Calculate coefficients of rolling regressions with dependent variables in the columns

Question

I'm hoping to calculate the coefficients (the intercepts in particular) of rolling regressions. There are many dependent variables. Some of them (Y1 and Y2) are shown below. Each of them is regressed with independent variables X1 and X2. Additionally, Y1 and Y2 both have NAs in different periods. The data is a time series with monthly interval. The rolling window is 6.

This is my code:

rr <- rollapply(df, width = 6,  
                  FUN = function(z) coef(lm(Y1~ X1+X2, 
                          data = as.data.frame(z))),
                by.column = FALSE, align = "right")

However, the problem of this code is that
1) it only deals with one independent variable (Y1 in this case) at once,
2) it gave the same coefficients for all rolling regressions. I assume the existence of NA messed up the rolling regression?
I would greatly appreciate if someone can shed some lights. Thanks.

This is a sample data.

Date        Y1    Y2    X1      X2
1/1/2009    NA  1.51    0.02    0.75
2/1/2009    NA  -0.38   0.01    0.59
3/1/2009    NA  1.54    0.02    0.96
4/1/2009    NA  1.78    0.01    0.92
5/1/2009    NA  0.94    0.02    0.02
6/1/2009    NA  1.37    0.01    0.46
7/1/2009    NA  1.22    0.01    0.61
8/1/2009    NA  1.32    0.01    0.04
9/1/2009    NA  0.83    0.01    0.03
10/1/2009   NA  0.95    0.02    0.61
11/1/2009   NA  0.28    0.03    0.53
12/1/2009   NA  0.17    0.01    0.32
1/1/2010    1.71    NA  0.03    0.53
2/1/2010    0.39    NA  0.03    0.16
3/1/2010    0.11    NA  0.01    0.58
4/1/2010    1.25    NA  0.01    0.41
5/1/2010    0.57    NA  0.01    0.9
6/1/2010    0.48    NA  0.01    0.58
7/1/2010    0.16    NA  0.01    0.03
8/1/2010    0.37    NA  0.01    0.23
9/1/2010    0.31    NA  0.01    0.77
10/1/2010   0.63    NA  0.01    0.75
11/1/2010   0.61    NA  0.01    0.74
12/1/2010   0.91    NA  0.01    0.41

Answer 1

There are two problems:

1. passing a data.frame to rollapply causes it to be converted to a matrix and since one of the columns is character column the result will be a character matrix whereas numeric is what is needed. Use df[-1] or the code shown below.

2. lm does not accept a dependent variable having all NA values. Check for that and return NA in such cases.

Adding a few improvements:

• first convert the input to class zoo.
• define function getCoef to get the coefficients given the data and the left and right side of the formula
• define function roll to do the actual rollapply using rollapplyr
• lapply the function roll over c("Y1", "Y2") to produce a list of 2 zoo objects
• optionally Map fortify.zoo over L to give a list of data frames

Code:

z <- read.zoo(df, FUN = as.yearmon, format = "%m/%d/%Y")

getCoef <- function(z, lhs, rhs) {
  if (all(is.na(z[, lhs]))) NA
  else coef(lm(paste(lhs, "~", rhs), z))
}

roll <- function(z, lhs, rhs = "X1 + X2") {
  rollapplyr(z, 6, getCoef, by.column = FALSE, coredata = FALSE, lhs = lhs, rhs = rhs)
}

ynames <- c("Y1", "Y2")
L <- lapply(ynames, roll, z = z)

Optionally, for a list of data.frames:

Map(fortify.zoo, L)

Note: The input df in reproducible form is:

Lines <- "Date    Y1  Y2  X1  X2
1/1/2009    NA  1.51    0.02    0.75
2/1/2009    NA  -0.38   0.01    0.59
3/1/2009    NA  1.54    0.02    0.96
4/1/2009    NA  1.78    0.01    0.92
5/1/2009    NA  0.94    0.02    0.02
6/1/2009    NA  1.37    0.01    0.46
7/1/2009    NA  1.22    0.01    0.61
8/1/2009    NA  1.32    0.01    0.04
9/1/2009    NA  0.83    0.01    0.03
10/1/2009   NA  0.95    0.02    0.61
11/1/2009   NA  0.28    0.03    0.53
12/1/2009   NA  0.17    0.01    0.32
1/1/2010    1.71    NA  0.03    0.53
2/1/2010    0.39    NA  0.03    0.16
3/1/2010    0.11    NA  0.01    0.58
4/1/2010    1.25    NA  0.01    0.41
5/1/2010    0.57    NA  0.01    0.9
6/1/2010    0.48    NA  0.01    0.58
7/1/2010    0.16    NA  0.01    0.03
8/1/2010    0.37    NA  0.01    0.23
9/1/2010    0.31    NA  0.01    0.77
10/1/2010   0.63    NA  0.01    0.75
11/1/2010   0.61    NA  0.01    0.74
12/1/2010   0.91    NA  0.01    0.41"
df <- read.table(text = Lines, header = TRUE)