Set seed and variables. Please assume all everything is this section is given and unalterable.
library(zoo)
set.seed(123)
a <- zoo(rnorm(10), order.by = as.Date(50:60))
b <- zoo(rnorm(10), order.by = as.Date(50:60))
c <- zoo(rnorm(10), order.by = as.Date(50:60))
lags <- c(1,3,1)
variables <- c("a","c","b")
I want to create an adequate zoo object that picks variables from the list "variables" in that order and applies lags from "lags". This is my desired output (including column names):
a.l1 c.l3 b.l1
20/02/1970 NA NA NA
21/02/1970 -0.56047565 NA 1.2240818
22/02/1970 -0.23017749 NA 0.3598138
23/02/1970 1.55870831 -1.0678237 0.4007715
24/02/1970 0.07050839 -0.2179749 0.1106827
25/02/1970 0.12928774 -1.0260044 -0.5558411
26/02/1970 1.71506499 -0.7288912 1.7869131
27/02/1970 0.46091621 -0.6250393 0.4978505
28/02/1970 -1.26506123 -1.6866933 -1.9666172
01/03/1970 -0.68685285 0.837787 0.7013559
02/03/1970 -0.44566197 0.1533731 -0.4727914
This is one the closest I could get to, but it doesn't work. The problem is somewhere in the "get" function I think.
lag(as.zoo(mget(variables)),lags-1)
Many thanks
mget(variables) is actually returning a list with one element per variable in variables, containing the vector of values in that variable.
You can get it into a structure usable by lag() by binding the elements of the list into columns using do.call("cbind", mget(variables)). As far as I know, it's actually unnecessary to wrap this in as.zoo().
To get the proper lags, you need -lags rather than lags-1.
Putting this together, you get:
lagged <- lag(do.call("cbind", mget(variables)), -lags)
That contains the 1 and 3 lags for each variable, so you'll have to do a bit of post-processing to get the format you want. The following should do it:
lagged <- lagged[, c("a.lag-1", "c.lag-3", "b.lag-1")]
colnames(lagged) <- c("a.l1", "c.l3", "b.l1")
Note though that since at 1970-02-20 all lags are NA, this row is excluded from the output.