Interpolate forcing adjustment factors — interp

This function interpolates 12 forcing adjustment factors (1 per month) by placing them at the 15th of the month then interpolating between the previous and next months values for every time step in between.

Usage

interp_fa(factors, month, day, hour)

Arguments

factors: 12 element vector of monthly adjustment factors
month: a vector of month values for each time step
day: a vector of day values for each time step
hour: a vector of hour values for each time step

Value

A vector matching the length of month, containing interpolated adjustment factors

Examples

d1 <- as.POSIXct("2001-01-01 00:00:00", tz = "UTC")
d2 <- as.POSIXct("2001-12-31 18:00:00", tz = "UTC")
dates <- seq.POSIXt(d1, d2, by = "6 hours")

month <- as.integer(format(dates, "%m"))
day <- as.integer(format(dates, "%d"))
hour <- as.integer(format(dates, "%H"))
factors <- c(.5, 2, 1, 1.5, 2, .5, 1, 2.5, 3, -1.5, 0, 1)

ifa <- interp_fa(factors, month, day, hour)

plot(dates, ifa, t = "l")
points(as.POSIXct(paste0("2001-", 1:12, "-15"), tz = "UTC"), factors, col = "red")