根据 0 到 100 之间的值，为 R 中的列提供正回报

如何解决根据 0 到 100 之间的值，为 R 中的列提供正回报

我正在尝试创建一个风险/回报图。回报是日均值。不是百分比。我想要它的百分比。当一只股票上涨 100% 时，它相当于一只股票下跌 50%。如果股价上涨 400%，就等于股价下跌 87.5%。

我正在尝试将我的正回报乘以以下，具体取决于每个回报的价值：

• 返回值在 0 到 49.99 之间，将返回值乘以 2
• 返回值在 50 到 74.99 之间，将返回值乘以 4
• 75 到 87.49 之间的返回值乘以 8 的返回值
• 返回值在 87.5 到 93.74 之间，将返回值乘以 16
• 返回 93.75 到 93.74 之间的返回值乘以 32 ...

以下是我无法执行的 if 语句的代码：

if (( 0 <= Poi[,2]) & (Poi[,2] <= 49.99999999)){
  Poi[,2] <- Poi[,2]*MyReturn*2
} else if (( 50 <= Poi[,2] <= 74.99999999)){
  Poi[,2]*MyReturn*4
} else if (( 75 <= Poi[,2] <= 87.49999999)){
  Poi[,2]*MyReturn*8
} else if (( 87.5 <= Poi[,2] <= 93.74999999)){
  Poi[,2]*MyReturn*16
} else if (( 93.75 <= Poi[,2] <= 96.87499999)){
  Poi[,2]*MyReturn*32
} else if (( 96.875 <= Poi[,2] <= 98.43749999)){
  Poi[,2]*MyReturn*64
} else if (( 98.4375 <= Poi[,2] <= 99.21874999)){
  Poi[,2]*MyReturn*128
} else if (( 99.21875 <= Poi[,2] <= 99.60937499)){
  Poi[,2]*MyReturn*256
} else if (( 99.609375 <= Poi[,2] <= 99.80468749)){
  Poi[,2]*MyReturn*512
} else if (( 99.8046875 <= Poi[,2] <= 99.90234374)){
  Poi[,2]*MyReturn*1024
} else if (( 99.90234375 <= Poi[,2] <= 99.951171865)){
  Poi[,2]*MyReturn*2048
} else if ( 99.951171875 <= Poi[,2]){
  Poi[,2]*MyReturn*4096
} else {
  Poi[,2]*MyReturn
}

---

dput(Poi):

structure(list(targetRisk = c(93.969862047238,76.9768091680639,63.5982003855003,54.5228767287575,46.0821549451386,37.6476764384589,29.2248472581225,20.8278052738397,12.5085917276091,4.66122739912144,1.70147647969736,4.34078819570542,7.31911601372062,10.4618350712519,13.948425299842,17.515722070763,21.228074686994,25.1663931325116,29.261108015602,33.5927110975208,38.2162769926765,43.116215831914,48.2106070287435,53.5716998572496),targetReturn = c(-42.7922615561753,-38.2096263740965,-33.6269911920178,-29.0443560099391,-24.4617208278605,-19.8790856457821,-15.2964504637031,-10.7138152816245,-6.13118009954602,-1.54854491746736,3.03409026461127,7.6167254466899,12.1993606287685,16.7819958108472,21.3646309929257,25.9472661750043,30.5299013570829,35.1125365391614,39.6951717212402,44.2778069033188,48.8604420853973,53.443077267476,58.0257124495546,62.6083466986961)),row.names = c("1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24"
),class = "data.frame")

解决方法

使用 findInterval 创建索引以将值替换为乘数，然后执行 *

可能更有效

out <- with(Poi,ifelse(targetReturn >=0,targetReturn * 2^(1:12)[findInterval(targetReturn,c(-Inf,49.99,74.99,87.49,93.74,98.43,99.21,99.60,99.8,99.90,99.95))],targetReturn))