根据匹配条件,对序列进行计数,直到下一次匹配

这里是一个使用dplyr包中的函数和数据中的rleid函数的解决方案。表包。dt2是最终输出。

# Load packages
library(dplyr)

# Process the data
dt2 <- dt %>%
  mutate(small_minus_1 = ifelse(roll.z.score.n3 < -1, 1, 0)) %>%
  mutate(Group = data.table::rleid(small_minus_1)) %>%
  mutate(Group = ifelse(Group %% 2 == 1, Group - 1, Group)) %>%
  group_by(Group) %>%
  mutate(sig.count = ifelse(is.na(small_minus_1), 0, 1:n())) %>%
  ungroup() %>%
  select(roll.z.score.n3, sig.count)

dt <- structure(list(roll.z.score.n3 = c(NA, NA, -1.135974424, 0.193311168, 
0.714285714, -1.148753543, -0.942160394, 0.695763683, 1.140646864, 
0.985196899, -0.768766574, -1.011293858, -0.516703612, -1.120897077, 
1.091089451, 0.968364052, 0.872871561, 1.099524999, 0.918397948
)), .Names = "roll.z.score.n3", row.names = c("1", "2", "3", 
"4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", 
"16", "17", "18", "19"), class = "data.frame")

a=sapply(dt,function(i) which(i< -1))
b=cbind.data.frame(dt,count=unlist(lapply(diff(c(1,a,nrow(dt)+1)),function(i) 1:i)))
b$count[is.na(b$roll.z.score.n3)]=0

a=with(dt,diff(c(1,which(roll.z.score.n3< -1),nrow(b)+1)))
b=cbind.data.frame(dt,count=unlist(sapply(a,function(i) 1:i)))
b$count[is.na(b$roll.z.score.n3)]=0

快速演示,不确定它是否是您想要的:

set.seed(63112)
x <- rnorm(20, -1, 1)
df <- data.frame(x = x, count = NA)

counter <- 0
seen_neg_one <- F
for (i in 1:nrow(df)) {
    if (df$x[i] < -1) {
        counter <- 1
        df$count[i] <- 1   
        seen_neg_one <- T
    } else if (is.na(df$x[i])) {
        df$count[i] <- 0
    } else {
        if (!seen_neg_one) {
            df$count[i] <- 0
        } else {
            counter <- counter + 1
            df$count[i] <- counter
        }
    }
}