = "data.frame", row.names =
> c("0.000", "0.005", "0.010", "0.020", "0.040", "0.080"))
>
> clr <- c("black", rainbow(5, start=.6, end=.1)) lambdaf <-
> c(expression(~widehat(beta)^OLS), ".005", ".01", ".02", ".04", ".08")
> op <- par(mar=c(4, 4, 1, 1) + 0.2, xpd=TRUE) with(pd, {plot(norm.beta,
> log.det, type="b",
>      cex.lab=1.25, pch=16, cex=1.5, col=clr,
>    xlab='shrinkage: ||b||',
>      ylab='variance: log |(Var(b)|)')
>      text(norm.beta, log.det, lambdaf, cex=1.25, pos=2)
>      text(min(norm.beta), max(log.det), "Variance vs. Shrinkage",
> cex=1.5, pos=4)
>      })
>
>
> # How to find the (x,y) positions for these values of lambda along the
> curve of log.det ~ norm.beta ?
> lambda.HKB <- 0.004275357
> lambda.LW <- 0.03229531
>
> --
> Michael Friendly     Email: friendly AT yorku DOT ca
> Professor, Psychology Dept.
> York University      Voice: 416 736-5115 x66249 Fax: 416 736-5814
> 4700 Keele Street    Web:   http://www.datavis.ca
> Toronto, ONT  M3J 1P3 CANADA
>
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