R递归条件嵌套有两个条件

Question

我看到了一些类似的问题，但还没有完全解决这个问题。希望没有错。

我有一个这样的DF:

Invoices<-c(20171100247, 20171100408, 20171200376,20171201052, 21609000088)
Oustanding.days<-c(15,85,96,251,123)
Quantile.low<-c(25,21,22,23,24)
Quantile.Medium<-c(45,65,85,93,74)
Quantile.top<-c(74,89,101,175,125)
Remittances<-c(25,47,5,7,2)
df<-cbind(Invoices,Oustanding.days,Quantile.low,Quantile.Medium,Quantile.top,Remittances)
df
        Invoices Oustanding.days Quantile.low Quantile.Medium Quantile.top Remittances
[1,] 20171100247              15           25              45           74          25
[2,] 20171100408              85           21              65           89          47
[3,] 20171200376              96           22              85          101           5
[4,] 20171201052             251           23              93          175           7
[5,] 21609000088             123           24              74          125           2

我想创建一个包含条件的 "Payment accuracy" 列，从这个意义上说：

如果汇款低于 5，那么我想线性分配精度：

1) df$Outstanding.days <60 -> 打印 "too early"

2) df$Outstanding.days >60 <90 -> 打印 "early"

3) df$Outstanding.days >90 -> 打印 "late"

如果汇款超过 5 我想用分位数分配它：

1) df$Outstanding.days < Quantile.low -> 打印 "too early"

2) df$Outstanding.days > Quantile.low & < Quantile.Medium -> 打印 "early"

3) df$Outstanding.days > Quantile.Medium & < Quantile.top -> 打印 "On date"

4) df$Outstanding.days > Quantile.top -> 打印 "late"

我正在尝试使用转换和嵌套条件

df.final<-transform(df,Payment.accuracy=( 
if (df$OutStandingDays <= df$Quantile.low) {print 
("too early")}
else (print ("NA"))))

但我做错了什么。

谢谢。

Answer 1

我添加了一个更正 df 的列（如果您想提取该列，以后会很容易），我使用了您最后的代码行：

 Invoices<-c(20171100247, 20171100408, 20171200376,20171201052, 21609000088)
 Oustanding_days<-c(15,85,96,251,123)
 Quantile_low<-c(25,21,22,23,24)
 Quantile_Medium<-c(45,65,85,93,74)
 Quantile_top<-c(74,89,101,175,125)
 Remittances<-c(25,47,5,7,2)
 df<-  cbind(Invoices,Oustanding_days,Quantile_low,Quantile_Medium,Quantile_top,Remittances)
 df <- as.data.frame(df)



 for (i in 1:length(df[,1])){
   if(df$Oustanding_days[i] <= df$Quantile_low[i]){
     df$final[i] <- print("too early")
   } else {
     df$final[i] <-print("NA")
   }
 }

通过该示例，您应该能够重现所需的所有条件。

祝你好运！

Answer 2

在此解决方案中，我根据 remittances 上的两个条件拆分数据，然后按行折叠。

library(tidyverse)

# First condition
df_less5 = df %>% filter(Remittances < 5)

df_less5 = df_less5 %>% 
  mutate(payment_accuracy = ifelse(Oustanding.days < 60, "too early",
                                   ifelse(Oustanding.days >60 & Oustanding.days <90, "early", "late")))

# Second condition
df_more5 = df %>% filter(Remittances > 5)

df_more5 = df_more5 %>% 
  mutate(payment_accuracy = ifelse(Oustanding.days < Quantile.low, "too early",
                                   ifelse(Oustanding.days > Quantile.low & Oustanding.days < Quantile.Medium, "early",
                                          ifelse(Oustanding.days > Quantile.Medium & Oustanding.days < Quantile.top, "on_date", 
                                                 ifelse(Oustanding.days > Quantile.top, "late", "other")))))


# new dataset
df_new = bind_rows(df_less5, df_more5)

给出以下输出：

 > df_new

  Invoices Oustanding.days Quantile.low Quantile.Medium Quantile.top Remittances payment_accuracy
1 21609000088             123           24              74          125           2             late
2 20171100247              15           25              45           74          25        too early
3 20171100408              85           21              65           89          47          on_date
4 20171201052             251           23              93          175           7             late

Answer 3

您可以为此使用 dplyr 和嵌套的 ifelse 语句。

请注意，像 >Quantile.low & < Quantile.Medium 这样的语句排除了它等于其中一个值的情况，您应该为此使用 <=。即它应该是 >=Quantile.low & < Quantile.Medium 或 >Quantile.low & <= Quantile.Medium。在下面的示例中，我假设了后一种选择。

df <- as.data.frame(df)   
library(dplyr)

df %>% mutate(x=ifelse(Remittances<5,
                     ifelse(Oustanding.days<=60,'too early',
                            ifelse(Oustanding.days>60 & Oustanding.days<=90,'early','late')),NA)) %>%
  mutate(x=ifelse(Remittances>=5,
                  ifelse(Oustanding.days<=Quantile.low,'too early',
                         ifelse(Oustanding.days>Quantile.low & Oustanding.days<=Quantile.Medium,'low',
                                ifelse(Oustanding.days>Quantile.Medium & Oustanding.days <= Quantile.top,'On date','late'))),x))

哪个returns:

     Invoices Oustanding.days Quantile.low Quantile.Medium Quantile.top Remittances         x
1 20171100247              15           25              45           74          25 too early
2 20171100408              85           21              65           89          47   On date
3 20171200376              96           22              85          101           5   On date
4 20171201052             251           23              93          175           7      late
5 21609000088             123           24              74          125           2      late

希望对您有所帮助！

Answer 4

我们可以使用 dplyr 包中的 case_when 来根据多个条件赋值。嵌套的 ifelse 语句或 for 循环有时可能过于复杂且难以阅读。

最后一行TRUE ~ NA_character_是指定NA到不满足以上任何条件的行

library(dplyr)

df2 <- df %>%
  mutate(`Payment accuracy` = case_when(
    Remittances < 5 & Outstanding.days < 60                            ~ "too early",
    Remittances < 5 & Outstanding.days >= 60 & Outstanding.days < 90   ~ "early",
    Remittances < 5 & Outstanding.days >= 90                           ~ "late",
    Remittances >= 5 & Outstanding.days < Quantile.low                 ~ "too early",
    Remittances >= 5 & Outstanding.days >= Quantile.low & 
      Outstanding.days < Quantile.Medium                               ~ "early",
    Remittances >= 5 & Outstanding.days >= Quantile.Medium & 
      Outstanding.days < Quantile.top                                  ~ "On date",
    Remittances >= 5 & Outstanding.days >= Quantile.top                ~ "late",
    TRUE                                                               ~ NA_character_
  ))
df2
#      Invoices Outstanding.days Quantile.low Quantile.Medium Quantile.top Remittances Payment accuracy
# 1 20171100247               15           25              45           74          25        too early
# 2 20171100408               85           21              65           89          47          On date
# 3 20171200376               96           22              85          101           5          On date
# 4 20171201052              251           23              93          175           7             late
# 5 21609000088              123           24              74          125           2             late

数据

请注意您的原始代码中有拼写错误，例如 Outstanding.days 和 Remittances。此外，您没有通过 cbind 创建数据框。您需要的功能是data.frame。 stringsAsFactors = FALSE 是为了确保列类型是字符，而不是因子。

Invoices<-c(20171100247, 20171100408, 20171200376,20171201052, 21609000088)
Outstanding.days<-c(15,85,96,251,123)
Quantile.low<-c(25,21,22,23,24)
Quantile.Medium<-c(45,65,85,93,74)
Quantile.top<-c(74,89,101,175,125)
Remittances<-c(25,47,5,7,2)
df <- data.frame(Invoices, Outstanding.days, Quantile.low, 
                 Quantile.Medium, Quantile.top, Remittances,
                 stringsAsFactors = FALSE)

R递归条件嵌套有两个条件

R recursive conditionals nested with two conditions

conditional

nested

r