使用data.table：

require(data.table) ## 1.9.2+
setDT(df)[ , list(idnum = idnum, month = seq(start, end, by = "month")), by = 1:nrow(df)]

# you may use dot notation as a shorthand alias of list in j:
setDT(df)[ , .(idnum = idnum, month = seq(start, end, by = "month")), by = 1:nrow(df)]

setDT将df转换为data.table。然后，对于每一行by = 1:nrow(df)，我们根据需要创建idnum和month。

使用dplyr：

test %>%
    group_by(idnum) %>%
    summarize(start=min(start),end=max(end)) %>%
    do(data.frame(idnum=.$idnum, month=seq(.$start,.$end,by="1 month")))

注意，这里我没有为每一行生成一个start和end之间的序列，而是为每一个idnum生成一个min(start)和max(end)之间的序列。

test %>%
    rowwise() %>%
    do(data.frame(idnum=.$idnum, month=seq(.$start,.$end,by="1 month")))

更新2

在purrr（0.3.0）和dplyr（0.8.0）的新版本中，可以使用map2完成此操作

library(dplyr)
library(purrr)
 test %>%
     # sequence of monthly dates for each corresponding start, end elements
     transmute(idnum, month = map2(start, end, seq, by = "1 month")) %>%
     # unnest the list column
     unnest %>% 
     # remove any duplicate rows
     distinct

更新

根据@阿南达·马赫托的评论

res1 <- melt(setNames(lapply(1:nrow(test), function(x) seq(test[x, "start"],
 test[x, "end"], by = "1 month")), test$idnum))

另外，

res2 <- setNames(do.call(`rbind`,
          with(test, 
          Map(`expand.grid`,idnum,
          Map(`seq`, start, end, by='1 month')))), c("idnum", "month"))

  head(res1)
 #  idnum      month
 #1    17 1993-01-01
 #2    17 1993-02-01
 #3    17 1993-03-01
 #4    17 1993-04-01
 #5    17 1993-05-01
 #6    17 1993-06-01

使用dplyr和tidyr为每行创建序列的一个选项可以是：

df %>%
 rowwise() %>%
 transmute(idnum,
           date = list(seq(start, end, by = "month"))) %>%
 unnest(date)

  idnum date      
   <int> <date>    
 1    17 1993-01-01
 2    17 1993-02-01
 3    17 1993-03-01
 4    17 1993-04-01
 5    17 1993-05-01
 6    17 1993-06-01
 7    17 1993-07-01
 8    17 1993-08-01
 9    17 1993-09-01
10    17 1993-10-01
# … with 26 more rows

或者使用分组ID创建序列：

df %>%
 group_by(idnum) %>%
 transmute(date = list(seq(min(start), max(end), by = "month"))) %>%
 unnest(date)

或者当目标是每个ID仅创建一个唯一序列时：

df %>%
 group_by(idnum) %>%
 summarise(start = min(start),
           end = max(end)) %>%
 transmute(date = list(seq(min(start), max(end), by = "month"))) %>%
 unnest(date)

   date      
   <date>    
 1 1993-01-01
 2 1993-02-01
 3 1993-03-01
 4 1993-04-01
 5 1993-05-01
 6 1993-06-01
 7 1993-07-01
 8 1993-08-01
 9 1993-09-01
10 1993-10-01
11 1993-11-01
12 1993-12-01

tidyverse答案
数据

df <- structure(list(idnum = c(17L, 17L, 17L), start = structure(c(8401, 
8401, 8401), class = "Date"), end = structure(c(8765, 8765, 8765
), class = "Date")), class = "data.frame", .Names = c("idnum", 
"start", "end"), row.names = c(NA, -3L))

回答和输出

library(tidyverse)
df %>%
  nest(start, end) %>%
  mutate(data = map(data, ~seq(unique(.x$start), unique(.x$end), 1))) %>%
  unnest(data)

# # A tibble: 365 x 2
   # idnum       data
   # <int>     <date>
 # 1    17 1993-01-01
 # 2    17 1993-01-02
 # 3    17 1993-01-03
 # 4    17 1993-01-04
 # 5    17 1993-01-05
 # 6    17 1993-01-06
 # 7    17 1993-01-07
 # 8    17 1993-01-08
 # 9    17 1993-01-09
# 10    17 1993-01-10
# # ... with 355 more rows

还有一种tidyverse方法是使用tidyr::expand：

library(dplyr, warn = FALSE)
library(tidyr)

df |> 
  mutate(
    row = row_number()
  ) |> 
  group_by(row) |> 
  expand(idnum, date = seq(start, end, "month")) |> 
  ungroup() |> 
  select(-row)
#> # A tibble: 36 × 2
#>    idnum date      
#>    <int> <date>    
#>  1    17 1993-01-01
#>  2    17 1993-02-01
#>  3    17 1993-03-01
#>  4    17 1993-04-01
#>  5    17 1993-05-01
#>  6    17 1993-06-01
#>  7    17 1993-07-01
#>  8    17 1993-08-01
#>  9    17 1993-09-01
#> 10    17 1993-10-01
#> # … with 26 more rows