根据订单删除重复项

Question

In R, I'm looking to remove any instances after the first two b and c after each a (please note the numbering). I've got the following:

1   a
2   b
3   c
4   a
5   b
6   c
7   a
8   b
9   c
10  b
11  c
12  a
13  b
14  c
15  c

I'm looking to reduce it to:

1   a
2   b
3   c
4   a
5   b
6   c
7   a
8   b
9   c
12  a
13  b
14  c

I'm trying to do this within a dplyr pipe if possible. Any ideas?

Answer 1

How about this?

d <- data.frame(lets = c("a", "b", "c", "a", "b", "c", "a", "b", "c", "b", "c", "a", "b", "c", "c"))
d %>% 
  mutate(lag1 = lag(lets),
         lag2 = lag(lag1)) %>% 
  filter(is.na(lag2) |
         !(lets == lag1 | lets == lag2 | lag1 == lag2)) %>% 
  select(lets)

   lets
1     a
2     b
3     c
4     a
5     b
6     c
7     a
8     b
9     c
10    a
11    b
12    c

Answer 2

One possible solution:

df = read.table(text="1   a
                2   b
                3   c
                4   a
                5   b
                6   c
                7   a
                8   b
                9   c
                10  b
                11  c
                12  a
                13  b
                14  c
                15  c",header=F)

library(dplyr)
df %>% mutate(x=cumsum(V2=='a')) %>%
  group_by(x) %>% 
  filter(!duplicated(V2)) %>% 
  ungroup() %>%
  select(-x)

Output:

# A tibble: 12 x 2
      V1     V2
   <int> <fctr>
 1     1      a
 2     2      b
 3     3      c
 4     4      a
 5     5      b
 6     6      c
 7     7      a
 8     8      b
 9     9      c
10    12      a
11    13      b
12    14      c

Note that this removes all duplicated elements every time after an a is encountered. If you only want to remove duplicated b 's and c 's, consider : filter(!(duplicated(V2) & (V2=='b' | V2=='c')))