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Ben
Asked: 2025-02-19 16:32:51 +0800 CST

删除多个向量中的重复项

  • 772

我想删除多个向量中的所有重复项，不留下任何重复项。例如，对于这些向量：

a <- c("dog", "fish", "cow")
b <- c("dog", "horse", "mouse")
c <- c("cat", "sheep", "mouse")

预期结果是：

a <- c("fish", "cow")
b <- c("horse")
c <- c("cat", "sheep")

有没有办法实现这一点，而无需连接向量并再次拆分它们？

7 个回答

  1. Best Answer

    你或许可以这样做：

    vec <- c(a, b, c)
sapply(list(a, b, c), function(x) x[!x %in% vec[duplicated(vec)]])

[[1]]
[1] "fish" "cow" 

[[2]]
[1] "horse"

[[3]]
[1] "cat"   "sheep"

    如果需要全局环境中的单独变量，可以添加lst()from tibble：

    vec <- c(a, b, c)
l <- sapply(lst(a, b, c), function(x) x[!x %in% vec[duplicated(vec)]])
list2env(l, envir = .GlobalEnv)
    • 12

  2. 给定列表中的数据，例如lst <- list(a = a, b = b, c = c)，你可以尝试

    • 选项 1
    > unstack(subset(stack(lst), ave(seq_along(values), values, FUN = length) == 1))
$a
[1] "fish" "cow"

$b
[1] "horse"

$c
[1] "cat"   "sheep"
    • 选项 2
    > lapply(seq_along(lst), \(k) setdiff(lst[[k]], unlist(lst[-k])))
[[1]]
[1] "fish" "cow"

[[2]]
[1] "horse"

[[3]]
[1] "cat"   "sheep"
    • 选项 3
    > v <- names(which(table(unlist(lst)) == 1))

> lapply(lst, intersect, v)
$a
[1] "fish" "cow"

$b
[1] "horse"

$c
[1] "cat"   "sheep"
    • 8

  3. 还有另一种可能性collapse::fduplicated(x, all = TRUE)。与基本 R 的 不同duplicated，此函数允许您包含所有出现多次的值：

    lst <- list(a = a, b = b, c = c)
unstack(subset(stack(lst), !collapse::fduplicated(values, all = TRUE)))

# $a
# [1] "fish" "cow" 
# 
# $b
# [1] "horse"
# 
# $c
# [1] "cat"   "sheep"

    对长度为 10 的 100 个元素的列表进行基准测试：我使用的答案collapse是最快的（显示相对时间）。@Friede 的基本 R 答案同样快。

      expression     min  median itr/sec mem_alloc n_itr
1     tmfmnk    5.51    5.95  445.47     44.72    10
2       Tic1    2.87    3.06  879.08      2.12    10
3       Tic2   27.05   26.28   98.60     59.35    10
4       Tic3    4.43    4.28  504.72      2.78    10
5     jay.sf 2931.20 2785.01    1.00   5925.16    10
6     Edward   28.03   27.67   98.86     56.71    10
7       Maël    1.00    1.00 2699.28      1.00    10
8     Friede    1.03    1.00 2568.27      1.25    10

    代码：

    lst <- lapply(setNames(as.list(replicate(100, sample(combn(letters, m = 2, paste, collapse = ""), size = 10, replace = TRUE), simplify = FALSE)), paste0('A', 1:100)), c)
vec <- unlist(lst, use.names = FALSE)

bench::mark(
  tmfmnk = sapply(lst, function(x) x[!x %in% vec[duplicated(vec)]]),
  Tic1 = unstack(subset(stack(lst), ave(seq_along(values), values, FUN = length) == 1)),
  Tic2 = lapply(seq_along(lst), \(k) setdiff(lst[[k]], unlist(lst[-k]))),
  Tic3 = {v <- names(which(table(unlist(lst)) == 1))
  lapply(lst, intersect, v)},
  jay.sf = outer(seq_along(lst), seq_along(lst), Vectorize(\(i, j) setdiff(lst[[i]], unlist(lst[-j])))) |>
    diag(),
  Edward = lapply(seq_along(lst), \(x) lst[[x]][!lst[[x]] %in% unlist(lst[setdiff(seq_along(lst)[-x], x)])]),
  Maël = unstack(subset(stack(lst), !collapse::fduplicated(values, all = TRUE))),
  Friede = unstack(subset(stack(lst), !duplicated(values) & !duplicated(values, fromLast=TRUE))),
  check = FALSE,
  iterations = 10,
  relative = TRUE
)
    • 8

  4. 答题聚会迟到了。

    基地R，做!duplicated()两次。

    unstack(subset(stack(l), !duplicated(values) & !duplicated(values, fromLast=TRUE)))

    $a
[1] "fish" "cow" 

$b
[1] "horse"

$c
[1] "cat"   "sheep"

    这避免了*apply-函数，Vectorize()（即mapply()）和outer()。

    数据

    l = list(a = c("dog", "fish", "cow"), b = c("dog", "horse", "mouse"), c = c("cat", "sheep", "mouse"))
    • 8 
  5. lst <- list(a,b,c)

lapply(seq_along(lst), \(x) lst[[x]][!lst[[x]] %in% unlist(lst[setdiff(seq_along(lst)[-x], x)])])

[[1]]
[1] "fish" "cow" 

[[2]]
[1] "horse"

[[3]]
[1] "cat"   "sheep"

    此解决方案将重复项保留在同一个向量中，并且仅当重复项存在于多个向量中时才将其删除，如问题所述。例如，将函数应用于

    a <- c("dog", "fish", "dog")
b <- c("cow", "horse", "mouse")
c <- c("cat", "sheep", "mouse")

lst <- list(a,b,c); lst

    给出

    [[1]]
[1] "dog"  "fish" "dog" 

[[2]]
[1] "cow"   "horse"

[[3]]
[1] "cat"   "sheep"

    而其他答案则给出

    [[1]]
[1] "fish"

[[2]]
[1] "cow"   "horse"

[[3]]
[1] "cat"   "sheep"
    • 6

  6. 使用setdiff.给出结果outer。diag

    > lst <- list(a, b, c)
> outer(seq_along(lst), seq_along(lst), 
+       Vectorize(\(i, j) setdiff(lst[[i]], unlist(lst[-j])))) |>
+   diag()
[[1]]
[1] "fish" "cow" 

[[2]]
[1] "horse"

[[3]]
[1] "cat"   "sheep"
    • 6

  7. 如果“重复”的概念适用，这些向量实际上是一个数据集。你应该把它们放在一个数据结构中，并创建“整洁的数据”。我建议使用包 data.table，特别是如果你的数据集很大：

    library(data.table)
DT <- data.table(a, b, c)
DT <- melt(DT, measure.vars = 1:3)

    然后您可以轻松删除重复的值。

    DT[!duplicated(value) & !duplicated(value, fromLast = TRUE)]
#   variable  value
#     <fctr> <char>
#1:        a   fish
#2:        a    cow
#3:        b  horse
#4:        c    cat
#5:        c  sheep

    这种方法假设您的数据集不是太大以至于该variable列的内存需求不成问题。

    • 1

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