如何对 R 中没有 group_by 的列中的值进行重复数据删除？

Question

I have one column (a) with only two types of values: NULL and TRUE. I want to create 'a_deduplicated' column out of 'a', which is taking the first TRUE value and make the rest NULL. I do not have any other columns to group by it.

Could you please help me to write a code in R to get the 'a_deduplicated' column?

a	a_deduplicated
NA	NA
NA	NA
NA	NA
TRUE	TRUE
TRUE	NA
NA	NA
TRUE	TRUE
TRUE	NA
TRUE	NA
NA	NA
NA	NA
TRUE	TRUE
TRUE	NA
TRUE	NA
TRUE	NA

df to run in R:

df <- structure(list(a = c(NA, NA, NA, TRUE, TRUE, NA, TRUE, TRUE, TRUE, NA, NA, TRUE, TRUE, TRUE, TRUE), 
                      a_deduplicated = c(NA, NA, NA, TRUE, NA, NA, TRUE, NA, NA, NA, NA, TRUE, NA, NA, NA)), class = c("tbl_df","tbl", "data.frame"), 
 row.names = c(NA, -15L))

Answer 1

Similar to Mael's answer, but using the lag function

require(tidyverse)

df <- df %>%  
  mutate(
    new = ifelse(a & is.na(lag(a)), TRUE, NA)
  )

df
#> # A tibble: 15 × 3
#>    a     a_deduplicated new  
#>    <lgl> <lgl>          <lgl>
#>  1 NA    NA             NA   
#>  2 NA    NA             NA   
#>  3 NA    NA             NA   
#>  4 TRUE  TRUE           TRUE 
#>  5 TRUE  NA             NA   
#>  6 NA    NA             NA   
#>  7 TRUE  TRUE           TRUE 
#>  8 TRUE  NA             NA   
#>  9 TRUE  NA             NA   
#> 10 NA    NA             NA   
#> 11 NA    NA             NA   
#> 12 TRUE  TRUE           TRUE 
#> 13 TRUE  NA             NA   
#> 14 TRUE  NA             NA   
#> 15 TRUE  NA             NA

require(waldo)
#> Lade nötiges Paket: waldo
waldo::compare(df$a_deduplicated, df$new)
#> ✔ No differences

Answer 2

You can do:

transform(df, a_new = ifelse(a & is.na(diff(c(0, a))), T, NA))

      a a_deduplicated a_new
1    NA             NA    NA
2    NA             NA    NA
3    NA             NA    NA
4  TRUE           TRUE  TRUE
5  TRUE             NA    NA
6    NA             NA    NA
7  TRUE           TRUE  TRUE
8  TRUE             NA    NA
9  TRUE             NA    NA
10   NA             NA    NA
11   NA             NA    NA
12 TRUE           TRUE  TRUE
13 TRUE             NA    NA
14 TRUE             NA    NA
15 TRUE             NA    NA

Answer 3

Solution using data.table :

setDT(df)
df[, flag:=shift(a,1,type="lag")]
df[, dedup:=ifelse(is.na(flag) & a == TRUE,a,NA)]
df


       a a_deduplicated flag dedup
 1:   NA             NA   NA    NA
 2:   NA             NA   NA    NA
 3:   NA             NA   NA    NA
 4: TRUE           TRUE   NA  TRUE
 5: TRUE             NA TRUE    NA
 6:   NA             NA TRUE    NA
 7: TRUE           TRUE   NA  TRUE
 8: TRUE             NA TRUE    NA
 9: TRUE             NA TRUE    NA
10:   NA             NA TRUE    NA
11:   NA             NA   NA    NA
12: TRUE           TRUE   NA  TRUE
13: TRUE             NA TRUE    NA
14: TRUE             NA TRUE    NA
15: TRUE             NA TRUE    NA

Answer 4

df <- structure(list(a = c(NA, NA, NA, TRUE, TRUE, NA, TRUE, TRUE, TRUE, NA, NA, TRUE, TRUE, TRUE, TRUE), 
                      a_deduplicated = c(NA, NA, NA, TRUE, NA, NA, TRUE, NA, NA, NA, NA, TRUE, NA, NA, NA)), class = c("tbl_df","tbl", "data.frame"), 
 row.names = c(NA, -15L))

library(tidyverse)
df %>% 
  group_by(grp = data.table::rleid(a)) %>% 
  mutate(res = if_else(row_number() == 1, a, NA), .keep = c("unused")) %>% 
  select(-grp) %>% 
  ungroup()

#> # A tibble: 15 x 3
#>      grp a_deduplicated res  
#>    <int> <lgl>          <lgl>
#>  1     1 NA             NA   
#>  2     1 NA             NA   
#>  3     1 NA             NA   
#>  4     2 TRUE           TRUE 
#>  5     2 NA             NA   
#>  6     3 NA             NA   
#>  7     4 TRUE           TRUE 
#>  8     4 NA             NA   
#>  9     4 NA             NA   
#> 10     5 NA             NA   
#> 11     5 NA             NA   
#> 12     6 TRUE           TRUE 
#> 13     6 NA             NA   
#> 14     6 NA             NA   
#> 15     6 NA             NA

library(data.table)
setDT(df)[, res := ifelse(seq(.N) == 1, a, NA), by = rleid(a)][]
#>        a a_deduplicated  res
#>  1:   NA             NA   NA
#>  2:   NA             NA   NA
#>  3:   NA             NA   NA
#>  4: TRUE           TRUE TRUE
#>  5: TRUE             NA   NA
#>  6:   NA             NA   NA
#>  7: TRUE           TRUE TRUE
#>  8: TRUE             NA   NA
#>  9: TRUE             NA   NA
#> 10:   NA             NA   NA
#> 11:   NA             NA   NA
#> 12: TRUE           TRUE TRUE
#> 13: TRUE             NA   NA
#> 14: TRUE             NA   NA
#> 15: TRUE             NA   NA

^{Created on 2022-08-15 by the reprex package (v2.0.1)}