data.table (version 1.18.2.1)

frolladapt: Adapt rolling window to irregularly spaced time series

Description

Helper function to generate adaptive window size based on the irregularly spaced time series index, to be passed as n argument to adaptive froll function (or N argument to adaptive frollapply). Experimental.

Usage

frolladapt(x, n, align="right", partial=FALSE, give.names=FALSE)

Value

When length(n)==1L then integer vector (adaptive window size) of length of x. Otherwise a list of length(n) having integer vectors (adaptive window sizes) of length of x for each window size provided in n.

Arguments

x

Integer. Must be sorted with no duplicates or missing values. Other objects with numeric storage (including most commonly Date and POSIXct) will be coerced to integer, which, for example, in case of POSIXct means truncating to whole seconds. It does not support vectorized input.

n

Integer, positive, rolling window size. Up to n values nearest to each value of x, with distance in the units of x and according to the window implied by align, are included in each rolling aggregation window. Thus when x is a POSIXct, n are seconds, and when x is a Date, n are days. It supports vectorized input, then it needs to be a vector.

align

Character, default "right". Other alignments have not yet been implemented.

partial

Logical, default FALSE. Should the rolling window size(s) provided in n be trimmed to available observations? For details see froll.

give.names

Logical, default FALSE. When TRUE, names are automatically generated corresponding to names of n. If answer is an integer vector, then the argument is ignored, see examples.

Details

Argument n allows multiple values to generate multiple adaptive windows, unlike x, as mixing different time series would make no sense.

See Also

froll, frollapply

Examples

idx = as.Date("2022-10-23") + c(0,1,4,5,6,7,9,10,14)
dt = data.table(index=idx, value=seq_along(idx))
dt
dt[, n3 := frolladapt(index, n=3L)]
dt
dt[, rollmean3 := frollmean(value, n3, adaptive=TRUE)]
dt
dt[, n3p := frolladapt(index, n=3L, partial=TRUE)]
dt[, rollmean3p := frollmean(value, n3p, adaptive=TRUE)]
dt

n34 = frolladapt(idx, c(small=3, big=4), give.names=TRUE)
n34
dt[, frollmean(value, n34, adaptive=TRUE, give.names=TRUE)]