r - 重复行，以便所有列保持不变，但一列依次变大

Question

使用 R 我当前的表如下所示:

C1    C2                          C3
1     2011-02-01 04:30:00         4
2     2011-02-01 04:45:00         3
3     2011-02-01 05:00:00         5
4     2011-02-01 05:15:00         6

我希望它看起来像这样:

C1    C2                          C3       C4
1     2011-02-01 04:30:00         4        2011-02-01 04:30:00
2     2011-02-01 04:30:00         4        2011-02-01 04:35:00
3     2011-02-01 04:30:00         4        2011-02-01 04:40:00
4     2011-02-01 04:45:00         3        2011-02-01 04:45:00
5     2011-02-01 04:45:00         3        2011-02-01 04:50:00 
6     2011-02-01 04:45:00         3        2011-02-01 04:55:00
7     2011-02-01 05:00:00         5        2011-02-01 05:00:00
8     2011-02-01 05:00:00         5        2011-02-01 05:05:00

等等。等等，基本上只是想制作另一列，该列以五分钟的间隔上升，但与 C2 中的间隔匹配。 我正在考虑类似rep()函数的东西，但这意味着C2中的间隔总是一致的，但它们可能不是一致的。我真的在寻找能够根据 C2 中的间隔进行五分钟间隔的东西。

对此问题的任何帮助或反馈将不胜感激。谢谢

Answer 1

我们可以使用map2通过获取Datetime转换后的'C2的seq影响来创建一个list列'，length 由“C3”的相应元素按 5 分钟间隔指定，并取消嵌套 list 列

library(tidyverse)
df1 %>% 
  mutate(C4 = map2(lubridate::ymd_hms(C2), C3, ~ seq(.x, length.out = .y, by = '5 min'))) %>% 
  unnest
#  C1                  C2 C3                  C4
#1   1 2011-02-01 04:30:00  4 2011-02-01 04:30:00
#2   1 2011-02-01 04:30:00  4 2011-02-01 04:35:00
#3   1 2011-02-01 04:30:00  4 2011-02-01 04:40:00
#4   1 2011-02-01 04:30:00  4 2011-02-01 04:45:00
#5   2 2011-02-01 04:45:00  3 2011-02-01 04:45:00
#6   2 2011-02-01 04:45:00  3 2011-02-01 04:50:00
#7   2 2011-02-01 04:45:00  3 2011-02-01 04:55:00
#8   3 2011-02-01 05:00:00  5 2011-02-01 05:00:00
#9   3 2011-02-01 05:00:00  5 2011-02-01 05:05:00
#10  3 2011-02-01 05:00:00  5 2011-02-01 05:10:00
#11  3 2011-02-01 05:00:00  5 2011-02-01 05:15:00
#12  3 2011-02-01 05:00:00  5 2011-02-01 05:20:00
#13  4 2011-02-01 05:15:00  6 2011-02-01 05:15:00
#14  4 2011-02-01 05:15:00  6 2011-02-01 05:20:00
#15  4 2011-02-01 05:15:00  6 2011-02-01 05:25:00
#16  4 2011-02-01 05:15:00  6 2011-02-01 05:30:00
#17  4 2011-02-01 05:15:00  6 2011-02-01 05:35:00
#18  4 2011-02-01 05:15:00  6 2011-02-01 05:40:00

---

或者使用base R中的Map，获取日期时间序列的列表，其逻辑与上面相同。通过根据“lst1”的长度复制行序列来扩展原始数据集，并创建新列“C4”

lst1 <- Map(function(x, y) seq(x, length.out = y, by = '5 min'),
    as.POSIXct(df1$C2), df1$C3)
df2 <- df1[rep(seq_len(nrow(df1)), lengths(lst1)),]
df2$C4 <- do.call(c, lst1)
row.names(df2) <- NULL

---

如果条件基于“C2”的下一个值

df1 %>% 
   mutate(C4 = map2(ymd_hms(C2), lubridate::ymd_hms(lead(C2, default = last(C2))),
        seq, by = '5 min')) %>%
   unnest %>% 
   group_by(C1) %>% 
   slice(-1)
# A tibble: 9 x 4
# Groups:   C1 [3]
#     C1 C2                     C3 C4                 
#  <int> <chr>               <int> <dttm>             
#1     1 2011-02-01 04:30:00     4 2011-02-01 04:35:00
#2     1 2011-02-01 04:30:00     4 2011-02-01 04:40:00
#3     1 2011-02-01 04:30:00     4 2011-02-01 04:45:00
#4     2 2011-02-01 04:45:00     3 2011-02-01 04:50:00
#5     2 2011-02-01 04:45:00     3 2011-02-01 04:55:00
#6     2 2011-02-01 04:45:00     3 2011-02-01 05:00:00
#7     3 2011-02-01 05:00:00     5 2011-02-01 05:05:00
#8     3 2011-02-01 05:00:00     5 2011-02-01 05:10:00
#9     3 2011-02-01 05:00:00     5 2011-02-01 05:15:00

或使用data.table中的方法的类似选项

library(data.table)
setDT(df1)[, C2 := as.POSIXct(C2)][, C4 := list(Map(seq, 
   MoreArgs = list(by = '5 min'), C2, shift(C2, type = 'lead',
      fill = last(C2))))][, unnest(.SD)][, .SD[-1], by = C1]

数据

df1 <- structure(list(C1 = 1:4, C2 = c("2011-02-01 04:30:00", "2011-02-01 04:45:00", 
"2011-02-01 05:00:00", "2011-02-01 05:15:00"), C3 = c(4L, 3L, 
 5L, 6L)), class = "data.frame", row.names = c(NA, -4L))