R语言dataframe操作：删除delivery=0且下一行delivery=1的对应行

# 补全后的完整原始数据
df <- data.frame(
  client_id = c(7, 7, 7, 17, 17, 17, 19, 19, 19,19,19), 
  delivery = c(0,0,1,0,0,1,0,0,1,1,1), 
  order_id = c("snshw","nsnsjjw","smssjjw","jsnn","ksksues","ncbf","nsngw","manhsm","opltkg","snsjue","msmssje"), 
  which_order = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 4, 5)
)

方案1：dplyr实现（逻辑清晰易读） #

核心用lead()函数取当前行的下一行delivery值做判断，默认按client_id分组处理避免跨用户的相邻行判断错误：

library(dplyr)

df_result <- df %>%
  group_by(client_id) %>%
  # 过滤掉"当前delivery为0且下一行delivery为1"的行，默认最后一行下一行值设为0避免误删
  filter(!(delivery == 0 & lead(delivery, default = 0) == 1)) %>%
  ungroup()

如果确认不需要严格按client_id分组，直接删掉group_by(client_id)和对应的ungroup()行即可，逻辑同样生效，执行结果和你给出的预期完全一致。

方案2：base R实现（无需加载第三方包） #

用ave()函数按client_id分组做条件判断：

# 生成标记列：TRUE代表需要删除的行
df$to_drop <- ave(df$delivery, df$client_id, FUN = function(x) {
  x == 0 & c(tail(x, -1), 0) == 1
})

df_result <- df[df$to_drop == FALSE, setdiff(names(df), "to_drop")]

内容的提问来源于stack exchange，提问作者Zofia Smoleń