R/age_length.R

In jasondubois/spopmodel: Methods for and Application of Fish Population Dynamics Model

# ******************************************************************************
# Created: 02-Aug-2018
# Author:  J. DuBois
# Contact: jason.dubois@wildlife.ca.gov
# Purpose: This file contains functions or methods to create age-length key or
#          to get age frequency distribution.
# Source:
# ******************************************************************************

#' Make an age-length key with lengths as rows and ages as columns.
#'
#' @description \code{MakeALKey} creates an age-length key according
#'    to length frequency per age. Resulting frequency converted to
#'    proportion for application to entire length dataset.
#'
#' @param data A dataframe with length and age fields (columns).
#' @param len Field name in data containing numeric lengths.
#' @param age Field name in data containing nummeric ages.
#' @param lenBreaks How lengths are binned. See \code{\link{cut}}
#'    for more information. (Supplied to \code{breaks} in \code{cut}.)
#' @param breakLabs NULL (default), currently not used.
#'
#' @return Age-length key (length[rows] x age[columns]) with proportion
#'    or 0 in each cell. Return object is of class "table" with
#' \code{dimnames(<ALK>)} providing \code{Bin} and \code{Age} list items.
#' @export
#'
#' @note In the example below, some length bins contain no aged sample.
#'    One can adjust the "by" argument in \code{seq} as a possible fix.
#'    Or one needs to age fish within those length bins.
#'
#' @examples
#' # check range(trammel_catch[["FL"]])
#' len_breaks <- seq(from = 50, to = 220, by = 5)
#' MakeALKey(data = trammel_catch, len = FL, age = Age, lenBreaks = len_breaks)
MakeALKey <- function(data, len, age, lenBreaks, breakLabs = NULL) {

  # get length and age columns
  l <- as.character(substitute(len))
  a <- as.character(substitute(age))

  b <- CreateLenBins(len = data[[l]], lenBreaks = lenBreaks, numericBin = TRUE)

  alk <- prop.table(table(b, data[[a]], dnn = c("Bin", "Age")), margin = 1)

  alk
}
# end MakeALKey

#' Assign age to each fish based on extant age-length data.
#'
#' @description \code{AgeEach} inspired by \code{FSA::alkIndivAge}. Here
#' \code{AgeEach} output is more tailored for use within this package.
#' \code{AgeEach} handles age assignment at the length bin level, so
#'    instance of an age-length key is not required.
#'
#' @param data A dataframe with length and age fields (columns).
#' @param len Field name in data containing numeric lengths.
#' @param age Field name in data containing nummeric ages.
#' @param lenBreaks lenBreaks How lengths are binned. See \code{\link{cut}}
#'    for more information. (Supplied to \code{breaks} in \code{cut}.)
#' @param sex (default NULL) Currently not used.
#'
#' @note Print method for convenience of output.
#'
#' @return A list with age & bin diagnostics along with
#'    a numeric vector of ages, with length = \code{nrow(data)}.
#' @export
#'
#' @examples
#' # check range(trammel_catch[["FL"]])
#' len_breaks <- seq(from = 50, to = 220, by = 5)
#' AgeEach(data = trammel_catch, len = FL, age = Age, lenBreaks = len_breaks)
AgeEach <- function(data, len, age, lenBreaks, sex = NULL) {

  # get length and age columns
  l <- as.character(substitute(len))
  a <- as.character(substitute(age))

  # create length bins
  b <- CreateLenBins(len = data[[l]], lenBreaks = lenBreaks)

  # to check ages in each length bin
  age_split <- split(data[[a]], f = b, drop = FALSE)

  bin_check <- vapply(age_split, FUN = CheckBin, FUN.VALUE = character(1L))

  # only report values NOT NA
  bin_check <- bin_check[!is.na(bin_check)]

  # order does not reconcile with order in lengths in data need to figure out
  # order [order(b)] below does not work

  # unlist(lapply(age_split, FUN = CheckAge), use.names = FALSE)[order(b)]
  a <- unsplit(value = lapply(age_split, FUN = CheckAge), f = b, drop = FALSE)

  out <- list(
    BinDiagnostics = list(
      BinCheck = bin_check,
      CountNA = sum(is.na(a)),
      CountTotal = length(a)
    ),
    Ages = a
  )

  class(out) <- "AgeEach"

  out
}
# end AgeEach

#' @describeIn AgeEach \code{print} method (See notes).
#' @param x Object of class \code{AgeEach}.
#' @param ... Currently not used.
#' @export
print.AgeEach <- function(x, ...) {

  # get diagnostics
  bd <- x[["BinDiagnostics"]]

  out <- sprintf(
    fmt = "Number aged = %s, with %s NA.", # \n Bin Check:\n %s",
    bd[["CountTotal"]] - bd[["CountNA"]],
    bd[["CountNA"]] #,
    # bd[["BinCheck"]]
  )

  # print(out)
  cat(out, "\n", "\n", "Bins below empty or all NA:\n", sep = "")
  print(noquote(bd[["BinCheck"]]))
}
# end print.AgeEach

#' @keywords internal
#' @rdname spopmodel-internals
CheckAge <- function(age) {

  # TODO: test this function
  # TODO: finalize what to do with length(age) == 0
  # TODO: how to treat bins with no ages ??
  # TODO: nail down whole & sum_rem (still issues with double or integer values)
  # TODO: nail down replace = T or replace = F
  # TODO: age some input verification
  # TODO: set ouput to integer ??
  # TODO: add set seed component here or in AgeEach

  # age vector result of splitting age field on length bins so length(age) = 0
  # means no lengths (& thus ages) are in that specific bin return numeric(0)

  # various scenarios of incoming age vector -----------------------------------
  # age ==> (1) numeric(0) OR (2) length = 1 OR (3) length > 1
  #
  #         (4) all numbers (i.e., no NA values) OR ...
  #         (5) all NA OR ...
  #         (6) mixture of numbers & NA values
  #
  # if (1) return numeric(0) [length bin is empty]
  # if (2) if !NA see (4) else see (5)
  # if (3) see (4) OR (5) OR (6)
  #
  # if (4) use assigned age irrespective of length(age)
  # if (5) cannot do much; advise re-binning or if possible ageing
  #        some fish in bin [x, y)
  # if (6) length(age) must be > 1 so go through process using
  #        known ages (& proportions) to "age the NAs"
  #
  # processing scenario (6) **********************
  #
  # freq <- frequency (as proportion) of age variable
  # get column names of `freq` (these are the ages as character vector)
  #
  # (a) if length(freq) = 1 then set all NA
  #     values to column names of `freq` as there is only one age
  #     value for this length bin
  #
  # (b) if number NA values = 1 then
  #     a <- sample(column names of `freq`, size = 1, prop = `freq`)
  #     age[which is na] <- as.numeric(a)
  #     return age
  #
  # (c) set `freq` * sum(all NA values of age)
  #     perform integer division ...
  #     whole <- `freq` * sum(all NA values of age) %/% 1
  #
  #     get remainder using modulus
  #     remainder <- `freq` * sum(all NA values of age) %% 1
  #
  #     if all of whole = 0 then
  #       a <- sample(column names of `freq`,
  #                   size = sum(remainder),
  #                   prop = `freq`)
  #       age[which is na] <- as.numeric(a)
  #
  #       the reality with this scenario is that sum(remainder) should
  #       really be 1 if all of whole = 0 (I believe)
  #
  #     else
  #       create a character vector of ages using names of `whole` &
  #       values of `whole`
  #       new_ages <- rep(dimnames(whole)[[1]], times = whole)
  #
  #       then sample from these new ages
  #       new_ages <- sample(new_ages, size = length(new_ages), replace = T)
  #
  #       recalculate number remaining
  #       remainder <- remainder - length(new_ages)
  #
  #       sample again & combine with `new_ages` vector
  #       new_ages <- c(new_ages, sample(new_ages, size = sum_rem,
  #                                      replace = TRUE))
  #       age[which is NA] <- as.numeric(new_ages)
  #       return age
  #
  # code here ------------------------------------------------------------------

  # age is essentially empty
  if (length(age) == 0) return(numeric(length = 0L))

  # check for NAs within age vector
  is_na <- is.na(age)

  # check point: irrespective of length(age), we note if all values are aged OR
  # all values are NA; essentially not much we need to or can do here except
  # return age ---  we may issue a warning about return NA values, but we can
  # handle this in AgeEach()
  if (all(is_na) || all(!is_na)) return(age)

  # check point: we know at this point length(age) must be > 1 AND age is a
  # mixture of NA & numeric values

  # get proportion per age ignoring (for now) NA values
  freq <- prop.table(table(age, dnn = NULL))

  # get names of `freq` --- essentially age values as character
  char_ages <- dimnames(freq)[[1]]

  # if there is only one unique age in `age` then we have no other choice but to
  # assign all NA values to this age (i.e., no need for random sampling)
  if (length(char_ages) == 1) {
    age[is_na] <- as.numeric(char_ages)
    return(age)
  }

  # check point: we know at this stage we have > 1 unique age, now we need to
  # check the number of NA values; here we check if `age` contains 1 NA value
  num_not_aged <- sum(is_na)

  if (num_not_aged == 1) {
    a <- sample(char_ages, size = 1, prob = freq)
    age[is_na] <- as.numeric(a)
    return(age)
  }

  # check point: at this point we have > 1 unique age & > 1 NA value

  # essentially like using an al-key multiplying the count by proportions
  age_assign <- freq * num_not_aged

  # integer division to get complete age assignments
  whole <- age_assign %/% 1
  # modulo to get remainder (to be handled later)
  sum_rem <- sum(age_assign %% 1)
  # sum_rem <- ceiling(sum_rem)

  # the logic is that if all whole = 0 then sum_rem should equal the number not
  # aged thus we can just use num_not_aged as our size
  if (all(whole == 0)) {
    a <- sample(char_ages, size = num_not_aged, replace = TRUE, prob = freq)
    age[is_na] <- as.numeric(a)
    # print("all whole = 0")
    return(age)
  }

  # a needs "re"-sampling here before proceeding in case sum_rem != 0
  a <- rep(char_ages, times = whole)
  a <- sample(a, size = length(a), replace = TRUE)

  # in this case length(a) should equal number not aged
  if (sum_rem == 0) {
    # a <- sample(a, size = length(a), replace = TRUE)
    age[is_na] <- as.numeric(a)
    # print("sum rem = 0")
    return(age)
  }

  # NEED TO TEST LOGIC BELOW
  # check point:  here all(whole == 0) is FALSE AND sum_rem > 0

  # not sure if recalculation is necessary
  # recalculate number remaining
  # sum_rem <- length(a) - sum_rem     # WRONG

  # if (sum_rem > length(a)) {
  #   sum_rem <- sum_rem - length(a)   # WRONG
  # }

  # sum_rem <- max(sum_rem, length(a)) # WRONG

  # this *should be* the proper recalculation for sum_rem
  sum_rem <- num_not_aged - length(a)  # CORRECT

  # print calls for debugging (uncomment as needed)

  # print(a)
  # print(sum_rem)

  # combine 'a' with samples on the remainder
  a <- c(a, sample(char_ages, size = sum_rem, replace = TRUE, prob = freq))

  age[is_na] <- as.numeric(a)

  # print(a)
  # print(age_assign %% 1)
  # print("made it to the end")

  age
}
# end CheckAge