faceting multiple lm models

[Alexander Shenkin]

Hi folks,

I'm trying to facet a bunch of dotplots along with their lm fits.
Currently, I loop over the various subsets of the data to create a list
of lm fits. I can facet the dot plots, but I'm not sure how to get the
fits from from the list of lm-objects in there.

pseudo-code:
fit_list = list()
for (i in species) {
sub_data = subset(alldata, species == i)
fit_list[i] = lm(growth ~ size, data=sub_data)
}

ggplot(data = alldata, aes(x=size, y=growth)) + geom_point() +
facet_wrap(~ species)

The above gets me the faceted dot plots. I don't think geom_smooth
handles lists of lm fits. Any suggestions are greatly appreciated!

Thanks,
Allie

[R. Michael Weylandt]

If I understand you right, you want the fit line for each facet to
depend only on the points of that facet: I think what you are looking
for is

ggplot(alldata, aes(x = size, y = growth)) + geom_point() +
stat_smooth(method = "lm") + facet_wrap( ~ species)

E.g.,

data(diamonds)
ggplot(diamonds, aes(x = carat, y = price)) + geom_point() +
stat_smooth(method = "lm") + facet_wrap(~ cut, ncol = 1)

Note that in each facet, the resulting line is different.

Hope this helps,
Michael

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[Alexander Shenkin]

Thanks Michael - that was exactly what I was looking for. Perhaps my
question gets filed in the "duh" category.

Somehow I get paranoid about fitting lm myself, and then using ggplot to
plot the lines. For example, if there's some obscure difference in
subsetting, or some problem, etc. In the case that, for whatever
reason, I want to use my manually-fitted lm's for plotting, is there any
way to go about that in ggplot?

thanks again,
allie

[baptiste auguie]

You need to construct a data.frame with the same facetting variable,

data(diamonds)
ggplot(diamonds) + geom_point(aes(x = carat, y = price)) +

facet_wrap(~ cut, ncol = 1)

library(plyr)
fits = ddply(diamonds, .(cut), function(d) coeff = coef(lm(price~carat,data=d)))
last_plot() + geom_abline(aes(intercept=`(Intercept)`,slope =
carat),col="blue",data=fits)

HTH,

b.