根据一系列连续的时间数据计算日期(R)

Question

我有一列连续的时间(小时、分钟、秒(，间隔不相等，持续了好几天。

例如：

library(lubridate)    
df <- data.frame(Data  = c(1:10),
Time = hms(c("10:00:00","15:38:44","22:12:37",
"23:59:00","00:07:28","04:56:00",
"08:01:25","12:10:54","16:08:43",
"20:44:44")))

我想创建一个包含日期和时间的新列，假设第一个数据点是在2020年1月1日10:00:00获取的。因此，例如，数据点8将得到"0"；2020年1月2日12:10:54"；。

我解决这个问题的尝试不值得在这里发布，有人有什么建议吗？

非常感谢！

Answer 1

这有点棘手，但这里有一个解决方案：

require(tidyverse)    
require(lubridate)
dataf <- tibble(Data  = c(1:16),
Time = hms(c("10:00:00","15:38:44","22:12:37",
"23:59:00","00:07:28","04:56:00",
"08:01:25","12:10:54","16:08:43",
"20:44:44","00:07:28","04:56:00",
"08:01:25","12:10:54","16:08:43",
"20:44:44")))
dataf_2 <- dataf %>% 
mutate(Data = Data + 1,
Time_n = as.numeric(Time)) %>% 
select(Data, Time_n)
dataf %>%
left_join(dataf_2, by = c("Data")) %>%
replace_na(list(Time_n = 0)) %>% 
mutate(starting_date = dmy("01/01/2020")) %>% 
mutate(change_day = if_else(as.numeric(Time) >= Time_n, 0, 1)) %>% 
arrange(Data) %>% 
mutate(cumsum=cumsum(change_day),
t = paste(hour(Time), minute(Time), second(Time))) %>% 
mutate(final_date = ymd_hms(paste(starting_date + days(cumsum), t))) %>% 
select(Data, Time, final_date)

它将产生这样的输出：

# A tibble: 16 x 3
Data Time        final_date         
<dbl> <Period>    <dttm>             
1     1 10H 0M 0S   2020-01-01 10:00:00
2     2 15H 38M 44S 2020-01-01 15:38:44
3     3 22H 12M 37S 2020-01-01 22:12:37
4     4 23H 59M 0S  2020-01-01 23:59:00
5     5 7M 28S      2020-01-02 00:07:28
6     6 4H 56M 0S   2020-01-02 04:56:00
7     7 8H 1M 25S   2020-01-02 08:01:25
8     8 12H 10M 54S 2020-01-02 12:10:54
9     9 16H 8M 43S  2020-01-02 16:08:43
10    10 20H 44M 44S 2020-01-02 20:44:44

Answer 2

这应该有效：

df %>% 
mutate(
lag = lag(Time),
diff = Time - lag,
add_days = if_else(is.na(diff), 0, if_else(diff < 0, 1, 0)),
add_days_cum = cumsum(add_days),
date = ymd("2020-01-01") + days(add_days_cum) + Time
)