寻求一种更简洁的方法来避免嵌套的 if 语句（使用 sapply）

Question

我有2个数据框

lookup_table <- data.frame(Country = c("UK","France", "Germany"), A = c(0,0,1), B = c(1,6,7), C = c(4,8,9))
set.seed(123) # for being reproducible
df <-  data.frame(Country = c("UK","UK","France","France","Germany","Germany","Germany","France","UK"), Values =  runif(9, 1, 10))

我想在 df 中有一个第 3 列，它根据第 2 列中的值和国家分配 class。

类似于下面的内容，但值不应固定：它们应取决于查找中的值和国家/地区 table

Class <- function(x) { 
  if(x > 0 & x <= 1) y <- "A"
  if(x > 1 & x <= 4) y <- "B"
  if(x > 4) y <- "C"
  return(y)
}

df$Class <- sapply(df$Values,Class)

提前感谢您的帮助

Answer 1

我们可以在 'lookup_table' 和 'df' 之间做一个 join on 'Country'，melt 到 'long' 格式。正如@zx8754 评论的那样，使用按 'Country' 分组的 cut 函数（或 findInterval 获取数字索引，使用它来获取相应的 'variable'，将其分配为 'newVar'

library(data.table)
d1 <- melt(setDT(lookup_table)[df, on = "Country"], id.var = c("Country", "Values"))[,
          newVar:=unique(variable)[findInterval(Values, unique(value))], Country]

对感兴趣的列进行子集化并获得 unique 个值

unique(d1[, c("Country", "Values", "newVar"), with = FALSE])
#   Country   Values newVar
#1:      UK 3.588198      B
#2:      UK 8.094746      C
#3:  France 4.680792      A
#4:  France 8.947157      C
#5: Germany 9.464206      C
#6: Germany 1.410008      A
#7: Germany 5.752949      A
#8:  France 9.031771      C
#9:      UK 5.962915      C

Answer 2

另一个选项：

df <- merge(df, lookup_table, by='Country', all.x=T)

df$Class <- 'A'                         # default
df$Class <- with(df, replace(Class, Values > B & Values <= C, 'B'))
df$Class <- with(df, replace(Class, Values > C, 'C'))
df
#  Country   Values A B C Class
#1  France 2.371120 0 6 8     A
#2  France 6.155804 0 6 8     B
#3  France 5.635268 0 6 8     A
#4 Germany 9.661230 1 7 9     C
#5 Germany 6.412292 1 7 9     A
#6 Germany 3.148534 1 7 9     A
#7      UK 4.661493 0 1 4     C
#8      UK 6.933073 0 1 4     C
#9      UK 4.623160 0 1 4     C

您可以从结果中删除任何不需要的列。

Answer 3

这是以 R 为基数的结果：

dfa<-merge(lookup_table,df)
Class <- function(x) { 
  if(x[5] > x[2] & x[5] <= x[3]) y <- "A"
  if(x[5] > x[3] & x[5] <= x[4]) y <- "B"
  if(x[5] > x[4]) y <- "C"
  return(y)
}
dfa$Class <- sapply(1:nrow(dfa),function(ri)Class(dfa[ri,]))
dfa[,-c(2:4)]

> dfa[,-c(2:4)]
  Country   Values Class
1  France 4.680792     A
2  France 8.947157     C
3  France 9.031771     C
4 Germany 1.410008     A
5 Germany 5.752949     A
6 Germany 9.464206     C
7      UK 3.588198     B
8      UK 8.094746     C
9      UK 5.962915     C

Answer 4

这是 dplyr 解决方案。

library(dplyr)
df %>%
  inner_join(lookup_table, by = "Country") %>%
  mutate(Class = ifelse(Values > A & Values < B, "A", 
                 ifelse(Values > B & Values < C, "B", 
                 ifelse(Values > C, "C", "Not_found"))))

在管道的末尾添加 select(-c(A,B,C)) 以获得更清晰的输出 data.frame。作为这种方法的一个额外好处，任何不在范围内的值都将被标记为 "Not_found".

Answer 5

如果您更改 lookup_table 的形式并指定间隔，则可以使用 non-equi join 从 data.table 的开发版本 v1.9.7 轻松执行此任务(Installation instructions):

require(data.table) #v1.9.7+
setDT(df)[lookup, Class := i.Class, on = .(Country, Values > value1, Values <= value2)]
#    Country   Values Class
# 1:      UK 3.588198     B
# 2:      UK 8.094746     C
# 3:  France 4.680792     A
# 4:  France 8.947157     C
# 5: Germany 9.464206     C
# 6: Germany 1.410008     A
# 7: Germany 5.752949     A
# 8:  France 9.031771     C
# 9:      UK 5.962915     C

## i.Class refers to Class from i argument = lookup$Class

其中 lookup 从 lookup_table 构造如下：

setDT(lookup_table)[, D := Inf]
lookup = lookup_table[, .(Country, 
                          Class = rep(c("A", "B", "C"), each=.N), 
                          value1 = c(A, B, C), 
                          value2 = c(B, C, D))]
#    Country Class value1 value2
# 1:      UK     A      0      1
# 2:  France     A      0      6
# 3: Germany     A      1      7
# 4:      UK     B      1      4
# 5:  France     B      6      8
# 6: Germany     B      7      9
# 7:      UK     C      4    Inf
# 8:  France     C      8    Inf
# 9: Germany     C      9    Inf

寻求一种更简洁的方法来避免嵌套的 if 语句（使用 sapply）

Seeking a cleaner way to avoid nested if statements (with sapply)

if-statement

r

sapply