用 1 填充列直到找到值，然后用 0 重复填充直到再次找到值

Question

我有以下数据框：

miniDF1 <- data.frame(Pred = c("A","A","B","A","B","B","C","A","B","C","A","A","A","A","B","A","C","B"))

    Pred
1     A
2     A
3     B
4     A
5     B
6     B
7     C
8     A
9     B
10    C
11    A
12    A
13    A
14    A
15    B
16    A
17    C
18    B

我正在尝试创建一个用 1 填充的新列，直到在 Pred 中找到“C”，然后用 0 填充新列，直到找到下一个“C”，并如此重复直到结束DF。我尝试了以下方法：

miniDF1 <- miniDF1 %>%
  mutate(Outcome = ifelse(str_detect(Pred, "C"), 1, 0)) %>% 
  fill(Outcome, .direction = 'up')

     Pred Outcome
1     A       0
2     A       0
3     B       0
4     A       0
5     B       0
6     B       0
7     C       1
8     A       0
9     B       0
10    C       1
11    A       0
12    A       0
13    A       0
14    A       0
15    B       0
16    A       0
17    C       1
18    B       0

但这只是将 1 放在有“C”所在的同一行中。

这是预期的样子：

miniDF2 <- data.frame(Pred = c("A","A","B","A","B","B","C","A","B","C","A","A","A","A","B","A","C","B"),
                     Outcome = c(1,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,0))

     Pred Outcome
1     A       1
2     A       1
3     B       1
4     A       1
5     B       1
6     B       1
7     C       1
8     A       0
9     B       0
10    C       0
11    A       1
12    A       1
13    A       1
14    A       1
15    B       1
16    A       1
17    C       1
18    B       0

我不知道如何让值相应地翻转，但我认为这就是我的代码的 fill(Outcome, .direction = 'up') 部分的目的。

Answer 1

miniDF1$Outcome <- cumsum(c(1, head(miniDF1$Pred == 'C', -1))) %% 2

miniDF1
   Pred Outcome
1     A       1
2     A       1
3     B       1
4     A       1
5     B       1
6     B       1
7     C       1
8     A       0
9     B       0
10    C       0
11    A       1
12    A       1
13    A       1
14    A       1
15    B       1
16    A       1
17    C       1
18    B       0

在 tidyverse：

library(dplyr)
miniDF1 %>%
  mutate(Outcome = cumsum(lag(Pred == 'C', default = TRUE)) %% 2)

Answer 2

看起来很麻烦，但确实需要，

i1 <- which(miniDF1$Pred == 'C')
dd <- data.frame(v1 = c(1, 0),v2 = c(i1[1], abs(diff(i1)), nrow(miniDF1)-max(i1)))

rep(dd$v1, dd$v2)
#[1] 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 0

也可以将其包装在一个函数中，

fun1 <- function(x, val){
  i1 <- which(x == val)
  dd <- data.frame(v1 = c(1, 0),
                   v2 = c(i1[1], abs(diff(i1)), nrow(miniDF1)-max(i1)))
  return(rep(dd$v1, dd$v2))
}

miniDF1$outcome <- fun1(miniDF1$Pred, 'C')

#   Pred outcome
#1     A       1
#2     A       1
#3     B       1
#4     A       1
#5     B       1
#6     B       1
#7     C       1
#8     A       0
#9     B       0
#10    C       0
#11    A       1
#12    A       1
#13    A       1
#14    A       1
#15    B       1
#16    A       1
#17    C       1
#18    B       0

Answer 3

你可以试试rleidlike

miniDF1$Outcome <- ifelse(data.table::rleid(miniDF1$Pred == "C") %% 4 %in% c(1,2), 1, 0)
miniDF1

   Pred Outcome
1     A       1
2     A       1
3     B       1
4     A       1
5     B       1
6     B       1
7     C       1
8     A       0
9     B       0
10    C       0
11    A       1
12    A       1
13    A       1
14    A       1
15    B       1
16    A       1
17    C       1
18    B       0

解释

为了方便比较，我们试试miniDF1$x <- rleid(miniDF1$Pred == "C")。

   Pred Outcome x
1     A       1 1
2     A       1 1
3     B       1 1
4     A       1 1
5     B       1 1
6     B       1 1
7     C       1 2
8     A       0 3
9     B       0 3
10    C       0 4
11    A       1 5
12    A       1 5
13    A       1 5
14    A       1 5
15    B       1 5
16    A       1 5
17    C       1 6
18    B       0 7

如果出现C 和下一个，您可以看到rleid 的结果发生变化。另外，当出现C 时，您需要从1 切换到0。

这意味着，如果x 是1 and 2 或3 and 4 或...。需要有1、0、1 等等。所以我将 x 除以 4 并得到重复的 1 and 2/ 3 and 0 值。