diff options
Diffstat (limited to 'R')
| -rw-r--r-- | R/ID.R | 86 |
1 files changed, 43 insertions, 43 deletions
@@ -1,49 +1,49 @@ -#' ID method -#' -#' This function implements a least squares estimation method of R0 due to Fisman et al. (PloS One, 2013). See details for implementation notes. -#' -#'The method is based on a straightforward incidence decay model. The estimate of R0 is the value which minimizes the sum of squares between observed case counts and cases counts 'expected' under the model. -#' -#' This method is based on an approximation of the SIR model, which is most valid at the beginning of an epidemic. The method assumes that the mean of the serial distribution (sometimes called the serial interval) is known. The final estimate can be quite sensitive to this value, so sensitivity testing is recommended. Users should be careful about units of time (e.g. are counts observed daily or weekly?) when implementing. -#' -#' @param NT Vector of case counts -#' @param mu Mean of the serial distribution (needs to match case counts in time units; for example, if case counts are weekly and the serial distribution has a mean of seven days, then \code{mu} should be set to one) -#' -#' @return \code{ID} returns a list containing the following components: \code{Rhat} is the estimate of R0 and \code{inputs} is a list of the original input variables \code{NT, mu}. -#' -#' @examples -#' -#' ## ===================================================== ## -#' ## Illustrate on weekly data ## -#' ## ===================================================== ## -#' -#' NT <- c(1, 4, 10, 5, 3, 4, 19, 3, 3, 14, 4) -#' ## obtain Rhat when serial distribution has mean of five days -#' res1 <- ID(NT=NT, mu=5/7) -#' res1$Rhat -#' ## obtain Rhat when serial distribution has mean of three days -#' res2 <- ID(NT=NT, mu=3/7) -#' res2$Rhat -#' -#' ## ========================================================= ## -#' ## Compute Rhat using only the first five weeks of data ## -#' ## ========================================================= ## -#' -#' -#' res3 <- ID(NT=NT[1:5], mu=5/7) # serial distribution has mean of five days -#' res3$Rhat -#' @export -#' +#' ID method
+#'
+#' This function implements a least squares estimation method of R0 due to Fisman et al. (PloS One, 2013). See details for implementation notes.
+#'
+#'The method is based on a straightforward incidence decay model. The estimate of R0 is the value which minimizes the sum of squares between observed case counts and cases counts 'expected' under the model.
+#'
+#' This method is based on an approximation of the SIR model, which is most valid at the beginning of an epidemic. The method assumes that the mean of the serial distribution (sometimes called the serial interval) is known. The final estimate can be quite sensitive to this value, so sensitivity testing is recommended. Users should be careful about units of time (e.g. are counts observed daily or weekly?) when implementing.
+#'
+#' @param NT Vector of case counts
+#' @param mu Mean of the serial distribution (needs to match case counts in time units; for example, if case counts are weekly and the serial distribution has a mean of seven days, then \code{mu} should be set to one, if case counts are daily and the serial distribution has a mean of seven days, then \code{mu} should be set to seven)
+#'
+#' @return \code{ID} returns a list containing the following components: \code{Rhat} is the estimate of R0 and \code{inputs} is a list of the original input variables \code{NT, mu}.
+#'
+#' @examples
+#'
+#' ## ===================================================== ##
+#' ## Illustrate on weekly data ##
+#' ## ===================================================== ##
+#'
+#' NT <- c(1, 4, 10, 5, 3, 4, 19, 3, 3, 14, 4)
+#' ## obtain Rhat when serial distribution has mean of five days
+#' res1 <- ID(NT=NT, mu=5/7)
+#' res1$Rhat
+#' ## obtain Rhat when serial distribution has mean of three days
+#' res2 <- ID(NT=NT, mu=3/7)
+#' res2$Rhat
+#'
+#' ## ========================================================= ##
+#' ## Compute Rhat using only the first five weeks of data ##
+#' ## ========================================================= ##
+#'
+#'
+#' res3 <- ID(NT=NT[1:5], mu=5/7) # serial distribution has mean of five days
+#' res3$Rhat
+#' @export
+#'
ID <- function(NT, mu){
- - NT <- as.numeric(NT) - TT <- length(NT) - s <- (1:TT)/mu - y <- log(NT)/s - +
+ NT <- as.numeric(NT)
+ TT <- length(NT)
+ s <- (1:TT)/mu
+ y <- log(NT)/s
+
R0_ID <- exp(sum(y)/TT)
- +
return(list=c(Rhat=R0_ID, inputs=list(NT=NT, mu=mu)))
}
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