[R] plots on log="y" scale with smooths
David Winsemius
dwinsemius at comcast.net
Thu Sep 18 03:02:51 CEST 2014
On Sep 17, 2014, at 4:34 PM, Michael Friendly wrote:
> In the following example, I am trying to plot a response on a log
> scale, and add one or more smoothed
> curves, e.g., lowess and abline. In base graphics, I'd like to do
> this using log="y", so that the Y axis is
> spaced on the log scale, but labeled as the actual response values.
> Using ggplot2, I'm using
> scale_y_log10 to achieve the same purpose. However, my attempts to
> add the smooths differ
> considerably, so I must be missing something.
>
> Here's the data I'm working with for one example:
>
> data("CodParasites", package = "countreg")
> ## omit NAs in response & predictor
> CodParasites <- subset(CodParasites, !(is.na(intensity) |
> is.na(length)))
> ## plot only positive values of intensity
> CPpos <- subset(CodParasites, intensity>0)
>
> Here's the base graphics plot. The abline() is clearly wrong and
> the lowess smooth looks too low.
> How does one meld plots using log="y" with such additional plot
> annotations ?
>
> plot(jitter(intensity) ~ length, data = CPpos, log = "y")
> with(CPpos, lines(lowess(length, log(intensity)), col="red", lwd=2) )
> abline(lm(log(intensity) ~ length, data=CPpos))
This is what I would have expected to be a more accurate plot of the
estimated central tendencies:
with(CPpos, lines(lowess(length, exp(log(intensity))), col="red",
lwd=2) )
lines( CPpos$length, exp( lm(log(intensity) ~ length, data=CPpos)
$fitted) )
Just because you have changed the scaling of the plot axis does not
mean that you would plot log()-ed values.
>
> Here's an attempt at a ggplot2 version, that actually looks more
> reasonable, but I'm not sure that it
> is correct:
>
> library(ggplot2)
> ggplot(CPpos, aes(x=length, y=intensity)) +
> geom_jitter(position=position_jitter(height=.1), alpha=0.25) +
> scale_y_log10(breaks=c(1,2,5,10,20,50,100, 200)) +
> stat_smooth(method="loess", color="red", size=1.5) +
> stat_smooth(method="lm")
The loess line is somewhat different but the lm() prediction is the
same as I expected.
--
David.
David Winsemius, MD
Alameda, CA, USA
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