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R/benefitcost.R
In iemisc: Irucka Embry's Miscellaneous Functions
#' Benefit-Cost Ratio (Engineering Economics)
#'
#' Compute the benefit-cost ratio between two alternatives
#'
#' Benefit is expressed as
#'
#' \deqn{Benefit = AB\left[\frac{\left(1 + i\right)^n - 1}{i\left(1 + i\right)^n}\right]}
#'
#' \describe{
#' \item{\emph{Benefit}}{the present equivalent benefit}
#' \item{\emph{AB}}{the annual benefit}
#' \item{\emph{i}}{the "effective interest rate" per year}
#' \item{\emph{n}}{the number of years}
#' }
#'
#'
#'
#' Cost is expressed as
#'
#' \deqn{Cost = PC + OM\left[\frac{\left(1 + i\right)^n - 1}{i\left(1 + i\right)^n}\right] - S\left[\frac{1}{\left(1 + i\right)^n}\right]}
#'
#' \describe{
#' \item{\emph{Cost}}{the present equivalent cost}
#' \item{\emph{PC}}{the present or initial cost}
#' \item{\emph{OM}}{the annual operations & maintenance cost}
#' \item{\emph{S}}{the salvage value}
#' \item{\emph{i}}{the "effective interest rate" per year}
#' \item{\emph{n}}{the number of years}
#' }
#'
#'
#'
#' Benefit-Cost ratio is expressed as
#'
#' \deqn{BC = \frac{B_2 - B_1}{C_2 - C_1} \geq 1}
#'
#' \describe{
#' \item{\emph{BC}}{the present equivalent cost}
#' \item{\emph{\eqn{B_1}}}{the benefit for alternative 1}
#' \item{\emph{\eqn{B_2}}}{the benefit for alternative 2}
#' \item{\emph{\eqn{C_1}}}{the cost for alternative 1}
#' \item{\emph{\eqn{C_2}}}{the cost for alternative 2}
#' }
#'
#'
#'
#' @param ic1 numeric vector that contains the initial cost for option 1
#' @param n1 numeric vector that contains the useful life (years) for option 1
#' @param ac1 numeric vector that contains the annual cost [operations &
#' maintenance (O&M)] for option 1
#' @param i1 numeric vector that contains the effective interest rate per
#' period as a percent for option 1
#' @param ab1 numeric vector that contains the annual benefits for option 1
#' @param salvage1 numeric vector that contains the salvage value for option 1
#' @param option1 character vector that contains the option name for
#' option 1
#' @param ic2 numeric vector that contains the initial cost for option 2
#' @param n2 numeric vector that contains the useful life (years) for option 2
#' @param ac2 numeric vector that contains the annual cost [operations &
#' maintenance (O&M)] for option 2
#' @param i2 numeric vector that contains the effective interest rate per
#' period as a percent for option 2
#' @param ab2 numeric vector that contains the annual benefits for option 2
#' @param salvage2 numeric vector that contains the salvage value for option 2
#' @param option2 character vector that contains the option name for
#' option 2
#' @param table character vector that contains the table output format
#' (ptable, rtable, or both)
#'
#'
#' @return \code{\link[data.table]{data.table}} with character vectors with the monetary
#' values having thousands separator in a pretty table (ptable) & message with
#' the best option, data.frame with numeric vectors without the thousands
#' separator in regular table (rtable) & a message with the best option, or both
#' options combined in a list
#'
#'
#'
#'
#' @references
#' \enumerate{
#' \item Michael R. Lindeburg, PE, \emph{EIT Review Manual}, Belmont, California: Professional Publications, Inc., 1996, page 14-2, 14-4.
#' \item William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling, \emph{Engineering Economy}, Fourteenth Edition, Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2009, page 133, 142, 442-443, 452-453.
#' }
#'
#'
#'
#'
#' @author Irucka Embry
#'
#'
#'
#' @encoding UTF-8
#'
#'
#'
#'
#'
#'
#'
#' @examples
#'
#' library(iemisc)
#'
#' # Example from Lindeburg Reference text (page 14-4)
#' benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000, i1 = 10,
#' salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000, i2 = 10,
#' salvage2 = 10000, option1 = "A", option2 = "B", table = "rtable")
#'
#'
#' # This is useful for saving the results as the named data.frame rtable
#' rtable <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
#' i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
#' i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "rtable")
#'
#' rtable
#'
#'
#'
#' # This is useful for saving the results as the named data.frame ptable
#' ptable <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
#' i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
#' i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "ptable")
#'
#' ptable
#'
#'
#'
#' # This is useful for saving the results as the named list of 2 data.frames
#' # called both
#' both <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
#' i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
#' i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "both")
#'
#' both
#'
#'
#'
#' # Example 10-8 from the Sullivan Reference text (page 452-453)
#' project <- benefitcost(ic1 = 750000, n1 = 35, ac1 = 120000, ab1 = 245000,
#' i1 = 9, salvage1 = 0, ic2 = 625000, n2 = 25, ac2 = 110000, ab2 = 230000,
#' i2 = 9, salvage2 = 0, option1 = "Project I", option2 = "Project II",
#' table = "rtable")
#'
#' project
#'
#'
#'
#'
#' @importFrom data.table data.table setnames setattr
#' @importFrom assertthat assert_that
#' @importFrom checkmate qtest
#' @importFrom round round_r3
#'
#' @export
benefitcost <- function (ic1, n1, ac1, ab1, i1, salvage1, ic2, n2, ac2, ab2, i2, salvage2, option1, option2, table = c("ptable", "rtable", "both")) {
table <- table
checks <- c(ic1, n1, ac1, ab1, i1, salvage1, ic2, n2, ac2, ab2, i2, salvage2)
# Check
assert_that(!any(qtest(checks, "N+(,)") == FALSE), msg = "Either ic1, n1, ac1, ab1, i1, salvage1, ic2, n2, ac2, ab2, i2, salvage2 is NA, NaN, Inf, -Inf, empty, or a string. Please try again.")
# only process with finite values and provide an error message if the check fails
assert_that(!any(qtest(option1, "S==1") == FALSE), msg = "option1 is not a character vector of length 1. Please try again.")
# only process with a single character vector and provide an error message if the check fails
assert_that(!any(qtest(option2, "S==1") == FALSE), msg = "option2 is not a character vector of length 1. Please try again.")
# only process with a single character vector and provide an error message if the check fails
assert_that(qtest(table, "S==1"), msg = "There is not a table option or more than 1 table option. Please specify either 'ptable', 'rtable', or 'both'.")
# only process with the correct length of table variable and provide an error message if the check fails
assert_that(isTRUE(any(c("ptable", "rtable", "both") %in% table)), msg = "The table option has not been identified correctly as either ptable, rtable, or both. Please try again.")
# only process with a specified frequency and provide a stop warning if not
# option 1
# cost 1
i1 <- i1 / 100
cost1 <- ic1 + (ac1 * (((1 + i1) ^ n1 - 1) / (i1 * ((1 + i1) ^ n1)))) - (salvage1 * ((1 / (1 + i1)) ^ n1))
# benefit 1
benefit1 <- ab1 * (((1 + i1) ^ n1 - 1) / (i1 * ((1 + i1) ^ n1)))
b_c1 <- benefit1 / cost1
# option 2
# cost 2
i2 <- i2 / 100
cost2 <- ic2 + (ac2 * (((1 + i2) ^ n2 - 1) / (i2 * ((1 + i2) ^ n2)))) - (salvage2 * ((1 / (1 + i2)) ^ n2))
# benefit 2
benefit2 <- ab2 * (((1 + i2) ^ n2 - 1) / (i2 * ((1 + i2) ^ n2)))
b_c2 <- benefit2 / cost2
# Benefit-Cost ratio of Option 2 to Option 1
b_c_rank <- (benefit2 - benefit1) / (cost2 - cost1)
b_c_rank_choose <- ifelse(b_c_rank >= 1, paste0("choose", " ", option2, "."), paste0("choose", " ", option1, "."))
if (table == "ptable") {
ptable <- data.table(c(NA_character_, "Benefit", "Cost", "Benefit-Cost Ratio"), c(option1, formatC(benefit1, big.mark = ",", format = "f", digits = 2), formatC(cost1, big.mark = ",", format = "f", digits = 2), round_r3(b_c1, d = 2)), c(option2, formatC(benefit2, big.mark = ",", format = "f", digits = 2), formatC(cost2, big.mark = ",", format = "f", digits = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- as.character(ptable[1, ])
ptable <- ptable[-1, ]
cat("\n", paste("The Benefit-Cost ratio of", option2, "to", option1, "is", round_r3(b_c_rank, d = 2), "thus", b_c_rank_choose), "\n\n")
# code block below modified from data.table function
setattr(ptable, "col.names", setnames(ptable, col.names))
setattr(ptable, "class", c("data.table", "data.frame"))
ptable
} else if (table == "rtable") {
rtable <- data.table(c("Benefit", "Cost", "Benefit-Cost Ratio"), c(round_r3(benefit1, d = 2), round_r3(cost1, d = 2), round_r3(b_c1, d = 2)), c(round_r3(benefit2, d = 2), round_r3(cost2, d = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- c(NA_character_, option1, option2)
cat("\n", paste("The Benefit-Cost ratio of", option2, "to", option1, "is", round(b_c_rank, digits = 2), "thus", b_c_rank_choose), "\n\n")
# code block below modified from data.table function
setattr(rtable, "col.names", setnames(rtable, col.names))
setattr(rtable, "class", c("data.table", "data.frame"))
rtable
} else if (table == "both") {
# ptable
ptable <- data.table(c(NA_character_, "Benefit", "Cost", "Benefit-Cost Ratio"), c(option1, formatC(benefit1, big.mark = ",", format = "f", digits = 2), formatC(cost1, big.mark = ",", format = "f", digits = 2), round_r3(b_c1, d = 2)), c(option2, formatC(benefit2, big.mark = ",", format = "f", digits = 2), formatC(cost2, big.mark = ",", format = "f", digits = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- as.character(ptable[1, ])
ptable <- ptable[-1, ]
cat("\n", paste("The Benefit-Cost ratio of", option2, "to", option1, "is", round(b_c_rank, digits = 2), "thus", b_c_rank_choose), "\n\n")
# code block below modified from data.table function
setattr(ptable, "col.names", setnames(ptable, col.names))
setattr(ptable, "class", c("data.table", "data.frame"))
ptable
# rtable
rtable <- data.table(c("Benefit", "Cost", "Benefit-Cost Ratio"), c(round(benefit1, digits = 2), round_r3(cost1, d = 2), round_r3(b_c1, d = 2)), c(round_r3(benefit2, d = 2), round_r3(cost2, d = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- c(NA_character_, option1, option2)
# code block below modified from data.table function
setattr(rtable, "col.names", setnames(rtable, col.names))
setattr(rtable, "class", c("data.table", "data.frame"))
rtable
return(list(rtable, ptable))
}
}
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#' Benefit-Cost Ratio (Engineering Economics)
#'
#' Compute the benefit-cost ratio between two alternatives
#'
#' Benefit is expressed as
#'
#' \deqn{Benefit = AB\left[\frac{\left(1 + i\right)^n - 1}{i\left(1 + i\right)^n}\right]}
#'
#' \describe{
#' \item{\emph{Benefit}}{the present equivalent benefit}
#' \item{\emph{AB}}{the annual benefit}
#' \item{\emph{i}}{the "effective interest rate" per year}
#' \item{\emph{n}}{the number of years}
#' }
#'
#'
#'
#' Cost is expressed as
#'
#' \deqn{Cost = PC + OM\left[\frac{\left(1 + i\right)^n - 1}{i\left(1 + i\right)^n}\right] - S\left[\frac{1}{\left(1 + i\right)^n}\right]}
#'
#' \describe{
#' \item{\emph{Cost}}{the present equivalent cost}
#' \item{\emph{PC}}{the present or initial cost}
#' \item{\emph{OM}}{the annual operations & maintenance cost}
#' \item{\emph{S}}{the salvage value}
#' \item{\emph{i}}{the "effective interest rate" per year}
#' \item{\emph{n}}{the number of years}
#' }
#'
#'
#'
#' Benefit-Cost ratio is expressed as
#'
#' \deqn{BC = \frac{B_2 - B_1}{C_2 - C_1} \geq 1}
#'
#' \describe{
#' \item{\emph{BC}}{the present equivalent cost}
#' \item{\emph{\eqn{B_1}}}{the benefit for alternative 1}
#' \item{\emph{\eqn{B_2}}}{the benefit for alternative 2}
#' \item{\emph{\eqn{C_1}}}{the cost for alternative 1}
#' \item{\emph{\eqn{C_2}}}{the cost for alternative 2}
#' }
#'
#'
#'
#' @param ic1 numeric vector that contains the initial cost for option 1
#' @param n1 numeric vector that contains the useful life (years) for option 1
#' @param ac1 numeric vector that contains the annual cost [operations &
#' maintenance (O&M)] for option 1
#' @param i1 numeric vector that contains the effective interest rate per
#' period as a percent for option 1
#' @param ab1 numeric vector that contains the annual benefits for option 1
#' @param salvage1 numeric vector that contains the salvage value for option 1
#' @param option1 character vector that contains the option name for
#' option 1
#' @param ic2 numeric vector that contains the initial cost for option 2
#' @param n2 numeric vector that contains the useful life (years) for option 2
#' @param ac2 numeric vector that contains the annual cost [operations &
#' maintenance (O&M)] for option 2
#' @param i2 numeric vector that contains the effective interest rate per
#' period as a percent for option 2
#' @param ab2 numeric vector that contains the annual benefits for option 2
#' @param salvage2 numeric vector that contains the salvage value for option 2
#' @param option2 character vector that contains the option name for
#' option 2
#' @param table character vector that contains the table output format
#' (ptable, rtable, or both)
#'
#'
#' @return \code{\link[data.table]{data.table}} with character vectors with the monetary
#' values having thousands separator in a pretty table (ptable) & message with
#' the best option, data.frame with numeric vectors without the thousands
#' separator in regular table (rtable) & a message with the best option, or both
#' options combined in a list
#'
#'
#'
#'
#' @references
#' \enumerate{
#' \item Michael R. Lindeburg, PE, \emph{EIT Review Manual}, Belmont, California: Professional Publications, Inc., 1996, page 14-2, 14-4.
#' \item William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling, \emph{Engineering Economy}, Fourteenth Edition, Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2009, page 133, 142, 442-443, 452-453.
#' }
#'
#'
#'
#'
#' @author Irucka Embry
#'
#'
#'
#' @encoding UTF-8
#'
#'
#'
#'
#'
#'
#'
#' @examples
#'
#' library(iemisc)
#'
#' # Example from Lindeburg Reference text (page 14-4)
#' benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000, i1 = 10,
#' salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000, i2 = 10,
#' salvage2 = 10000, option1 = "A", option2 = "B", table = "rtable")
#'
#'
#' # This is useful for saving the results as the named data.frame rtable
#' rtable <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
#' i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
#' i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "rtable")
#'
#' rtable
#'
#'
#'
#' # This is useful for saving the results as the named data.frame ptable
#' ptable <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
#' i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
#' i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "ptable")
#'
#' ptable
#'
#'
#'
#' # This is useful for saving the results as the named list of 2 data.frames
#' # called both
#' both <- benefitcost(ic1 = 300000, n1 = 10, ac1 = 45000, ab1 = 150000,
#' i1 = 10, salvage1 = 0, ic2 = 400000, n2 = 10, ac2 = 35000, ab2 = 200000,
#' i2 = 10, salvage2 = 10000, option1 = "A", option2 = "B", table = "both")
#'
#' both
#'
#'
#'
#' # Example 10-8 from the Sullivan Reference text (page 452-453)
#' project <- benefitcost(ic1 = 750000, n1 = 35, ac1 = 120000, ab1 = 245000,
#' i1 = 9, salvage1 = 0, ic2 = 625000, n2 = 25, ac2 = 110000, ab2 = 230000,
#' i2 = 9, salvage2 = 0, option1 = "Project I", option2 = "Project II",
#' table = "rtable")
#'
#' project
#'
#'
#'
#'
#' @importFrom data.table data.table setnames setattr
#' @importFrom assertthat assert_that
#' @importFrom checkmate qtest
#' @importFrom round round_r3
#'
#' @export
benefitcost <- function (ic1, n1, ac1, ab1, i1, salvage1, ic2, n2, ac2, ab2, i2, salvage2, option1, option2, table = c("ptable", "rtable", "both")) {
table <- table
checks <- c(ic1, n1, ac1, ab1, i1, salvage1, ic2, n2, ac2, ab2, i2, salvage2)
# Check
assert_that(!any(qtest(checks, "N+(,)") == FALSE), msg = "Either ic1, n1, ac1, ab1, i1, salvage1, ic2, n2, ac2, ab2, i2, salvage2 is NA, NaN, Inf, -Inf, empty, or a string. Please try again.")
# only process with finite values and provide an error message if the check fails
assert_that(!any(qtest(option1, "S==1") == FALSE), msg = "option1 is not a character vector of length 1. Please try again.")
# only process with a single character vector and provide an error message if the check fails
assert_that(!any(qtest(option2, "S==1") == FALSE), msg = "option2 is not a character vector of length 1. Please try again.")
# only process with a single character vector and provide an error message if the check fails
assert_that(qtest(table, "S==1"), msg = "There is not a table option or more than 1 table option. Please specify either 'ptable', 'rtable', or 'both'.")
# only process with the correct length of table variable and provide an error message if the check fails
assert_that(isTRUE(any(c("ptable", "rtable", "both") %in% table)), msg = "The table option has not been identified correctly as either ptable, rtable, or both. Please try again.")
# only process with a specified frequency and provide a stop warning if not
# option 1
# cost 1
i1 <- i1 / 100
cost1 <- ic1 + (ac1 * (((1 + i1) ^ n1 - 1) / (i1 * ((1 + i1) ^ n1)))) - (salvage1 * ((1 / (1 + i1)) ^ n1))
# benefit 1
benefit1 <- ab1 * (((1 + i1) ^ n1 - 1) / (i1 * ((1 + i1) ^ n1)))
b_c1 <- benefit1 / cost1
# option 2
# cost 2
i2 <- i2 / 100
cost2 <- ic2 + (ac2 * (((1 + i2) ^ n2 - 1) / (i2 * ((1 + i2) ^ n2)))) - (salvage2 * ((1 / (1 + i2)) ^ n2))
# benefit 2
benefit2 <- ab2 * (((1 + i2) ^ n2 - 1) / (i2 * ((1 + i2) ^ n2)))
b_c2 <- benefit2 / cost2
# Benefit-Cost ratio of Option 2 to Option 1
b_c_rank <- (benefit2 - benefit1) / (cost2 - cost1)
b_c_rank_choose <- ifelse(b_c_rank >= 1, paste0("choose", " ", option2, "."), paste0("choose", " ", option1, "."))
if (table == "ptable") {
ptable <- data.table(c(NA_character_, "Benefit", "Cost", "Benefit-Cost Ratio"), c(option1, formatC(benefit1, big.mark = ",", format = "f", digits = 2), formatC(cost1, big.mark = ",", format = "f", digits = 2), round_r3(b_c1, d = 2)), c(option2, formatC(benefit2, big.mark = ",", format = "f", digits = 2), formatC(cost2, big.mark = ",", format = "f", digits = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- as.character(ptable[1, ])
ptable <- ptable[-1, ]
cat("\n", paste("The Benefit-Cost ratio of", option2, "to", option1, "is", round_r3(b_c_rank, d = 2), "thus", b_c_rank_choose), "\n\n")
# code block below modified from data.table function
setattr(ptable, "col.names", setnames(ptable, col.names))
setattr(ptable, "class", c("data.table", "data.frame"))
ptable
} else if (table == "rtable") {
rtable <- data.table(c("Benefit", "Cost", "Benefit-Cost Ratio"), c(round_r3(benefit1, d = 2), round_r3(cost1, d = 2), round_r3(b_c1, d = 2)), c(round_r3(benefit2, d = 2), round_r3(cost2, d = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- c(NA_character_, option1, option2)
cat("\n", paste("The Benefit-Cost ratio of", option2, "to", option1, "is", round(b_c_rank, digits = 2), "thus", b_c_rank_choose), "\n\n")
# code block below modified from data.table function
setattr(rtable, "col.names", setnames(rtable, col.names))
setattr(rtable, "class", c("data.table", "data.frame"))
rtable
} else if (table == "both") {
# ptable
ptable <- data.table(c(NA_character_, "Benefit", "Cost", "Benefit-Cost Ratio"), c(option1, formatC(benefit1, big.mark = ",", format = "f", digits = 2), formatC(cost1, big.mark = ",", format = "f", digits = 2), round_r3(b_c1, d = 2)), c(option2, formatC(benefit2, big.mark = ",", format = "f", digits = 2), formatC(cost2, big.mark = ",", format = "f", digits = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- as.character(ptable[1, ])
ptable <- ptable[-1, ]
cat("\n", paste("The Benefit-Cost ratio of", option2, "to", option1, "is", round(b_c_rank, digits = 2), "thus", b_c_rank_choose), "\n\n")
# code block below modified from data.table function
setattr(ptable, "col.names", setnames(ptable, col.names))
setattr(ptable, "class", c("data.table", "data.frame"))
ptable
# rtable
rtable <- data.table(c("Benefit", "Cost", "Benefit-Cost Ratio"), c(round(benefit1, digits = 2), round_r3(cost1, d = 2), round_r3(b_c1, d = 2)), c(round_r3(benefit2, d = 2), round_r3(cost2, d = 2), round_r3(b_c2, d = 2)), stringsAsFactors = FALSE)
col.names <- c(NA_character_, option1, option2)
# code block below modified from data.table function
setattr(rtable, "col.names", setnames(rtable, col.names))
setattr(rtable, "class", c("data.table", "data.frame"))
rtable
return(list(rtable, ptable))
}
}
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