在R中按条件为累积和补0（data.table实现）

首先修正初始的data.table创建代码（ID值需要加引号，否则R会视为未定义变量）：

library(data.table)
DT <- data.table(
  ID = c("A","A","A","A","B","B","B","C","C","C","C"),
  date = c("2008-01-01","2009-01-01","2010-01-01","2011-01-01","1978-01-01","1982-01-01","1985-01-01","2001-01-01","2011-01-01","2015-01-01","2019-01-01"),
  Type = c(1,2,2,3,3,1,1,2,2,1,3),
  Days = c(18,333,26,57,48,10,17,212,55,64,18)
)

接下来用简洁的代码实现需求：对每个ID，分别计算Type1、Type2、Type3的累积和（非对应Type时按0处理）：

# 确保数据按ID和日期排序（若已排序可省略，但显式排序更稳妥）
DT <- DT[order(ID, date)]

# 批量生成三个累积和列
DT[, c(paste0("cumdays", 1:3)) := lapply(1:3, function(t) {
  cumsum(ifelse(Type == t, Days, 0))
}), by = ID]

代码说明 #

• 使用lapply(1:3, ...)批量处理三个Type，避免重复编写代码
• ifelse(Type == t, Days, 0)将非对应Type的Days替换为0，从根源上避免NA
• cumsum(...)按ID分组计算累积和，确保每个ID内的累积逻辑独立
• paste0("cumdays",1:3)自动生成目标列名，保持代码简洁

执行后得到的结果与期望输出一致：

ID       date Type Days cumdays1 cumdays2 cumdays3
 1:  A 2008-01-01    1   18       18        0        0
 2:  A 2009-01-01    2  333       18      333        0
 3:  A 2010-01-01    2   26       18      359        0
 4:  A 2011-01-01    3   57       18      359       57
 5:  B 1978-01-01    3   48        0        0       48
 6:  B 1982-01-01    1   10       10        0       48
 7:  B 1985-01-01    1   17       27        0       48
 8:  C 2001-01-01    2  212        0      212        0
 9:  C 2011-01-01    2   55        0      267        0
10:  C 2015-01-01    1   64       64      267        0
11:  C 2019-01-01    3   18       64      267       18

内容的提问来源于stack exchange，提问作者Hong