Regression in R with loops - r

Question

I need to run a simple regression using Lm() in R. Its simple because I have only one independent variable. However the catch is that I need to test this independent variable for a number of dependents which are columns in a data frame.
So basically I have one common X and numerous Y's for which i need to extract the intercept and slope and store them all in a data frame.
In excel this is possible with the intercept and slope functions and then dragging across columns. I need something in R that would basically do the same, I could of course run separate regressions , but the requirement is that I need to run all of them in one loop and store estimates of intercept and slopes together for each.
Im still learning R and any help on this would be great. Thanks :)

Answer 1

The lmList function in package nlme was designed for this.
Let's use the iris dataset as an example:
DF <- iris[, 1:4]
# Sepal.Length Sepal.Width Petal.Length Petal.Width
#1 5.1 3.5 1.4 0.2
#2 4.9 3.0 1.4 0.2
#3 4.7 3.2 1.3 0.2
#4 4.6 3.1 1.5 0.2
#5 5.0 3.6 1.4 0.2
#6 5.4 3.9 1.7 0.4
#...
First we have to reshape it. We want Sepal.Length as the dependent and the other columns as predictors in this example.
library(reshape2)
DF <- melt(DF, id.vars = "Sepal.Length")
# Sepal.Length variable value
#1 5.1 Sepal.Width 3.5
#2 4.9 Sepal.Width 3.0
#3 4.7 Sepal.Width 3.2
#4 4.6 Sepal.Width 3.1
#5 5.0 Sepal.Width 3.6
#6 5.4 Sepal.Width 3.9
#...
Now we can do the fits.
library(nlme)
mods <- lmList(Sepal.Length ~ value | variable,
data = DF, pool = FALSE)
We can now extract intercept and slope for each model.
coef(mods)
# (Intercept) value
#Sepal.Width 6.526223 -0.2233611
#Petal.Length 4.306603 0.4089223
#Petal.Width 4.777629 0.8885803
And get the usual t-table:
summary(mods)
# Call:
# Model: Sepal.Length ~ value | variable
# Data: DF
#
# Coefficients:
# (Intercept)
# Estimate Std. Error t value Pr(>|t|)
# Sepal.Width 6.526223 0.47889634 13.62763 6.469702e-28
# Petal.Length 4.306603 0.07838896 54.93890 2.426713e-100
# Petal.Width 4.777629 0.07293476 65.50552 3.340431e-111
# value
# Estimate Std. Error t value Pr(>|t|)
# Sepal.Width -0.2233611 0.15508093 -1.440287 1.518983e-01
# Petal.Length 0.4089223 0.01889134 21.646019 1.038667e-47
# Petal.Width 0.8885803 0.05137355 17.296454 2.325498e-37
Or the R-squared values:
summary(mods)$r.squared
#[1] 0.01382265 0.75995465 0.66902769
However, if you need something more efficient, you can use package data.table together with lm's workhorse lm.fit:
library(data.table)
setDT(DF)
DF[, setNames(as.list(lm.fit(cbind(1, value),
Sepal.Length)[["coefficients"]]),
c("intercept", "slope")), by = variable]
# variable intercept slope
#1: Sepal.Width 6.526223 -0.2233611
#2: Petal.Length 4.306603 0.4089223
#3: Petal.Width 4.777629 0.8885803
And of course the R.squared values of these models are just the squared Pearson correlation coefficients:
DF[, .(r.sq = cor(Sepal.Length, value)^2), by = variable]
# variable r.sq
#1: Sepal.Width 0.01382265
#2: Petal.Length 0.75995465
#3: Petal.Width 0.66902769