根据两列中的组合删除“重复”行

Question

我有这个示例 data.frame：

df1 <- data.frame(v1 = c('A', 'B', 'C', 'D', 'E', 'F', 'G', 'H'),
                  v2 = c('B', 'A', 'D', 'C', 'F', 'E', 'H', 'G'),
                  value = c(1.12, 1.12, 12.52, 12.52, 3.19, 3.19, 12.52, 12.52))
> df1
  v1 v2 value
1  A  B  1.12
2  B  A  1.12
3  C  D 12.52
4  D  C 12.52
5  E  F  3.19
6  F  E  3.19
7  G  H 12.52
8  H  G 12.52

row 1

中的A和B等组合对我来说与B和A等组合相同，其中

value

列中的值也相同。如何删除出于我的目的而重复的行？

预期结果：

df2 <- data.frame(v1 = c('A', 'C', 'E', 'G'),
                  v2 = c('B', 'D', 'F', 'H'),
                  value = c(1.12, 12.52, 3.19, 12.52))

> df2
  v1 v2 value
1  A  B  1.12
2  C  D 12.52
3  E  F  3.19
4  G  H 12.52

Answer 1

这个想法是考虑 v1 和 v2 可以互换。

df1 <- data.frame(v1 = c('A', 'B', 'C', 'D', 'E', 'F', 'G', 'H'),
                  v2 = c('B', 'A', 'D', 'C', 'F', 'E', 'H', 'G'),
                  value = c(1.12, 1.12, 12.52, 12.52, 3.19, 3.19, 12.52, 12.52))

### with tidyverse:

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
library(purrr)

df2 <- df1 %>%
        mutate(combination = pmap_chr(list(v1, v2), ~ paste(sort(c(..1, ..2)), collapse = ","))) %>%
        filter(!duplicated(combination)) %>%
        select(-combination)

df2
#>   v1 v2 value
#> 1  A  B  1.12
#> 2  C  D 12.52
#> 3  E  F  3.19
#> 4  G  H 12.52

### Base R:
df2 <- df1[!duplicated(t(apply(df1[, c("v1", "v2")], 1, sort))), ]

df2
#>   v1 v2 value
#> 1  A  B  1.12
#> 3  C  D 12.52
#> 5  E  F  3.19
#> 7  G  H 12.52

^{创建于 2023-12-24，使用 reprex v2.0.2}

Answer 2

您可以按元素

c

连接

v1

和

v2

的min和max，看看它们是否是

duplicated

。

> df1 <- structure(list(v1 = c("A", "B", "C", "D", "E", "F", "G", "H"), 
+                       v2 = c("B", "A", "D", "C", "F", "E", "H", "G"), value = c(1.12, 
+                                                                                 1.12, 12.52, 12.52, 3.19, 3.19, 12.52, 12.52)), class = "data.frame", row.names = c(NA, 
+                                                                                                                                                                     -8L))
> df1[!with(df1, mapply(\(...) c(min(...), max(...)), v1, v2)) |> t() |> duplicated(), ]
  v1 v2 value
1  A  B  1.12
3  C  D 12.52
5  E  F  3.19
7  G  H 12.52

基准

速度相当快。

> dfB <- df1[sample.int(nrow(df1), 1e4, replace=TRUE), ]
> library(dplyr); library(purrr)
> microbenchmark::microbenchmark(
+   mapply=dfB[!with(dfB, mapply(\(...) c(min(...), max(...)), v1, v2)) |> t() |> duplicated(), ],
+   apply=dfB[!duplicated(t(apply(dfB[, c("v1", "v2")], 1, sort))), ],
+   tidy=dfB %>%
+     mutate(combination = pmap_chr(list(v1, v2), ~ paste(sort(c(..1, ..2)), collapse = ","))) %>%
+     filter(!duplicated(combination)) %>%
+     select(-combination),
+   check='equivalent'
+ )

$ Rscript --vanilla foo.R
Unit: milliseconds
   expr       min        lq      mean    median        uq      max neval cld
 mapply  69.62856  76.20623  83.45243  81.29934  89.74701 110.8039   100 a  
  apply 514.42432 559.80302 593.04269 594.35724 620.06086 728.9266   100  b 
   tidy 339.91324 368.55064 389.65171 386.97757 410.54651 485.9414   100   c

---

数据：

> dput(df1)
structure(list(v1 = c("A", "B", "C", "D", "E", "F", "G", "H"), 
    v2 = c("B", "A", "D", "C", "F", "E", "H", "G"), value = c(1.12, 
    1.12, 12.52, 12.52, 3.19, 3.19, 12.52, 12.52)), class = "data.frame", row.names = c(NA, 
-8L))