Loop Through Data with Sequential Time Lags output Linear Regression Coefficients

Question

I have a dataset like so:

 set.seed(242)
 df<- data.frame(month=order(seq(1,20,1),decreasing=TRUE), 
 psit=sample(1:100,20,replace=TRUE),  var=sample(1:10,20,         


        

Answer 1

Try the dyn package which allows lm to process zoo and other time series objects:

library(dyn)

z <- read.zoo(df)
models <- lapply(1:3, function(i) dyn$lm(psit ~ lag(var, -i), tail(z, 12+i)))
sapply(models, function(x) summary(x)$r.squared)
## [1] 0.31209189 0.04923393 0.09995727

Note that typically if one uses lag k then one also includes all smaller values of k as well. In that case:

models <- lapply(1:3, function(i) dyn$lm(psit ~ lag(var, -(1:i)), tail(z, 12+i)))
do.call("anova", models)

giving:

Model 1: psit ~ lag(var, -(1:i))
Model 2: psit ~ lag(var, -(1:i))
Model 3: psit ~ lag(var, -(1:i))
  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     10 8688.5                           
2      9 8221.7  1    466.73 0.4545 0.5192
3      8 8215.5  1      6.24 0.0061 0.9398