r - 在二进制列 r 中寻找模式

Question

我需要查找并计算在 3 个或更多连续零之后出现 1 的 ID。

这是我所拥有的一个例子:

#  ID   Jan  Feb Mar  Apr May Jun Jul Aug Sept Oct
#   1   0    0   0    1   0   0   1   1    1    0
#   2   0    0   0    0   0   0   1   0    0    0
#   3   0    0   0    0   0   0   0   0    0    1
#   4   1    0   0    1   0   1   0   1    0    1
#   5   0    0   1    0   0   1   1   0    0    1

c1<- c("ID","Jan","Feb", "Mar","Apr", "May","Jun", "Jul", "Aug", "Sept", "Oct")
c2<-  c(1,0,0,0,1,0,0,1,1,1,0)
c3<- c(2,0,0,0,0,0,0,1,0,0,0)
c4<- c(3,0,0,0,0,0,0,0,0,0,1)
c5<- c(4,1,0,0,1,0,1,0,1,0,1)
c6<- c(5,0,0,1,0,0,1,1,0,0,1)
BD<-data.frame(rbind(c2,c3,c4,c5,c6))
colnames(BD)<-c1

我期望的结果是这样的:

#  ID   Jan  Feb Mar  Apr May Jun Jul Aug Sept Oct
#   1   0    0   0    1   0   0   1   1    1    0
#   2   0    0   0    0   0   0   1   0    0    1
#   3   0    0   0    0   0   0   0   0    0    1

有人知道怎么做吗？谢谢!

Answer 1

您可以折叠成字符串并使用 grep() 来搜索模式。

k <- 3

grep(sprintf(paste0("%0", k + 1, "d"), 1), apply(d[-1], 1, paste, collapse=""))
# [1] 2 4 5 6 8

如果不需要后面的 1，您可以使用 rle()。

d
#     id Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
# c1   1   1   1   0   1   1   0   0   1   1   1   0   0
# c2   2   0   0   0   1   1   1   0   1   1   0   1   0
# c3   3   1   0   0   1   1   0   1   1   1   0   1   0
# c4   4   0   0   0   0   0   1   1   0   0   1   1   0
# c5   5   0   0   0   1   1   1   1   0   0   1   0   1
# c6   6   1   0   0   0   1   0   1   0   0   0   0   1
# c7   7   0   1   0   0   1   0   1   1   1   0   0   1
# c8   8   0   1   1   1   1   1   1   1   0   0   0   1
# c9   9   0   1   0   0   1   1   0   0   1   1   1   0
# c10 10   1   1   0   1   0   1   1   0   0   1   0   1

k <- 3
d$id[sapply(as.data.frame(t(d[-1])), function(x) any(rle(x)$lengths[rle(x)$values == 0] >= k))]
# [1] 2 4 5 6 8

数据:

set.seed(0)
d <- data.frame(id=1:10, 
                  `dimnames<-`(matrix(sample(0:1, 120, r=1), 10), 
                               list(paste0("c", 1:10), month.abb)))