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1 #' ID method
2 #'
3 #' This function implements a least squares estimation method of R0 due to Fisman et al. (PloS One, 2013).
4 #' See details for implementation notes.
5 #'
6 #' The method is based on a straightforward incidence decay model. The estimate of R0 is the value which
7 #' minimizes the sum of squares between observed case counts and cases counts 'expected' under the model.
8 #'
9 #' This method is based on an approximation of the SIR model, which is most valid at the beginning of an epidemic.
10 #' The method assumes that the mean of the serial distribution (sometimes called the serial interval) is known.
11 #' The final estimate can be quite sensitive to this value, so sensitivity testing is strongly recommended.
12 #' Users should be careful about units of time (e.g. are counts observed daily or weekly?) when implementing.
13 #'
14 #' @param NT Vector of case counts
15 #' @param mu Mean of the serial distribution (needs to match case counts in time units; for example, if case counts are
16 #' weekly and the serial distribution has a mean of seven days, then \code{mu} should be set to one, if case
17 #' counts are daily and the serial distribution has a mean of seven days, then \code{mu} should be set to seven)
18 #'
19 #' @return \code{ID} returns a list containing the following components: \code{Rhat} is the estimate of R0 and
20 #' \code{inputs} is a list of the original input variables \code{NT, mu}.
21 #'
22 #' @examples
23 #'
24 #' ## ===================================================== ##
25 #' ## Illustrate on weekly data ##
26 #' ## ===================================================== ##
27 #'
28 #' NT <- c(1, 4, 10, 5, 3, 4, 19, 3, 3, 14, 4)
29 #' ## obtain Rhat when serial distribution has mean of five days
30 #' res1 <- ID(NT=NT, mu=5/7)
31 #' res1$Rhat
32 #' ## obtain Rhat when serial distribution has mean of three days
33 #' res2 <- ID(NT=NT, mu=3/7)
34 #' res2$Rhat
35 #'
36 #' ## ========================================================= ##
37 #' ## Compute Rhat using only the first five weeks of data ##
38 #' ## ========================================================= ##
39 #'
40 #'
41 #' res3 <- ID(NT=NT[1:5], mu=5/7) # serial distribution has mean of five days
42 #' res3$Rhat
43 #'
44 #' @export
45 ID <- function(NT, mu) {
46 NT <- as.numeric(NT)
47 TT <- length(NT)
48 s <- (1:TT) / mu
49 y <- log(NT) / s
50
51 R0_ID <- exp(sum(y) / TT)
52
53 return(list=c(Rhat=R0_ID, inputs=list(NT=NT, mu=mu)))
54 }