Using Rollapply on two columns - r

Question

I'm trying to do something similar I was asking for here and unfortunately I cannot work it out.
This is my data frame (data), a time series of prices:
Date Price Vol
1998-01-01 200 0.3
1998-01-02 400 0.4
1998-01-03 600 -0.2
1998-01-04 100 0.1
...
1998-01-20 100 0.1
1998-01-21 200 -0.4
1998-01-21 500 0.06
....
1998-02-01 100 0.2
1998-02-02 200 0.4
1998-02-03 500 0.3
1998-02-04 100 0.1
etc.
I would like to tell R, to
take the 1st value of "Vol" and divide it by the 20th value of "Price", then
take the 2st value of "Vol" and divide it by the 21th value of "Price", then.
take the 3st value of "Vol" and divide it by the 22th value of "Price", then
etc.
In my other post, I was able to use this function to calculate a return over a holding period of 20 days:
> data.xts <- xts(data[, -1], data[, 1])
> hold <- 20
> f <- function(x) log(tail(x, 1)) - log(head(x, 1))
> data.xts$returns.xts <- rollapply(data.xts$Price, FUN=f,
width=hold+1, align="left", na.pad=T)
Is there a way to do something very similar for the problem stated above? So something like
f1 <- function(x,y) head(x, 1) / tail(y,1)
where x is "Vol" and y is "Price" and then apply "rollapply"?
Thank you very much
UPDATE: # Dr G:
Thanks for your suggestions. With a slight change, it did what I wanted!
data.xts <- xts(data[, -1], data[, 1])
hold <- 20
data.xts$quo <- lag(data.xts[,2], hold) / data.xts[,1]
Now my problem is, that the resulting data frame looks like this:
Date Price Vol quo
1 1998-01-01 200 0.3 NA
2 1998-01-02 400 0.4 NA
3 1998-01-03 600 -0.2 NA
4 1998-01-04 100 0.1 NA
...
21 1998-01-20 180 0.2 0.003
I know that there must be NA's as an outcome, but only for the last 20 observations, not the first 20 ones. The formula stated above calculates the correct values, however puts them starting at the 21st row instead of the first row. Do you know how I could change that?

Answer 1

Use by.column = FALSE in rollapply. In order to use the posted data we will divide the volume in the first row by the price in the 3rd row and so on for purposes of reproducible illustration:
library(zoo)
Lines <- "Date Price Vol
1998-01-01 200 0.3
1998-01-02 400 0.4
1998-01-03 600 -0.2
1998-01-04 100 0.1
1998-01-20 100 0.1
1998-01-21 200 -0.4
1998-01-21 500 0.06
1998-02-01 100 0.2
1998-02-02 200 0.4
1998-02-03 500 0.3
1998-02-04 100 0.1"
# read in and use aggregate to remove all but last point in each day.
# In reality we would replace textConnection(Lines) with something
# like "myfile.dat"
z <- read.zoo(textConnection(Lines), header = TRUE,
aggregate = function(x) tail(x, 1))
# divide Volume by the Price of the point 2 rows ahead using by.column = FALSE
# Note use of align = "left" to align with the volume.
# If we used align = "right" it would align with the price.
rollapply(z, 3, function(x) x[1, "Vol"] / x[3, "Price"], by.column = FALSE,
align = "left")
# and this is the same as rollapply with align = "left" as above
z$Vol / lag(z$Price, 2)
# this is the same as using rollapply with align = "right"
lag(z$Vol, -2) / z$Price
By the way, note that zoo uses the same convention for the sign of lag as does R but xts uses the opposite convention so if you convert the above to xts you will have to negate the lags.

Answer 2

It's actually easier than that. Just do this:
data.xts <- xts(data[, -1], data[, 1])
hold <- 20
returns.xts = data.xts[,2] / lag(data.xts[,1], hold)
Actually for this using zoo instead of xts would work as well:
data.zoo<- zoo(data[, -1], data[, 1])
hold <- 20
returns.zoo = data.zoo[,2] / lag(data.zoo[,1], -hold)
Only thing that changes is the sign of the lags (zoo convention is different than xts)

Answer 3

You just need to use
data.xts$quo <- data.xts[,2] / lag( data.xts[,1], -hold)