如何将data.table中指定序列的多行替换为单行？

没问题，我来帮你搞定这个序列匹配和替换的需求！下面我会分别给出data.table和data.frame两种实现方案，完全贴合你的预期结果。

基于data.table的解决方案 #

首先先确认原始数据：

library(data.table)
DT <- data.table(id=c(1,1,1,1,2,2,2,2), 
                 place = c("a","b","c","d","a","b","d","e"), 
                 seq = c(1,2,3,4,1,2,3,4))
setorder(DT, id, seq)

核心思路 #

针对每个id分组，先识别出连续的"b","c","d"序列，标记出需要替换的"b"和"c"行，然后移除这两行，添加一行place="z"的记录（沿用"b"行的seq值），最后重新排序。

具体代码 #

# 按id分组，获取下一个、下下个位置的place值，用于匹配序列
DT[, `:=`(
  next_place = shift(place, type = "lead"),
  next_next_place = shift(place, n = 2, type = "lead")
), by = id]

# 标记出属于目标序列的b行（后面跟着c和d）
DT[, is_target_b := place == "b" & next_place == "c" & next_next_place == "d"]
# 标记出属于目标序列的c行（前面是b，后面是d）
DT[, is_target_c := place == "c" & shift(place) == "b" & next_place == "d"]

# 生成最终结果：保留非目标行 + 添加替换后的z行
result <- rbind(
  # 保留不需要替换的行（排除目标b和c）
  DT[!is_target_b & !is_target_c, .(id, place, seq)],
  # 添加替换后的z行，取目标b行的信息
  DT[is_target_b, .(id, place = "z", seq)]
)

# 重新按id和seq排序，得到预期格式
setorder(result, id, seq)
print(result)

运行后输出和你给出的DT.tobe完全一致：

id place seq
1:  1     a   1
2:  1     z   2
3:  1     d   4
4:  2     a   1
5:  2     b   2
6:  2     d   3
7:  2     e   4

基于data.frame（dplyr）的解决方案 #

如果你习惯用data.frame和dplyr语法，也可以用下面的实现：

library(dplyr)

DT_df <- as.data.frame(DT)

result_df <- DT_df %>%
  group_by(id) %>%
  # 标记目标序列的行
  mutate(
    next_place = lead(place),
    next_next_place = lead(place, n = 2),
    is_target_b = place == "b" & next_place == "c" & next_next_place == "d",
    is_target_c = place == "c" & lag(place) == "b" & next_place == "d"
  ) %>%
  # 保留非目标行
  filter(!is_target_b & !is_target_c) %>%
  # 添加替换后的z行
  bind_rows(
    DT_df %>%
      group_by(id) %>%
      filter(place == "b" & lead(place) == "c" & lead(place, n=2) == "d") %>%
      mutate(place = "z") %>%
      select(id, place, seq)
  ) %>%
  # 重新排序
  arrange(id, seq) %>%
  ungroup()

print(result_df)

这个方案的逻辑和data.table版本完全一致，输出结果也相同。

内容的提问来源于stack exchange，提问作者User800701