如果窗口在CLARME或集团之外，则有条件领先滞后滚动总和返回NA

我需要获取y列中每个1的两侧x列的值和.y中每次出现1之前和之后，窗口分别为1:4和5:8.但是，如果窗口超出了pdf的限制，则返回0.这是一个问题，因为0可能是有效的结果.完整数据具有多个组(id)，但希望此repex足以帮助创建可扩展的解决方案.

我以前从未使用过zoo，因此无法弄清楚如何合并na.rm = TRUE并返回NA，因为窗口完全在DF范围之外.解决方案不必使用zoo，但如果可能的话，我更喜欢基于动词的答案.

Data and packages:

library(dplyr)
library(zoo)

set.seed(1)
df <- data.frame(id = rep("A", 40),
                 x = sample(0:3, 40, replace = TRUE),
                 y = 0)

df[c(2, 8, 9, 30, 33, 39), "y"] <- 1

What I tried:

w <- 4

df %>%
  mutate(bf2 = ifelse(y == 1, rollapply(lag(x, 5), width = w, sum, fill = NA, align = "right", partial = TRUE, na.rm = TRUE), NA),
         bf1 = ifelse(y == 1, rollapply(lag(x, 1), width = w, sum, fill = NA, align = "right", partial = TRUE, na.rm = TRUE), NA),
         af1 = ifelse(y == 1, rollapply(lead(x, 1), width = w, sum, fill = NA, align = "left", partial = TRUE, na.rm = TRUE), NA),
         af2 = ifelse(y == 1, rollapply(lead(x, 5), width = w, sum, fill = NA, align = "left", partial = TRUE, na.rm = TRUE), NA)) %>%
  filter(y == 1)

  id x y bf2 bf1 af1 af2
1  A 3 1   0   0   3   6
2  A 2 1   5   3   6   1
3  A 1 1   5   5   5   2
4  A 1 1   2   1   5   7
5  A 0 1   1   3   8   4
6  A 1 1   5   7   1   0

Desired output:

  id x y bf2 bf1 af1 af2
1  A 3 1  NA   0   3   6
2  A 2 1   5   3   6   1
3  A 1 1   5   5   5   2
4  A 1 1   2   1   5   7
5  A 0 1   1   3   8   4
6  A 1 1   5   7   1  NA