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嵌套ifelse在R中,带有“conditionhaslength>1”警告

if-statement dataframe r

Bubbles · 技术社区 · 1 年前

我有以下数据框架,并希望强加一些规则来生成新的值:

column_1<-c("B","B","B","B","A","A","A","A")
column_2<-c("L","L","B","B","L","L","B","B")
column_3<-c("c","b","c","b","c","b","c","b")
column_4<-rep(1)
column_5<-rep(0.5)
column_6<-rep(2)

df_test<-data.frame(column_1,column_2,column_3,column_4,column_5,column_6)
#concatenating column 1,2,3 as a flag, naming it column 7
df_test$column_7<-paste0(df_test$column_1,df_test$column_2,df_test$column_3)

> df_test
  column_1 column_2 column_3 column_4 column_5 column_6 column_7
1        B        L        c        1      0.5        2      BLc
2        B        L        b        1      0.5        2      BLb
3        B        B        c        1      0.5        2      BBc
4        B        B        b        1      0.5        2      BBb
5        A        L        c        1      0.5        2      ALc
6        A        L        b        1      0.5        2      ALb
7        A        B        c        1      0.5        2      ABc
8        A        B        b        1      0.5        2      ABb

#rule 1: if column_7=="BLc|BBb", then df_test$result will be equal to column 4 +column 5 +column_6
#rule 2: if column_7=="BLb|BBc", then df_test$result will be equal to column 4 +column 5 -column_6
#rule 3: if column_7=="ALc|ABb", then df_test$result will be equal to column 4 -column 5 -column_6
#rule 4: if column_7=="ABc|ALb", then df_test$result will be equal to column 4 -column 5 +column_6

要转换规则,我有以下片段:

df_test$result<-if(df_test$column_7=="BLc|BBb"){
  df_test$column_4+df_test$column_5+df_test$column_6
}else{
  if(df_test$column_7=="BLb|BBc"){
    df_test$column_4+df_test$column_5-df_test$column_6
  }else{
    if(df_test$column_7=="ALc|ABb"){
      df_test$column_4-df_test$column_5-df_test$column_6
    } else{
      if(df_test$column_7=="ABc|ALb"){
        df_test$column_4-df_test$column_5+df_test$column_6
      }
    }
  }
}

Warning messages:
1: In if (df_test$column_7 == "BLc|BBb") { :
  the condition has length > 1 and only the first element will be used
2: In if (df_test$column_7 == "BLb|BBc") { :
  the condition has length > 1 and only the first element will be used
3: In if (df_test$column_7 == "ALc|ABb") { :
  the condition has length > 1 and only the first element will be used
4: In if (df_test$column_7 == "ABc|ALb") { :
  the condition has length > 1 and only the first element will be used

> print(df_test$result)
NULL

在这里,我假设我的嵌套方式,如果其他地方出了问题,你能给我一些提示吗?

如果一切顺利,预计会看到以下内容:

> data.frame(desired_column_8)
  desired_column_8
1              3.5
2             -0.5
3             -0.5
4              3.5
5             -1.5
6              2.5
7              2.5
8             -1.5

不管怎样,感谢大家花时间查看这篇文章。

更新1:尝试dplyr::case_when

library(dplyr)
df_test<-df_test %>%
  mutate(column_8=case_when(column_7=="BLc|BBb" ~df_test$column_4+df_test$column_5+df_test$column_6,
                            column_7=="BLb|BBc" ~df_test$column_4+df_test$column_5-df_test$column_6,
                            column_7=="ALc|ABb" ~df_test$column_4-df_test$column_5-df_test$column_6,
                            column_7=="ABc|ALb"~df_test$column_4-df_test$column_5+df_test$column_6))


> df_test
  column_1 column_2 column_3 column_4 column_5 column_6 column_7 column_8
1        B        L        c        1      0.5        2      BLc       NA
2        B        L        b        1      0.5        2      BLb       NA
3        B        B        c        1      0.5        2      BBc       NA
4        B        B        b        1      0.5        2      BBb       NA
5        A        L        c        1      0.5        2      ALc       NA
6        A        L        b        1      0.5        2      ALb       NA
7        A        B        c        1      0.5        2      ABc       NA
8        A        B        b        1      0.5        2      ABb       NA
>

1 回复 | 直到 1 年前

jblood94 1 年前

使用的选项 match 。它可能比嵌套更具性能 ifelse s

codes <- c("BLc", "BBb", "BLb", "BBc", "ABc", "ALb", "ALc", "ABb")
transform(
  df_test,
  result = {
    i <- (match(column_7, codes) - 1)%/%2
    column_4 + column_5*(1 - 2*(i%/%2)) + column_6*(1 - 2*(i%%2))
  }
)
#>   column_1 column_2 column_3 column_4 column_5 column_6 column_7 result
#> 1        B        L        c        1      0.5        2      BLc    3.5
#> 2        B        L        b        1      0.5        2      BLb   -0.5
#> 3        B        B        c        1      0.5        2      BBc   -0.5
#> 4        B        B        b        1      0.5        2      BBb    3.5
#> 5        A        L        c        1      0.5        2      ALc   -1.5
#> 6        A        L        b        1      0.5        2      ALb    2.5
#> 7        A        B        c        1      0.5        2      ABc    2.5
#> 8        A        B        b        1      0.5        2      ABb   -1.5

B. Christian Kamgang 1 年前

您可以使用regex解决您的问题,如下所示:

library(dplyr)
library(stringr)

df_test %>%
  mutate(column_8=case_when(str_detect(column_7, "BLc|BBb") ~ column_4 + column_5 + column_6,
                            str_detect(column_7, "BLb|BBc") ~ column_4 + column_5 - column_6,
                            str_detect(column_7, "ALc|ABb") ~ column_4 - column_5 - column_6,
                            str_detect(column_7, "ABc|ALb") ~ column_4 - column_5 + column_6))

  column_1 column_2 column_3 column_4 column_5 column_6 column_7 column_8
1        B        L        c        1      0.5        2      BLc      3.5
2        B        L        b        1      0.5        2      BLb     -0.5
3        B        B        c        1      0.5        2      BBc     -0.5
4        B        B        b        1      0.5        2      BBb      3.5
5        A        L        c        1      0.5        2      ALc     -1.5
6        A        L        b        1      0.5        2      ALb      2.5
7        A        B        c        1      0.5        2      ABc      2.5
8        A        B        b        1      0.5        2      ABb     -1.5