R 计算字母表在特定字母之前出现的次数

我想在每个Id级别出现第一个“C”之前计算“I”的出现次数。我尝试过这段代码，但可以计算列中的所有“I”。 我试过的代码

library(plyr)
Impres = ddply(df, .(Id), summarize, No_of_I_before_First_C = length(which(Character == "I")))

样本数据

Id  Character
1     I
1     I
1     C
1     I
2     I
2     C

输出应该是这样的

Id  Count_Of_I_before_First_C
1     2
2     1

这里有一个想法

first1 <- function(x, letter){
           which(x == letter)[1]-1
           }

aggregate(Character ~ Id, df, first1, 'C')
#  Id Character
#1  1         2
#2  2         1

first1
结果:

# A tibble: 2 × 2
     Id Count_Of_I_before_First_C
  <dbl>                     <int>
1     1                         2
2     2                         1

#一个tible:2×2
在第一次之前的Id计数
1     1                         2
2     2                         1
以下是
数据。表
解决方案：

library(data.table)
dt <- data.table(Id = c(1,1,1,1,2,2), Character = c('I', 'I', 'C', 'I', 'I', 'C'))
dt[, cnt.c := cumsum(Character == "C"), by = Id]
res <- dt[cnt.c == 0, .(Count_Of_I_before_First_C = length(Character)), by = Id]

库（data.table）
dt也许：

library(dplyr)

rlei <- function(x) {
  r <- rle(x)
  I <- which(r$values=="I")
  C <- which(r$values=="C")
  r$lengths[which(I<C)][1]
}

group_by(df, Id) %>% 
  summarise(Count_Of_I_before_First_C=rlei(.$Character))

库（dplyr）
rlei这将是相当大的缓慢dataset@Bulat我只是跟着问题的
agregate
标签走（即没有包裹）。我知道
dplyr
和
data.table
都有更有效的方法
df %>% 
  group_by(Id) %>% 
  summarise(Count_Of_I_before_First_C = foo(Character))

# A tibble: 2 × 2
     Id Count_Of_I_before_First_C
  <dbl>                     <int>
1     1                         2
2     2                         1

library(data.table)
dt <- data.table(Id = c(1,1,1,1,2,2), Character = c('I', 'I', 'C', 'I', 'I', 'C'))
dt[, cnt.c := cumsum(Character == "C"), by = Id]
res <- dt[cnt.c == 0, .(Count_Of_I_before_First_C = length(Character)), by = Id]

library(dplyr)

rlei <- function(x) {
  r <- rle(x)
  I <- which(r$values=="I")
  C <- which(r$values=="C")
  r$lengths[which(I<C)][1]
}

group_by(df, Id) %>% 
  summarise(Count_Of_I_before_First_C=rlei(.$Character))