我有一个ID和时间戳的数据集。
每个ID通常每分钟有12个时间戳记。 我很难为每分钟为每个时间戳添加5秒的间隔。
Actual Data Desired Format
ID Time ID Time
4466 12/1/14 19:56 4466 12/1/14 19:56:00
4466 12/1/14 19:56 4466 12/1/14 19:56:05
4466 12/1/14 19:56 4466 12/1/14 19:56:10
4466 12/1/14 19:56 4466 12/1/14 19:56:15
4466 12/1/14 19:56 4466 12/1/14 19:56:20
4466 12/1/14 19:56 4466 12/1/14 19:56:25
4466 12/1/14 19:56 4466 12/1/14 19:56:30
4466 12/1/14 19:56 4466 12/1/14 19:56:35
4466 12/1/14 19:56 4466 12/1/14 19:56:40
4466 12/1/14 19:56 4466 12/1/14 19:56:45
4466 12/1/14 19:56 4466 12/1/14 19:56:50
4466 12/1/14 19:56 4466 12/1/14 19:56:55
1136 3/23/15 23:00 1136 3/23/15 23:00:00
1136 3/23/15 23:00 1136 3/23/15 23:00:05
1136 3/23/15 23:00 1136 3/23/15 23:00:10
1136 3/23/15 23:01 1136 3/23/15 23:01:00
1136 3/23/15 23:01 1136 3/23/15 23:01:05
1136 3/23/15 23:01 1136 3/23/15 23:01:10
1136 3/23/15 23:01 1136 3/23/15 23:01:15
1136 3/23/15 23:01 1136 3/23/15 23:01:20
1136 3/23/15 23:01 1136 3/23/15 23:01:25
1136 3/23/15 23:01 1136 3/23/15 23:01:30
1136 3/23/15 23:01 1136 3/23/15 23:01:35
1136 3/23/15 23:01 1136 3/23/15 23:01:40
1136 3/23/15 23:01 1136 3/23/15 23:01:45
1136 3/23/15 23:01 1136 3/23/15 23:01:50
1136 3/23/15 23:01 1136 3/23/15 23:01:55
答案 0 :(得分：1)
我们可以使用setDT(df1)执行此操作。将'data.frame'转换为'data.table'（sprintf），按'ID'和'时间'分组，使用library(data.table)
setDT(df1)[, NewTime := sprintf("%s:%02d", Time,
seq(0, 55, by = 5)[seq_len(.N)]) , .(ID, Time)]
df1
# ID Time NewTime
# 1: 4466 12/1/14 19:56 12/1/14 19:56:00
# 2: 4466 12/1/14 19:56 12/1/14 19:56:05
# 3: 4466 12/1/14 19:56 12/1/14 19:56:10
# 4: 4466 12/1/14 19:56 12/1/14 19:56:15
# 5: 4466 12/1/14 19:56 12/1/14 19:56:20
# 6: 4466 12/1/14 19:56 12/1/14 19:56:25
# 7: 4466 12/1/14 19:56 12/1/14 19:56:30
# 8: 4466 12/1/14 19:56 12/1/14 19:56:35
# 9: 4466 12/1/14 19:56 12/1/14 19:56:40
#10: 4466 12/1/14 19:56 12/1/14 19:56:45
#11: 4466 12/1/14 19:56 12/1/14 19:56:50
#12: 4466 12/1/14 19:56 12/1/14 19:56:55
#13: 1136 3/23/15 23:00 3/23/15 23:00:00
#14: 1136 3/23/15 23:00 3/23/15 23:00:05
#15: 1136 3/23/15 23:00 3/23/15 23:00:10
#16: 1136 3/23/15 23:01 3/23/15 23:01:00
#17: 1136 3/23/15 23:01 3/23/15 23:01:05
#18: 1136 3/23/15 23:01 3/23/15 23:01:10
#19: 1136 3/23/15 23:01 3/23/15 23:01:15
#20: 1136 3/23/15 23:01 3/23/15 23:01:20
#21: 1136 3/23/15 23:01 3/23/15 23:01:25
#22: 1136 3/23/15 23:01 3/23/15 23:01:30
#23: 1136 3/23/15 23:01 3/23/15 23:01:35
#24: 1136 3/23/15 23:01 3/23/15 23:01:40
#25: 1136 3/23/15 23:01 3/23/15 23:01:45
#26: 1136 3/23/15 23:01 3/23/15 23:01:50
#27: 1136 3/23/15 23:01 3/23/15 23:01:55
将每次5秒创建的序列粘贴'时间'到创建'NewTime'列
df1 <- structure(list(ID = c(4466L, 4466L, 4466L, 4466L, 4466L, 4466L,
4466L, 4466L, 4466L, 4466L, 4466L, 4466L, 1136L, 1136L, 1136L,
1136L, 1136L, 1136L, 1136L, 1136L, 1136L, 1136L, 1136L, 1136L,
1136L, 1136L, 1136L), Time = c("12/1/14 19:56", "12/1/14 19:56",
"12/1/14 19:56", "12/1/14 19:56", "12/1/14 19:56", "12/1/14 19:56",
"12/1/14 19:56", "12/1/14 19:56", "12/1/14 19:56", "12/1/14 19:56",
"12/1/14 19:56", "12/1/14 19:56", "3/23/15 23:00", "3/23/15 23:00",
"3/23/15 23:00", "3/23/15 23:01", "3/23/15 23:01", "3/23/15 23:01",
"3/23/15 23:01", "3/23/15 23:01", "3/23/15 23:01", "3/23/15 23:01",
"3/23/15 23:01", "3/23/15 23:01", "3/23/15 23:01", "3/23/15 23:01",
"3/23/15 23:01")), .Names = c("ID", "Time"), row.names = c(NA,
-27L), class = "data.frame")
price < 1 && price >-1 && price != 0答案 1 :(得分：1)
您可以使用row_number()包中的dplyr和library(dplyr)
df %>%
group_by(ID,Time ) %>%
mutate(NTime=sprintf("%s.%02d",Time, row_number(Time)*5))
：
npycurses.ActionFormV2答案 2 :(得分：0)
或者只是：
df <- data.frame(ID = 4466, Time = "12/1/14 19:56")
df2 <- cbind(df[rep(1 : nrow(df),each=12),], ntime = strptime (paste(df$Time, ":", seq(0, 55, 5),sep=""), "%m/%d/%y %H:%M:%S", tz= "UTC"))