r - 使用 "any"函数跨多个列的逻辑函数

Question

我想跨许多列运行逻辑运算(多个条件)。我写了一个运行良好的查询。但是，我想缩短我的代码，因为我必须编写多个查询。

我尝试使用“any”和“括号”来缩短查询。然而，第二个查询运行良好，但给了我不同的答案。 “任意”函数是否适用于多列？

这是我的条件 -

1. 任意列(B2 到 B5)具有 1 和 B1 <=2，则“无问题”
2. 任意一列(B2 到 B5)的值为 -99 且 B1 <=2，则“无问题”
3. B1 ==3，则“无问题”
4. 休息就是问题

<表类=“s-表”><标题>参与B1B2B3B4B5查询1查询2 <正文>3-1-1-1-1-1没问题没问题1-11-1-11没问题没问题1-1-1-1-1-1问题没问题2-111-11没问题没问题21111-1没问题没问题1-99-99-99-99-99没问题没问题

如果有人帮助我减少使用不同函数的代码行，我将不胜感激。

 mutate(Batch_v1, 
               case_when (
                 ((Batch_v1$B1 == 1 |  Batch_v1$B2 == 1 | Batch_v1$B3 == 1 | Batch_v1$B4 == 1 | Batch_v1$B5 == 1| Batch_v1$B6 == 1| Batch_v1$B7 == 1|Batch_v1$B8 == 1|Batch_v1$B9 == 1|Batch_v1$B10 == 1|Batch_v1$BOth == 1) & 
                    Batch_v1$Participate %in% c(1,2,-99))~"Noissue",
                 ((Batch_v1$B1 == -99 |  Batch_v1$B2 == -99 | Batch_v1$B3 == -99|Batch_v1$B4 == -99 |Batch_v1$B5 == -99|Batch_v1$B6 == -99|Batch_v1$B7 == -99|Batch_v1$B8 == 1|Batch_v1$B9 == -99|Batch_v1$B10 == -99|Batch_v1$BOth == -99) & 
                    Batch_v1$Participate %in% c(1,2,-99))~"Noissue",
                 Batch_v1$Participate ==3 ~ "Noissue",
                 TRUE ~ "Issue"))





mutate(Batch_v1, 
   case_when (
     ((any(Batch_v1[,2:6] == 1)) & Batch_v1$Participate %in% c(1,2,-99))~ "Noissue",
     ((any(Batch_v1[,2:6] == -99)) & Batch_v1$Participate %in% c(1,2,-99))~ "Noissue",
     Batch_v1$Participate ==3 ~ "Noissue",
     TRUE ~ "Issue"))

Answer 1

我们可以将 across 与 case_when 一起使用

library(dplyr)
df %>% 
    mutate(across(B2:B5, ~case_when(. == 1 & B1 <=2 ~ "Noissue",
                                    . == -99 & B1 <=2 ~ "Noissue",
                                    B1 == 3 ~ "Noissue",
                                    TRUE ~ "issue")
                  )
           )

输出:

  Participate    B1 B2      B3      B4      B5      Query1  Query2 
        <dbl> <dbl> <chr>   <chr>   <chr>   <chr>   <chr>   <chr>  
1           3    -1 issue   issue   issue   issue   Noissue Noissue
2           1    -1 Noissue issue   issue   Noissue Noissue Noissue
3           1    -1 issue   issue   issue   issue   Issue   Noissue
4           2    -1 Noissue Noissue issue   Noissue Noissue Noissue
5           2     1 Noissue Noissue Noissue issue   Noissue Noissue
6           1   -99 Noissue Noissue Noissue Noissue Noissue Noissue

数据:

df <- structure(list(Participate = c(3, 1, 1, 2, 2, 1), B1 = c(-1, 
-1, -1, -1, 1, -99), B2 = c(-1, 1, -1, 1, 1, -99), B3 = c(-1, 
-1, -1, 1, 1, -99), B4 = c(-1, -1, -1, -1, 1, -99), B5 = c(-1, 
1, -1, 1, -1, -99), Query1 = c("Noissue", "Noissue", "Issue", 
"Noissue", "Noissue", "Noissue"), Query2 = c("Noissue", "Noissue", 
"Noissue", "Noissue", "Noissue", "Noissue")), problems = structure(list(
row = 6L, col = "Query2", expected = "", actual = "embedded null", 
file = "'test'"), row.names = c(NA, -1L), class = c("tbl_df", 
"tbl", "data.frame")), class = c("spec_tbl_df", "tbl_df", "tbl", 
"data.frame"), row.names = c(NA, -6L))