使用“any”函数跨多个列的逻辑函数

Question

我想跨多列运行逻辑操作（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，因为我必须编写几个查询。

我尝试使用“any”和“brackets”来缩短查询。但是，第二个查询运行良好，但给了我不同的答案。 “任何”功能是否适用于多列？

这是我的条件-

1. 任何一列（B2 到 B5）都有 1 & B1
2. 任何一列（B2 到 B5）都有 -99 和 B1
3. B1 ==3，然后是“Noissue”
4. 休息就是问题

Participate	B1	B2	B3	B4	B5	Query1	Query2
3	-1	-1	-1	-1	-1	Noissue	Noissue
1	-1	1	-1	-1	1	Noissue	Noissue
1	-1	-1	-1	-1	-1	Issue	Noissue
2	-1	1	1	-1	1	Noissue	Noissue
2	1	1	1	1	-1	Noissue	Noissue
1	-99	-99	-99	-99	-99	Noissue	Noissue

如果有人帮助我减少使用不同功能的代码行，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

我们可以使用across 和case_when

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))

Answer 2

当我们必须在多列中逐行使用逻辑条件时，通常应该考虑两种主要方法。这些消除了对rowwise() 和Reduce() 的需要，可以使用lapply/map %>% Reduce/reduce 或复杂的case_when()statements。

-1) rowSums(condition)
-2)if_any() / if_all()

这个问题最适合if_any()的解决方案。

与if_any()

Batch_v1 %>% mutate(query3 = ifelse(if_any(B2:B5, ~.x %in% c(-99, 1)) & B1<=2,
                                    "Noissue", "Issue"))

与rowSums()

Batch_v1 %>% mutate(query3 = ifelse(rowSums(across(B2:B5, ~.x %in% c(-99, 1)))>0 & B1<=2,
                                    "Noissue", "Issue"))

输出

# A tibble: 6 x 9
  Participate    B1    B2    B3    B4    B5 Query1  Query2  query3 
        <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr>   <chr>   <chr>  
1           3    -1    -1    -1    -1    -1 Noissue Noissue Issue  
2           1    -1     1    -1    -1     1 Noissue Noissue Noissue
3           1    -1    -1    -1    -1    -1 Issue   Noissue Issue  
4           2    -1     1     1    -1     1 Noissue Noissue Noissue
5           2     1     1     1     1    -1 Noissue Noissue Noissue
6           1   -99   -99   -99   -99   -99 Noissue Noissue Noissue

这里有类似问题的一些很好的答案：
Rowwise logical operations with mutate() and filter() in R 在这里：
R - Remove rows from dataframe that contain only zeros in numeric columns, base R and pipe-friendly methods? 免责声明：我提出或回答了这些问题

Answer 3

你可以使用

library(dplyr)

Batch_v1 %>% 
  rowwise() %>%
  mutate(
    Query3 = case_when(
      any(B1:B5 == 1)   & Participate %in% c(1,2,-99) ~ "Noissue",
      any(B1:B5 == -99) & Participate %in% c(1,2,-99) ~ "Noissue",
      Participate == 3                                ~ "Noissue",
      TRUE                                            ~ "Issue"
      )
    )

返回

# A tibble: 6 x 9
# Rowwise: 
  Participate    B1    B2    B3    B4    B5 Query1  Query2  Query3 
        <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr>   <chr>   <chr>  
1           3    -1    -1    -1    -1    -1 Noissue Noissue Noissue
2           1    -1     1    -1    -1     1 Noissue Noissue Noissue
3           1    -1    -1    -1    -1    -1 Issue   Noissue Issue  
4           2    -1     1     1    -1     1 Noissue Noissue Noissue
5           2     1     1     1     1    -1 Noissue Noissue Noissue
6           1   -99   -99   -99   -99   -99 Noissue Noissue Noissue

第二个代码的主要问题是函数

any(Batch_v1[,2:6] == 1)

我们来看看

Batch_v1[,2:6] == 1

#>         B1    B2    B3    B4    B5
#> [1,] FALSE FALSE FALSE FALSE FALSE
#> [2,] FALSE  TRUE FALSE FALSE  TRUE
#> [3,] FALSE FALSE FALSE FALSE FALSE
#> [4,] FALSE  TRUE  TRUE FALSE  TRUE
#> [5,]  TRUE  TRUE  TRUE  TRUE FALSE
#> [6,] FALSE FALSE FALSE FALSE FALSE

所以Batch_v1[,2:6] == 1 返回一个布尔值的data.frame。如果此 data.frame 内的值的 any 为 TRUE，则在此 data.frame 上应用 any 将返回 TRUE。这显然不是你想要的行为。 使用rowwise() 会强制应用any...嗯...每行。

注意：在tidyverse-pipe 中，如果您要引用正在使用的当前对象，则不希望使用Batch_v1$B1。 Batch_v1$B1 例如指的是原始 Batch_v1，没有进行任何转换。在这种情况下，没有真正的区别，但一般情况下您不应该依赖它。