Using lapply on a list of models - r

Question

I have generated a list of models, and would like to create a summary table.
As and example, here are two models:
x <- seq(1:10)
y <- sin(x)^2
model1 <- lm(y ~ x)
model2 <- lm(y ~ x + I(x^2) + I(x^3))
and two formulas, the first generating the equation from components of formula
get.model.equation <- function(x) {
x <- as.character((x$call)$formula)
x <- paste(x[2],x[1],x[3])
}
and the second generating the name of model as a string
get.model.name <- function(x) {
x <- deparse(substitute(x))
}
With these, I create a summary table
model.list <- list(model1, model2)
AIC.data <- lapply(X = model.list, FUN = AIC)
AIC.data <- as.numeric(AIC.data)
model.models <- lapply(X = model.list, FUN = get.model)
model.summary <- cbind(model.models, AIC.data)
model.summary <- as.data.frame(model.summary)
names(model.summary) <- c("Model", "AIC")
model.summary$AIC <- unlist(model.summary$AIC)
rm(AIC.data)
model.summary[order(model.summary$AIC),]
Which all works fine.
I'd like to add the model name to the table using get.model.name
x <- get.model.name(model1)
Which gives me "model1" as I want.
So now I apply the function to the list of models
model.names <- lapply(X = model.list, FUN = get.model.name)
but now instead of model1 I get X[[1L]]
How do I get model1 rather than X[[1L]]?
I'm after a table that looks like this:
Model Formula AIC
model1 y ~ x 11.89136
model2 y ~ x + I(x^2) + I(x^3) 15.03888

Answer 1

Do you want something like this?
model.list <- list(model1 = lm(y ~ x),
model2 = lm(y ~ x + I(x^2) + I(x^3)))
sapply(X = model.list, FUN = AIC)

Answer 2

I'd do something like this:
model.list <- list(model1 = lm(y ~ x),
model2 = lm(y ~ x + I(x^2) + I(x^3)))
# changed Reduce('rbind', ...) to do.call(rbind, ...) (Hadley's comment)
do.call(rbind,
lapply(names(model.list), function(x)
data.frame(model = x,
formula = get.model.equation(model.list[[x]]),
AIC = AIC(model.list[[x]])
)
)
)
# model formula AIC
# 1 model1 y ~ x 11.89136
# 2 model2 y ~ x + I(x^2) + I(x^3) 15.03888

Answer 3

Another option, with ldply, but see hadley's comment below for a more efficient use of ldply:
# prepare data
x <- seq(1:10)
y <- sin(x)^2
dat <- data.frame(x,y)
# create list of named models obviously these are not suited to the data here, just to make the workflow work...
models <- list(model1=lm(y~x, data = dat),
model2=lm(y~I(1/x), data=dat),
model3=lm(y ~ log(x), data = dat),
model4=nls(y ~ I(1/x*a) + b*x, data = dat, start = list(a = 1, b = 1)),
model5=nls(y ~ (a + b*log(x)), data=dat, start = setNames(coef(lm(y ~ log(x), data=dat)), c("a", "b"))),
model6=nls(y ~ I(exp(1)^(a + b * x)), data=dat, start = list(a=0,b=0)),
model7=nls(y ~ I(1/x*a)+b, data=dat, start = list(a=1,b=1))
)
library(plyr)
library(AICcmodavg) # for small sample sizes
# build table with model names, function, AIC and AICc
data.frame(cbind(ldply(models, function(x) cbind(AICc = AICc(x), AIC = AIC(x))),
model = sapply(1:length(models), function(x) deparse(formula(models[[x]])))
))
.id AICc AIC model
1 model1 15.89136 11.89136 y ~ x
2 model2 15.78480 11.78480 y ~ I(1/x)
3 model3 15.80406 11.80406 y ~ log(x)
4 model4 16.62157 12.62157 y ~ I(1/x * a) + b * x
5 model5 15.80406 11.80406 y ~ (a + b * log(x))
6 model6 15.88937 11.88937 y ~ I(exp(1)^(a + b * x))
7 model7 15.78480 11.78480 y ~ I(1/x * a) + b
It's not immediately obvious to me how to replace the .id with a column name in the ldply function, any tips?