Question

PresentValue <- function(interest.rate, number.periods, frequency) {

    if (interest.rate > 1) {interest.rate = interest.rate/100} else {"input interest.rate as a    whole number"}
    if (frequency < 1 || frequency > 12 ) {"frequency must be between 1 and 12"} else {interest.rate = interest.rate/frequency}

    (1+interest.rate)^number.periods  
}

上面的代码很简单，但我需要捕获用户输入错误。我看了tryCatch，但这对我没有意义。另外，我认为这不是我现在所需要的。

基本上，我需要做以下事情

首先，确保用户输入的利率为整数（即5％不是.05）其次，我需要“询问”用户如何按月支付费率（频率），每年等等。然后将利率除以频率。

奇怪的是，第一个条件只检查其他if语句何时被注释掉。

这个函数将是一个更大的函数集中的一个对象，所以我认为我需要将它们视为最终将与try catch一起使用的警告。最后，如果在利率误差上函数没有终止，只是要求用户正确输入利率，那就太好了。

提前为简单道歉但却找不到我想要的东西。

编辑添加一些conetxt

以下是更大的功能。这显示了凸度的表输出我需要将持续时间，然后是产量和价格表以及使用swerve输出到页面。这是关于结构性融资的教科书。所以它需要适合学生

Bond.Cash.Flow <-function(issue.date, start.date, end.date, coupon, principal, 
frequency,  price){

library(lubridate)
  issue.date <- as.Date(c(issue.date), "%m-%d-%Y")
  start.date <- as.Date(c(start.date), "%m-%d-%Y")
  end.date <- as.Date(c(end.date), "%m-%d-%Y")
  price = price/100


  # 30/360 day count calculation 
    d1 = day(issue.date)
    m1 = month(issue.date)
    y1 = year(issue.date)
    d2 = day(end.date)
    m2 = month(end.date)
    y2 = year(end.date)


 diff = (max(0, 30 - d1) + min(30, d2) + 360*(y2-y1) + 30*(m2-m1-1))/360
 ncashflows = diff * frequency
 cf.period = seq(1:ncashflows)
 pmtdate = seq(start.date, end.date, by = "6 months")

time.period = (cf.period * 6)/12

 couponincome = rep(coupon/frequency * principal, ncashflows)
 principalincome = rep(0,ncashflows)
 principalincome[ncashflows] = principal
cashflow = couponincome + principalincome


  # Yield to maturity
     irr <- function(rate, time.period, cashflow, principal, price){
      pv = cashflow * 1/(1+rate)^time.period
      proceeds = principal * price
      sum(pv) - proceeds
  }

ytm = uniroot(irr, interval = c(lower = -.20, upper = .20), tol =.000000001, 
time.period = time.period, cashflow = cashflow, principal = principal, price = price)$root

 ytm.vec = c(rep(ytm,ncashflows))
 pv.factor = 1/(1+ytm.vec)^time.period
 pv.cashflow = cashflow*pv.factor
 pv.price = pv.cashflow /(principal * (price/100))
 pv.period = pv.price * time.period
 cvx.time = time.period*(time.period + 1)
 cf.cvx = (cashflow/(1+ytm)^(time.period + 2))/(principal * (price/100))
 cf.cvx.period = cf.cvx * cvx.time

cashflow.table <- data.frame(Period = cf.period, Time = time.period, 
Cashflow = cashflow,  PVFactor = pv.factor,PV = pv.cashflow, PV.Price = pv.price, 
pv.period = pv.period, cvx.time = cvx.time, cf.cvx = cf.cvx, 
cf.cvx.period = cf.cvx.period)


cashflow.table
(sum(pv.period) / (price * 100))/( 1+ (ytm/frequency))
.5 * ((sum(cf.cvx.period)/(price * 100)))
print(xtable(cashflow.table, digits = 4))
}

Bond.Cash.Flow <-function(issue.date, start.date, end.date, coupon, principal, 
frequency,  price){

library(lubridate)
  issue.date <- as.Date(c(issue.date), "%m-%d-%Y")
  start.date <- as.Date(c(start.date), "%m-%d-%Y")
  end.date <- as.Date(c(end.date), "%m-%d-%Y")
  price = price/100

＃30/360天数计算 d1 =天（issue.date） m1 =月（issue.date） y1 =年（issue.date） d2 =天（end.date） m2 =月（end.date） y2 =年（end.date）

 diff = (max(0, 30 - d1) + min(30, d2) + 360*(y2-y1) + 30*(m2-m1-1))/360
 ncashflows = diff * frequency
 cf.period = seq(1:ncashflows)
 pmtdate = seq(start.date, end.date, by = "6 months")

time.period = (cf.period * 6)/12

 couponincome = rep(coupon/frequency * principal, ncashflows)
 principalincome = rep(0,ncashflows)
 principalincome[ncashflows] = principal
cashflow = couponincome + principalincome

＃到期收益率 irr＆lt; - function（rate，time.period，cashflow，principal，price）{ pv = cashflow * 1 /（1 + rate）^ time.period 收益=本金*价格 sum（pv） - 收益 }

ytm = uniroot(irr, interval = c(lower = -.20, upper = .20), tol =.000000001, 
time.period = time.period, cashflow = cashflow, principal = principal, price = price)$root

 ytm.vec = c(rep(ytm,ncashflows))
 pv.factor = 1/(1+ytm.vec)^time.period
 pv.cashflow = cashflow*pv.factor
 pv.price = pv.cashflow /(principal * (price/100))
 pv.period = pv.price * time.period
 cvx.time = time.period*(time.period + 1)
 cf.cvx = (cashflow/(1+ytm)^(time.period + 2))/(principal * (price/100))
 cf.cvx.period = cf.cvx * cvx.time

cashflow.table <- data.frame(Period = cf.period, Time = time.period, 
Cashflow = cashflow,  PVFactor = pv.factor,PV = pv.cashflow, PV.Price = pv.price, 
pv.period = pv.period, cvx.time = cvx.time, cf.cvx = cf.cvx, 
cf.cvx.period = cf.cvx.period)


cashflow.table
(sum(pv.period) / (price * 100))/( 1+ (ytm/frequency))
.5 * ((sum(cf.cvx.period)/(price * 100)))
print(xtable(cashflow.table, digits = 4))

}

Answer 1

目前尚不清楚您想要做什么，因为您没有给出函数调用的精确上下文。所以我的回答将是不完整的。

请为您的参数指定默认值，尤其是您有多个要验证的参数的默认值。默认参数在R中使用延迟评估，并且实际上是该语言的强大功能。
使用stop和/或warning抛出错误/警告并在最终调用上下文中捕获它以向用户发送人工消息。
看一下match.call（初学者很难）来有效调用你的函数。

这是使用stop的示例：

PresentValue <- function(interest.rate, number.periods=1, frequency=1) {
if (missing(interest.rate))
  stop("Need to specify interest.rate as number between 0 and 1 for calculations.")
if (!is.numeric(interest.rate)  )
  stop("No numeric  interest.rate specified.")
if (interest.rate <0 | interest.rate > 1)
  stop("No valid  interest.rate specified.")
 ## you do the same thing for other arguments

}

一些测试：

PresentValue()
Error in PresentValue() : 
  Need to specify interest.rate as number between 0 and 1 for calculations.
> PresentValue("a")
Error in PresentValue("a") : No numeric  interest.rate specified.
> PresentValue(5)
Error in PresentValue(5) : No valid  interest.rate specified.
> PresentValue(0.9)  ## normal use

R中的错误陷阱和逻辑

编辑添加一些conetxt

1 个答案: