ctbi.long.term: ctbi.long.term

In fritte2/ctbi: A Procedure to Clean, Decompose and Aggregate Timeseries

ctbi.long.term

Description

Calculate the long-term trend with a linear interpolation between the mean (or median) of bins defined between two consecutive centers. Bins defined between two consecutive sides are calculated as well to complete for missing values if they have neighbors. Bins without sufficient data are discarded.

Usage

ctbi.long.term(data0, n.bin.min, seq.bin.side, outliers.checked)

Arguments

data0	data.table with the columns x (time series), y (values), side.index (index associated with each bin defined between two consecutive centers) and index.bin (index associated with each bin defined between two consecutive sides)
n.bin.min	minimum number of points for a bin to be accepted
seq.bin.side	sequence of the sides of the bins
outliers.checked	boolean to indicate if the median (outliers.checked = FALSE) or the mean (outliers.checked = TRUE) should be used to calculate the long-term trend

Value

data0 with the long-term trend added (column long.term)

Examples

library(data.table)
x <- seq(from=as.Date('2001-01-01'),to=as.Date('2010-12-01'),by='1 month')
y <- 3*cos(2*pi*(0:(length(x)-1))/12)+runif(length(x))
outliers.checked <- TRUE
seq.bin.side <- seq(from=as.Date('2001-01-01'),to=as.Date('2011-01-01'),by='1 year')
seq.bin.center <- seq(from=as.Date('2001-06-01'),to=as.Date('2010-06-01'),by='1 year')
index.bin <- findInterval(x,seq.bin.side)
side.index <- findInterval(x,seq.bin.center)+0.5
n.bin.min <- 10 # minimum of 10 months of data for a bin to be accepted
data0 <- data.table(x=x,y=y,index.bin=index.bin,side.index=side.index)
data0.with.long.term <- ctbi.long.term(data0,n.bin.min,seq.bin.side,outliers.checked)

fritte2/ctbi documentation built on Feb. 7, 2023, 5:43 a.m.