根据查找表创建新变量

Question

我想在使用查找表的数据框上创建一个新变量。所以我有df1（数据帧）有Amount和Term。我需要创建一个新的变量“Premium”，使用查找表创建其值。

我尝试了ifelse功能，但它太繁琐了。以下是插图/示例

df1 <- data.frame(Amount, Term)
df1
#   Amount Term
# 1   2500   23
# 2   3600   30
# 3   7000   45
# 4  12000   50
# 5  16000   38

我需要使用下面的Premium Lookup表创建新的'Premium'变量。

                  Term          
Amount           0-24 Mos  25-36 Mos 37-48 Mos 49-60 Mos
0 - 5,000         133      163       175       186
5,001 - 10,000    191      213       229       249
10,001 - 15,000   229      252       275       306
15,001 - 20,000   600      615       625       719
20,001 - 25,000   635      645       675       786

所以保费的输出应该是。

df1
#   Amount Term Premium
# 1   2500   23     133
# 2   3600   30     163
# 3   7000   45     229
# 4  12000   50     306
# 5  16000   38     625

Answer 1

要获得您想要的结果，您需要组织表格并对数据进行分类。我提供了一个处理这种情况的潜在工作流程。希望这有用：

library(tidyverse)

df1 <- data.frame(
  Amount = c(2500L, 3600L, 7000L, 12000L, 16000L),
  Term = c(23L, 30L, 45L, 50L, 38L)
)

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
# functions for analysis ####
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

amount_tier_function <- function(x){

  case_when(x <= 5000   ~ "Tier_5000",
            x <= 10000  ~ "Tier_10000",
            x <= 15000  ~ "Tier_15000",
            x <= 20000  ~ "Tier_20000",
            TRUE        ~ "Tier_25000")
}


month_tier_function <- function(x){

  case_when(x <= 24   ~ "Tier_24",
            x <= 36   ~ "Tier_36",
            x <= 48   ~ "Tier_48",
            TRUE      ~ "Tier_60")
}

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
# Recut lookup table headings ####
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

lookup_df <- data.frame(stringsAsFactors=FALSE,
                amount_tier = c("Tier_5000", "Tier_10000", "Tier_15000", "Tier_20000",
                                "Tier_25000"),
                    Tier_24 = c(133L, 191L, 229L, 600L, 635L),
                    Tier_36 = c(163L, 213L, 252L, 615L, 645L),
                    Tier_48 = c(175L, 229L, 275L, 625L, 675L),
                    Tier_60 = c(186L, 249L, 306L, 719L, 786L)
             )

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#
# Join everything together ####
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

lookup_df_tidy <- lookup_df %>% 
  gather(mth_tier, Premium, - amount_tier)


df1 %>%
  mutate(amount_tier = amount_tier_function(Amount),
         mth_tier    = month_tier_function(Term)) %>%
  left_join(., lookup_df_tidy) %>%
  select(-amount_tier, -mth_tier)

Answer 2

数据

df1 <- structure(list(Amount    = c(2500L, 3600L, 7000L, 12000L, 16000L), 
                      Term      = c(23L, 30L, 45L, 50L, 38L)), 
                 class     = "data.frame",
                 row.names = c(NA, -5L))

lkp  <- structure(c(133L, 191L, 229L, 600L, 635L, 
                    163L, 213L, 252L, 615L, 645L, 
                    175L, 229L, 275L, 625L, 675L, 
                    186L, 249L, 306L, 719L, 786L), 
                  .Dim      = 5:4, 
                  .Dimnames = list(Amount = c("0 - 5,000", "5,001 - 10,000",
                                              "10,001 - 15,000", "15,001 - 20,000", 
                                              "20,001 - 25,000"),
                                   Term   = c("0-24 Mos", "25-36 Mos", "37-48 Mos", 
                                              "49-60 Mos")))

码

1. 使用列和行名称中的正则表达式首先创建月份和金额的上限（您没有以可重现的方式发布数据，因此此正则表达式可能需要根据您的真实查找表结构进行调整）：

   (month <- c(0, as.numeric(sub("\\d+-(\\d+) Mos$", 
                                 "\\1", 
                                 colnames(lkp)))))
   # [1]  0 24 36 48 60
   
   (amt   <- c(0, as.numeric(sub("^\\d+,*\\d* - (\\d+),(\\d+)$", 
                             "\\1\\2", 
                              rownames(lkp)))))
   # [1]     0  5000 10000 15000 20000 25000

2. 使用df1获取findInterval的每个元素的位置：

   (rows <- findInterval(df1$Amount, amt))
   # [1] 1 1 2 3 4
   (cols <- findInterval(df1$Term, month)) 
   # [1] 1 2 3 4 3

3. 使用这些索引来对查找矩阵进行子集化：

   df1$Premium <- lkp[cbind(rows, cols)]
   df1
   #   Amount Term Premium
   # 1   2500   23     133
   # 2   3600   30     163
   # 3   7000   45     229
   # 4  12000   50     306
   # 5  16000   38     625