r - 计算 R 中的所有重复值

Question

我有以下称为 x 的向量:

x <- c(1, 1, 4, 5, 4, 6, 1, 1)
x
#> [1] 1 1 4 5 4 6 1 1

我想计算所有重复值。在这种情况下，数字 1,1,1,1,4,4 是重复的，这意味着共有 6 个重复值。以下是一些尝试:

x <- c(1, 1, 4, 5, 4, 6, 1, 1)
# Wrong outputs
sum(duplicated(x))
#> [1] 4
sum(table(x)-1)
#> [1] 4
# Returns number of duplicated values in this case 1 and 4
nrow(data.frame(table(x))[data.frame(table(x))$Freq > 1,])
#> [1] 2

^{创建于 2022-12-08 reprex v2.0.2}

所以我想知道是否有人知道如何计算所有重复值而不是计算具有重复值的数量？

Answer 1

其他选项:

sum(Filter(\(z) z > 1, table(x)))
sum(setdiff(table(x), 1L))
sum(x %in% x[duplicated(x)])

最后一个显然是最快的，akrun 紧随其后:

bench::mark(
  sum(Filter(\(z) z > 1, table(x))),
  sum(setdiff(table(x), 1L)),
  sum(x %in% x[duplicated(x)]),
  sum(table(x)[names(table(x)) %in% x[duplicated(x)]]),
  sum(duplicated(x)|duplicated(x, fromLast = TRUE))
)
# # A tibble: 5 x 13
#   expression                                                min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result    memory              time                gc                   
#   <bch:expr>                                           <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> <list>    <list>              <list>              <list>               
# 1 sum(Filter(function(z) z > 1, table(x)))                 58us   67.5us    14335.    5.35KB     6.62  6499     3    453.4ms <int [1]> <Rprofmem [16 x 3]> <bench_tm [6,502]>  <tibble [6,502 x 3]> 
# 2 sum(setdiff(table(x), 1L))                             51.6us   60.9us    16046.        0B     6.56  7338     3    457.3ms <int [1]> <Rprofmem [0 x 3]>  <bench_tm [7,341]>  <tibble [7,341 x 3]> 
# 3 sum(x %in% x[duplicated(x)])                            2.8us    3.2us   294065.        0B     0    10000     0       34ms <int [1]> <Rprofmem [0 x 3]>  <bench_tm [10,000]> <tibble [10,000 x 3]>
# 4 sum(table(x)[names(table(x)) %in% x[duplicated(x)]])  102.1us  123.4us     7957.        0B     4.26  3737     2    469.6ms <int [1]> <Rprofmem [0 x 3]>  <bench_tm [3,739]>  <tibble [3,739 x 3]> 
# 5 sum(duplicated(x) | duplicated(x, fromLast = TRUE))     4.3us    4.9us   194347.        0B    19.4   9999     1     51.4ms <int [1]> <Rprofmem [0 x 3]>  <bench_tm [10,000]> <tibble [10,000 x 3]>

(免责声明:用这么小的数据分析代码真的是徒劳的……但我很好奇。)