R data.table按条件向前填充指定行数的ID值问题

我正在处理如下格式的数据集：

Date         ID    diff
    2020-01-01   NA    NA
    2020-01-02   NA    NA
    2020-01-03   NA    NA
    2020-01-04   NA    NA
    2020-01-05   1     NA
    2020-01-06   NA    NA
    2020-01-07   NA    NA
    2020-01-08   NA    NA
    2020-01-09   2     4
    2020-01-10   NA    NA
    2020-01-11   NA    NA
    2020-01-12   NA    NA
    2020-01-13   NA    NA
    2020-01-14   NA    NA
    2020-01-15   3     6
    2020-01-16   NA    NA
    2020-01-17   NA    NA
    2020-01-18   NA    NA
    2020-01-19   4     4
    2020-01-20   NA    NA

需求是将ID值向前填充diff列数值减1行，例如当ID为4、diff为4时，需将其前3行的ID填充为4，期望得到如下结果：

Date         ID    diff
    2020-01-01   NA    NA
    2020-01-02   NA    NA
    2020-01-03   NA    NA
    2020-01-04   NA    NA
    2020-01-05   1     NA
    2020-01-06   2     NA
    2020-01-07   2     NA
    2020-01-08   2     NA
    2020-01-09   2     4
    2020-01-10   3     NA
    2020-01-11   3     NA
    2020-01-12   3     NA
    2020-01-13   3     NA
    2020-01-14   3     NA
    2020-01-15   3     6
    2020-01-16   4     NA
    2020-01-17   4     NA
    2020-01-18   4     NA
    2020-01-19   4     4
    2020-01-20   NA    NA

我尝试使用lag函数，但无法实现连续行数的填充，请问有更合适的函数或方法吗？

方法1：tidyverse工具链实现（dplyr + tidyr） #

核心思路是先标记需要填充的行范围，再对区间内的ID批量赋值：

library(dplyr)
library(tidyr)

# 给数据集添加行号，方便定位填充区间
df <- df %>% mutate(row_num = row_number())

# 提取所有有有效ID和diff的行，计算出需要填充的起始行和结束行
fill_ranges <- df %>%
  filter(!is.na(ID) & !is.na(diff)) %>%
  mutate(start_row = row_num - (diff - 1),
         end_row = row_num) %>%
  select(start_row, end_row, ID)

# 遍历每个填充区间，给对应行的ID赋值
for (i in 1:nrow(fill_ranges)) {
  df$ID[fill_ranges$start_row[i]:fill_ranges$end_row[i]] <- fill_ranges$ID[i]
}

# 移除临时添加的行号列
df <- df %>% select(-row_num)

方法2：data.table高效实现（适合大数据集） #

data.table的区间赋值语法更简洁，处理大数据时性能更优：

library(data.table)

# 转换为data.table格式
setDT(df)
# 添加行号
df[, row_num := .I]

# 生成填充规则：起始行、结束行、对应的ID值
fill_rules <- df[!is.na(ID) & !is.na(diff), .(start = row_num - (diff - 1), end = row_num, ID = ID)]

# 遍历规则，对每个区间的ID进行赋值
for (rule in fill_rules) {
  df[row_num %between% c(rule$start, rule$end), ID := rule$ID]
}

# 删除临时行号列
df[, row_num := NULL]

方法3：纯基础R实现（无需第三方包） #

如果不想依赖任何外部包，用基础R也能完成：

# 添加行号列
df$row_num <- 1:nrow(df)

# 筛选出所有有有效ID和diff的行的索引
target_indices <- which(!is.na(df$ID) & !is.na(df$diff))

# 遍历每个目标行，计算填充范围并赋值
for (idx in target_indices) {
  current_id <- df$ID[idx]
  fill_length <- df$diff[idx] - 1
  # 确保起始行不小于1，避免越界
  start_idx <- max(1, idx - fill_length)
  df$ID[start_idx:idx] <- current_id
}

# 移除临时行号列
df$row_num <- NULL

内容的提问来源于stack exchange，提问作者bziggy