这类似于我最近问过的question，但有所不同。可以说我有以下数据：
library(tidyverse)
df <- structure(list(x = c("a", "a", "a", "a", "b", "b", "b", "b",
"b", "c", "c", "c", "c", "d", "d", "e", "e", "f", "g", "g", "g",
"g", "g", "g", "g", "g"), y = c(" free", " with", " sus", " sus",
" sus", " free", " free", " sus", " free", " with", " sus",
" free", " sus", " free", " free", " with", " sus", " sus",
" free", " sus", " sus", " sus", " sus", " free", " sus", " free"
), indicator = c(0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0,
0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0)), row.names = c(NA, -26L), class = c("tbl_df",
"tbl", "data.frame"))
df
x y indicator
<chr> <chr> <dbl>
1 a " free" 0
2 a " with" 0
3 a " sus" 1
4 a " sus" 0
5 b " sus" 1
6 b " free" 0
7 b " free" 0
8 b " sus" 1
9 b " free" 0
10 c " with" 0
11 c " sus" 1
12 c " free" 0
13 c " sus" 1
14 d " free" 0
15 d " free" 0
16 e " with" 0
17 e " sus" 1
18 f " sus" 1
19 g " free" 0
20 g " sus" 0
21 g " sus" 1
22 g " sus" 0
23 g " sus" 0
24 g " free" 0
25 g " sus" 1
26 g " free" 0
我想创建一个变量，如果indicator==1，我通过对变量x进行分组来搜索前一行和后一行，如果下一个和前一个不是{ {1}}等于sus。因此，如果下一个或最后一个free之前有一个with，则它的值不会为1。如果free且它在组的最后一行或第一行中，那么我假设indicator==1不在下一行或上一行，例如组with，b，c。我想要的输出是：
e我想要一种灵活的方法，它可以遍历许多行（在 x y indicator newvariable
<chr> <chr> <dbl> <dbl>
1 a " free" 0 0
2 a " with" 0 0
3 a " sus" 1 0
4 a " sus" 0 0
5 b " sus" 1 1
6 b " free" 0 0
7 b " free" 0 0
8 b " sus" 1 1
9 b " free" 0 0
10 c " with" 0 0
11 c " sus" 1 0
12 c " free" 0 0
13 c " sus" 1 1
14 d " free" 0 0
15 d " free" 0 0
16 e " with" 0 0
17 e " sus" 1 0
18 f " sus" 1 1
19 g " free" 0 0
20 g " sus" 0 0
21 g " sus" 1 1
22 g " sus" 0 0
23 g " sus" 0 0
24 g " free" 0 0
25 g " sus" 1 1
26 g " free" 0 0
之前可以有许多sus，并且像组free一样，每个组可以有多个indicator==1 ）。我在想以下内容，但我希望g和lag遍历许多先前和随后的行：
lead我认为我无法采用与上一个问题df %>%
group_by(x) %>%
mutate(newvariable = as.integer(indicator == 1 & lag(y[y != "sus"]) =='free' & lead(y[y != "sus"]) == 'free' ))
#taken idea from previous question
#mutate(newvariable = as.integer(last(y) == 'sus' & last(y[y != "sus"]) == 'with')
相同的方法，但是如果有人有想法请找相似的东西吗？也许last？
答案 0 :(得分：1)
这似乎可以完成工作。我使用的技巧是建立一个临时数据帧，在该数据帧中，连续的相同y值序列会折叠成一条记录。
df = df %>%
group_by(x) %>%
do({
# subgroup is constant in series of records with constant x and y
df_x = mutate(., i=row_number(), subgroup=cumsum(y!=lag(y, default="")))
df_subgroups = df_x %>%
distinct(subgroup, y) %>%
mutate(
prev_distinct_y = lag(y, default="free"),
next_distinct_y = lead(y, default="free")
)
df_x = df_x %>% left_join(df_subgroups)
df_x %>% mutate(fixed_indicator = 0 + (indicator==1 & prev_distinct_y!=" with" & next_distinct_y!=" with") )
})