1 files changed, 48 insertions, 48 deletions
diff --git a/R/ID.R b/R/ID.R
index cd991a4..0e3cc35 100644
--- a/R/ID.R
+++ b/R/ID.R
@@ -1,48 +1,48 @@
-#' 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 strongly 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. This 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 single value, the estimate of R0.
-#'
-#' @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
-#' ID(NT=NT, mu=5/7)
-#' ## obtain Rhat when serial distribution has mean of three days
-#' ID(NT=NT, mu=3/7)
-#'
-#' ## ========================================================= ##
-#' ## Compute Rhat using only the first five weeks of data      ##
-#' ## ========================================================= ##
-#'
-#' ID(NT=NT[1:5], mu=5/7) # serial distribution has mean of five days
-#'
-#' @export
-ID <- function(NT, mu) {
-    NT <- as.numeric(NT)
-    TT <- length(NT)
-    s <- (1:TT) / mu
-    y <- log(NT) / s
-
-    R0_ID <- exp(sum(y) / TT)
-
-    return(R0_ID)
-}
+#' 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 strongly 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. This 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 single value, the estimate of R0.
+#'
+#' @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
+#' ID(NT=NT, mu=5/7)
+#' ## obtain Rhat when serial distribution has mean of three days
+#' ID(NT=NT, mu=3/7)
+#'
+#' ## ========================================================= ##
+#' ## Compute Rhat using only the first five weeks of data      ##
+#' ## ========================================================= ##
+#'
+#' ID(NT=NT[1:5], mu=5/7) # serial distribution has mean of five days
+#'
+#' @export
+ID <- function(NT, mu) {
+    NT <- as.numeric(NT)
+    TT <- length(NT)
+    s <- (1:TT) / mu
+    y <- log(NT) / s
+
+    R0_ID <- exp(sum(y) / TT)
+
+    return(R0_ID)
+}