R: how to perform more complex calculations from a combn of a dataset? - r

Question

Right now, I have a combn from the built in dataset iris. So far, I have been guided into being able to find the coefficient of lm() of the pair of values.
myPairs <- combn(names(iris[1:4]), 2)
formula <- apply(myPairs, MARGIN=2, FUN=paste, collapse="~")
model <- lapply(formula, function(x) lm(formula=x, data=iris)$coefficients[2])
model
However, I would like to go a few steps further and use the coefficient from lm() to be used in further calculations. I would like to do something like this:
Coefficient <- lm(formula=x, data=iris)$coefficients[2]
Spread <- myPairs[1] - coefficient*myPairs[2]
library(tseries)
adf.test(Spread)
The procedure itself is simple enough, but I haven't been able to find a way to do this for each combn in the data set. (As a sidenote, the adf.test would not be applied to such data, but I'm just using the iris dataset for demonstration).
I'm wondering, would it be better to write a loop for such a procedure?

Answer 1

You can do all of this within combn.
If you just wanted to run the regression over all combinations, and extract the second coefficient you could do
fun <- function(x) coef(lm(paste(x, collapse="~"), data=iris))[2]
combn(names(iris[1:4]), 2, fun)
You can then extend the function to calculate the spread
fun <- function(x) {
est <- coef(lm(paste(x, collapse="~"), data=iris))[2]
spread <- iris[,x[1]] - est*iris[,x[2]]
adf.test(spread)
}
out <- combn(names(iris[1:4]), 2, fun, simplify=FALSE)
out[[1]]
# Augmented Dickey-Fuller Test
#data: spread
#Dickey-Fuller = -3.879, Lag order = 5, p-value = 0.01707
#alternative hypothesis: stationary
Compare results to running the first one manually
est <- coef(lm(Sepal.Length ~ Sepal.Width, data=iris))[2]
spread <- iris[,"Sepal.Length"] - est*iris[,"Sepal.Width"]
adf.test(spread)
# Augmented Dickey-Fuller Test
# data: spread
# Dickey-Fuller = -3.879, Lag order = 5, p-value = 0.01707
# alternative hypothesis: stationary

Answer 2

Sounds like you would want to write your own function and call it in your myPairs loop (apply):
yourfun <- function(pair){
fm <- paste(pair, collapse='~')
coef <- lm(formula=fm, data=iris)$coefficients[2]
Spread <- iris[,pair[1]] - coef*iris[,pair[2]]
return(Spread)
}
Then you can call this function:
model <- apply(myPairs, 2, yourfun)
I think this is the cleanest way. But I don't know what exactly you want to do, so I was making up the example for Spread. Note that in my example you get warning messages, since column Species is a factor.

Answer 3

A few tips: I wouldn't name things that you with the same name as built-in functions (model, formula come to mind in your original version).
Also, you can simplify the paste you are doing - see the below.
Finally, a more general statement: don't feel like everything needs to be done in a *apply of some kind. Sometimes brevity and short code is actually harder to understand, and remember, the *apply functions offer at best, marginal speed gains over a simple for loop. (This was not always the case with R, but it is at this point).
# Get pairs
myPairs <- combn(x = names(x = iris[1:4]),m = 2)
# Just directly use paste() here
myFormulas <- paste(myPairs[1,],myPairs[2,],sep = "~")
# Store the models themselves into a list
# This lets you go back to the models later if you need something else
myModels <- lapply(X = myFormulas,FUN = lm,data = iris)
# If you use sapply() and this simple function, you get back a named vector
# This seems like it could be useful to what you want to do
myCoeffs <- sapply(X = myModels,FUN = function (x) {return(x$coefficients[2])})
# Now, you can do this using vectorized operations
iris[myPairs[1,]] - iris[myPairs[2,]] * myCoeffs[myPairs[2,]]
If I am understanding right, I believe the above will work. Note that the names on the output at present will be nonsensical, you would need to replace them with something of your own design (maybe the values of myFormulas).