R/conv_lengths.R

In durtal/RcappeR: Tools to aid the analysis and handicapping of Thoroughbred Horse Racing

#' Convert lengths
#'
#' Convert lengths from character vector, containing values like "3/4" and "hd"
#' into a numeric vector.
#'
#' @details This function makes some assumptions, so may need tweaking over time.
#' For each string in a vector it is first split up by the spaces in it, so "1 3/4"
#' is split into "1" "3/4", if there are no spaces then it assesses the single
#' number/letters against a list of common length margins.  If the split vector
#' has length 2 then it assumes the first element is a whole number, and the
#' second is a fraction (like "3/4"), it calculates the total margin.  There is no
#' doubt for edge cases that this function doesn't cover which will cause errors,
#' the function was harder to write than initially thought, and will likely need
#' improvements as edge cases are encountered
#'
#' @param lengths character vector of lengths
#'
#' @export
#' @examples
#' \dontrun{
#' lengths <- c("0", "nse", "hd", "3/4", "1 1/2")
#' conv_len(lengths = lengths)
#' }
conv_len <- function(lengths) {

    lens <- sapply(lengths, function(x) {
        helper_len(x)
    })
    lens <- as.vector(lens)

    return(lens)
}

#' Helper function for conv_len
#'
#' @param length character string representing a margin in lengths
helper_len <- function(length) {

    lenlist <- list("0" = 0, "dh" = 0, "nse" = 0.02, "shd" = 0.05, "sh" = 0.05,
                    "hd" = 0.1, "nk" = 0.2, "1/4" = 0.25, "1/2" = 0.5, "3/4" = 0.75)

    if(nchar(length) == 0) {
        len <- 0
        return(len)
    }

    if(grepl(pattern = "[[:alpha:]]", length)) {
        len <- tolower(length)
        len <- lenlist[[len]]
        return(len)
    }

    if(grepl(pattern = "[[:digit:]]/[[:digit:]]", x = length)) {
        if(grepl(pattern = "\\s+|-", length)) {
            x <- unlist(strsplit(x = length, split = "\\s+|-"))
            frac <- x[2]
            frac <- lenlist[[frac]]

            len <- as.numeric(x[1]) + frac
            return(len)
        } else {
            len <- lenlist[[length]]
            return(len)
        }
    }

    if(grepl(pattern = "[[:digit:]]+(\\.[[:digit:]]+)?", x = length)) {
        len <- as.numeric(length)
        return(len)
    }

    len <- NA
    return(len)
}