使用regex

pattern <- "(\\d+).+\\b\\1\\b.*"
df[!grepl(pattern, do.call(paste, df)), ]

#      Var1 Var2  Var3
# 3 2  5  9 3  7 6  19

与dobur相同

df |> 
  rowwise() |> 
  filter(!grepl(pattern, paste0(c_across(everything()), collapse = " ")))

您可以像下面这样使用anyDuplicated尝试apply

> df[!apply(df, 1, \(x) anyDuplicated(scan(text = x, quiet = TRUE))), ]
     Var1 Var2  Var3
3 2  5  9 3  7 6  19

奇怪的数据结构。在这里，我将其转换为长格式，将字符串分隔为更长格式的行，重复数据删除，然后将其放回初始格式：

library(dplyr)
library(tidyr)
df |>
  mutate(
    row = row_number()
  ) |>
  pivot_longer(-row) |>
  separate_longer_delim(value, delim = stringr::regex(" +")) |>
  filter(!anyDuplicated(value), .by = row) |>
  summarize(value = paste(value, collapse = "  "), .by = name) |>
  pivot_wider(names_from = name, values_from = value)
# # A tibble: 1 × 3
#   Var1    Var2  Var3 
#   <chr>   <chr> <chr>
# 1 2  5  9 3  7  6  19

library(tidyverse)

df <- data.frame(
  Var1 = c(
    "1  2  3", "1  6",
    "2  5  9", "1  5  3"
  ),
  Var2 = c(
    "1  2", "1  6  0  5",
    "3  7", "1  5"
  ),
  Var3 = c("2  8", "1  3", "6  19", "1  3")
)
# take a string of numbers and spaces and get a vector of numbers
myfunc <- function(x) {
  parse_number(unlist(strsplit(x, split = " ", fixed = TRUE))) |>
    na.omit() |>
    as.integer()
}
# test 
myfunc("1 2 3") |> str()

  # calculate
  (df2 <- mutate(rowwise(df),
                 fstring = paste0(c_across(everything()), collapse = " "),
                 nums = list(myfunc(fstring))
  ))
  
  # analyse
  (keepvec <- map_lgl(df2$nums, \(x)!anyDuplicated(x)))
  
  # final_result
  df |> filter(keepvec)