if中嵌套的ifelse语句

时间：2018-12-19 12:02:20

标签： r if-statement cycle

我有一个包含3个变量的DF：

unix# VAGRANT_LOG=info vagrant up >vagrant.log
unix# VAGRANT_LOG=info vagrant up | tee vagrant.log

power-shell> vagrant up --debug &> vagrant.log
power-shell> vagrant up --debug | Tee-Object -FilePath ".\vagrant.log"

我需要计算以下算法： enter image description here

我提出了这个（不完整的）代码：

# Generate df
set.seed(101)
df <- data.frame("phd" = sample(c("yes", "no"), 100, replace = TRUE),
                 "age" = sample(c(23:45), 100, replace = TRUE),
                 "gre" = sample(c(130:170), 100, replace = TRUE))

我在编写此代码时遇到了麻烦。我指的是几个posts

5 个答案:

答案 0 :(得分：3)

我认为您对此过于复杂（或过度概念化）。逻辑不需要分支树，只需将三个选择粘贴到一个答案中即可。这是一个整齐的诗歌版本，可一步完成-3种选择。

library(dplyr)
df2<- df %>% mutate(phd.status = paste(if_else(phd =="yes", "PhD", "No_PhD"), 
                                       if_else(age < 30, "30yr", ""), 
                                       if_else(gre < 151 , "low score", "high score") ))

head(df2)
  phd age gre          phd.status
1 yes  39 132      PhD  low score
2 yes  34 166     PhD  high score
3 yes  32 153     PhD  high score
4 yes  33 132      PhD  low score
5 yes  43 132      PhD  low score
6 yes  27 169 PhD 30yr high score

答案 1 :(得分：1)

类似的事情应该起作用：

df$phd.status <- with(df, ifelse(phd == "yes",
    ifelse(age < 30,
        ifelse(gre < 151, "PhD 30yr low score", "PhD 30yr high score"),
        ifelse(gre < 151, "PhD 30yr low score", "PhD 30yr high score")),
    ifelse(gre < 151, "No PhD low score", "No PhD high score")))

答案 2 :(得分：1)

严格遵循图表时：

df$phd.state = ifelse(df$phd == "no", # Did you get Ph.d?
                   # No
                   ifelse(df$gre < 151, # GRE < 151?
                        #Yes
                        "No PhD low score",
                        #No
                        "No PhD high score"
                   ),
                   # Yes
                   ifelse(df$age < 30, # < 30 Yr?
                        #Yes
                        ifelse(df$gre < 151, # GRE < 151?
                            #Yes
                            "PhD 30yr low score",
                            #No
                            "PhD 30yr high score"
                        ),
                        #No
                        ifelse(df$gre < 151, # GRE < 151?
                            #Yes
                            "PhD +30yr low score",
                            #No
                            "PhD +30yr high score"
                        )
                    )
                )

答案 3 :(得分：1)

第一个ifelse()语句可能会很容易地将每个变量更改为图片。根据该算法，顺序并不重要，因此仅mutate(variable = ifelse())是合理的。

如果要使用给出的算法中的输出，可以在tidyr::unite()之后使用mutate()。通过sep = " "，在unite之后的三个之间将有一个空格。

library(tidyverse)
df %>% # your data
  mutate( # each ifelse
    phd = ifelse(phd == "yes", "PhD", "No PhD"),
    age = ifelse(age < 30, "30yr", "+30yr"),
    gre = ifelse(gre < 151, "low score", "high score")
  ) %>% 
  unite(col = status, sep = " ") # unite all three column to new status column
#>                      status
#> 1       PhD 30yr high score
#> 2       PhD 30yr high score
#> 3    No PhD +30yr low score
#> 4    No PhD +30yr low score
#> 5      PhD +30yr high score
#> 6       PhD +30yr low score
#> 7    No PhD 30yr high score
#> 8        PhD 30yr low score
#> 9   No PhD +30yr high score
#> 10   No PhD +30yr low score
#> 11    No PhD 30yr low score
#> 12   No PhD 30yr high score
#> 13  No PhD +30yr high score
#> 14   No PhD +30yr low score
#> 15      PhD +30yr low score
#> 16  No PhD +30yr high score
#> 17   No PhD +30yr low score
#> 18     PhD +30yr high score
#> 19     PhD +30yr high score
#> 20      PhD +30yr low score
#> 21   No PhD +30yr low score
#> 22    No PhD 30yr low score
#> 23       PhD 30yr low score
#> 24   No PhD +30yr low score
#> 25    No PhD 30yr low score
#> 26   No PhD +30yr low score
#> 27     PhD +30yr high score
#> 28      PhD +30yr low score
#> 29     PhD +30yr high score
#> 30   No PhD 30yr high score
#> 31      PhD +30yr low score
#> 32      PhD +30yr low score
#> 33       PhD 30yr low score
#> 34      PhD 30yr high score
#> 35   No PhD +30yr low score
#> 36  No PhD +30yr high score
#> 37      PhD +30yr low score
#> 38   No PhD +30yr low score
#> 39     PhD +30yr high score
#> 40   No PhD +30yr low score
#> 41      PhD +30yr low score
#> 42      PhD 30yr high score
#> 43   No PhD +30yr low score
#> 44      PhD +30yr low score
#> 45      PhD +30yr low score
#> 46       PhD 30yr low score
#> 47      PhD +30yr low score
#> 48   No PhD +30yr low score
#> 49   No PhD 30yr high score
#> 50      PhD +30yr low score
#> 51       PhD 30yr low score
#> 52      PhD +30yr low score
#> 53   No PhD +30yr low score
#> 54  No PhD +30yr high score
#> 55       PhD 30yr low score
#> 56   No PhD 30yr high score
#> 57  No PhD +30yr high score
#> 58  No PhD +30yr high score
#> 59   No PhD +30yr low score
#> 60   No PhD +30yr low score
#> 61   No PhD +30yr low score
#> 62  No PhD +30yr high score
#> 63  No PhD +30yr high score
#> 64   No PhD +30yr low score
#> 65   No PhD +30yr low score
#> 66  No PhD +30yr high score
#> 67       PhD 30yr low score
#> 68     PhD +30yr high score
#> 69   No PhD 30yr high score
#> 70  No PhD +30yr high score
#> 71      PhD +30yr low score
#> 72  No PhD +30yr high score
#> 73  No PhD +30yr high score
#> 74   No PhD +30yr low score
#> 75  No PhD +30yr high score
#> 76      PhD +30yr low score
#> 77     PhD +30yr high score
#> 78     PhD +30yr high score
#> 79   No PhD +30yr low score
#> 80   No PhD +30yr low score
#> 81   No PhD +30yr low score
#> 82   No PhD +30yr low score
#> 83   No PhD +30yr low score
#> 84    No PhD 30yr low score
#> 85      PhD +30yr low score
#> 86    No PhD 30yr low score
#> 87      PhD 30yr high score
#> 88      PhD 30yr high score
#> 89     PhD +30yr high score
#> 90       PhD 30yr low score
#> 91    No PhD 30yr low score
#> 92   No PhD 30yr high score
#> 93  No PhD +30yr high score
#> 94      PhD 30yr high score
#> 95      PhD +30yr low score
#> 96   No PhD +30yr low score
#> 97      PhD +30yr low score
#> 98     PhD +30yr high score
#> 99       PhD 30yr low score
#> 100     PhD +30yr low score

这些值就是图中的值。

答案 4 :(得分：0)

对二叉树的最终状态进行分类的一种简单方法是执行以下操作：

df2 <- df %>%
      mutate(phd=ifelse(phd=='yes', 100, 0),
             age=ifelse(age<30, 10, 0),
             gre=ifelse(gre<151, 1, 0),
             bucket = phd + age + gre
      )  %>%
      arrange(bucket)

存储桶值为您提供每种可能状态的输出。

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