如何用base R或dplyr高效按条件填充数据框reGrp列？

需求规则 #

reGrp列的取值逻辑如下：

• 当grpRnk≠2时，直接使用reSeq的值
• 当grpRnk=2且为首次出现时，使用reSeq的值
• 当grpRnk=2且为非首次出现时，继承上一行的reGrp值

数据示例 #

Element grpRnk reSeq reGrp   Explanation of reGrp values
1       B     NA   1.1   1.1   since grpRnk <> 2, use reSeq value 
2       R     NA   1.1   1.1   since grpRnk <> 2, use reSeq value  
3       R      2   2.0   2.0   since grpRnk = 2 and it is the first instance of grpRnk 2, use reSeq value 
4       R      2   3.0   2.0   since grpRnk = 2 and it is not the first instance of grpRnk 2, borrow the reGrp value from the row above
5       B     NA   1.2   1.2   since grpRnk <> 2, use reSeq value 
6       X      1   1.1   1.1   since grpRnk <> 2, use reSeq value 
7       X      1   1.2   1.2   since grpRnk <> 2, use reSeq value 

现有实现代码 #

library(dplyr)

data <- data.frame(
  Element = c("B","R","R","R","B","X","X"),
  grpRnk = c(NA,NA,2,2,NA,1,1),
  reSeq = c(1.1,1.1,2,3,1.2,1.1,1.2)
)

data %>% 
  mutate(reGrp = ifelse(grpRnk == 2 & is.na(lag(grpRnk)),reSeq,NA)) %>%
  mutate(reGrp = ifelse(is.na(reGrp) & grpRnk == lag(grpRnk) & grpRnk ==2,lag(reGrp),reGrp)) %>%
  mutate(reGrp = ifelse(is.na(reGrp),reSeq,reGrp))

优化方案 #

1. dplyr 单步逻辑实现 #

利用case_when定义初始值，结合fill函数完成连续值填充，仅需两次操作即可完成，避免重复改写同一列：

library(dplyr)
library(tidyr)

data %>%
  mutate(
    reGrp = case_when(
      # 非grpRnk=2的行直接取reSeq
      grpRnk != 2 ~ reSeq,
      # 首次出现的grpRnk=2行取reSeq
      grpRnk == 2 & (lag(grpRnk, default = 0) != 2) ~ reSeq,
      # 其余grpRnk=2的行暂设为NA
      TRUE ~ NA_real_
    )
  ) %>%
  # 向下填充连续grpRnk=2行的NA值
  fill(reGrp, .direction = "down")

2. base R 无依赖实现 #

通过分组标记和聚合函数直接完成赋值，无需额外包：

data_base <- data

# 生成连续grpRnk=2块的分组ID
grp_tags <- cumsum(!is.na(data_base$grpRnk) & data_base$grpRnk == 2 & 
                     (is.na(lag(data_base$grpRnk)) | lag(data_base$grpRnk) != 2))

# 按规则赋值reGrp
data_base$reGrp <- with(data_base, 
                        ifelse(grpRnk != 2, reSeq, ave(reSeq, grp_tags, FUN = function(x) x[1])))

内容的提问来源于stack exchange，提问作者Village.Idyot