如何将小时：分钟：秒转换为R中的十进制数

Question

根据我的最新问题

我想将日期格式的 data.frame 列转换为十进制数。但是，我想找到另一种方法，但使用 POSIXct 中的秒数。我们可以将此向量乘以 c(1, 1/60, 1/3600) 例如 02:20:00 将变为 2.333333 我可以将其计算到我的数据中吗？

# A tibble: 10 x 3
   trip_id start_time          end_time           
     <int> <dttm>              <dttm>             
 1       1 2020-10-05 11:11:36 2020-10-05 12:12:54
 2       2 2020-10-05 16:09:16 2020-10-05 20:00:42
 3       3 2020-10-05 09:16:33 2020-10-05 11:16:27
 4       4 2020-10-05 14:16:38 2020-10-05 14:37:38
 5       5 2020-10-05 13:08:16 2020-10-05 13:13:16
 6       6 2020-10-05 11:02:23 2020-10-05 13:04:16
 7       7 2020-10-05 13:15:19 2020-10-05 15:54:19
 8   56562 2020-10-09 11:05:25 2020-10-09 13:37:44
 9   56563 2020-10-09 14:11:30 2020-10-09 14:12:30
10   56564 2020-10-09 16:00:40 2020-10-09 16:46:58

data <- structure(list(trip_id = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 56562L,56563L, 56564L), start_time = structure(c(1601896296, 1601914156,1601889393, 1601907398, 1601903296, 1601895743, 1601903719, 1602241525,1602252690, 1602259240), tzone = "UTC", class = c("POSIXct","POSIXt")), end_time = structure(c(1601899974, 1601928042, 1601896587,1601908658, 1601903596, 1601903056, 1601913259, 1602250664, 1602252750,1602262018), tzone = "UTC", class = c("POSIXct", "POSIXt"))), row.names = c(NA,-10L), class = c("tbl_df", "tbl", "data.frame"))

提前致谢。

Answer 1

不确定我是否完全满足你的数学要求，但这对你有用吗？

library(dplyr)
data %>% 
   mutate(dec = as.numeric(strftime(start_time, format = '%H', tz = 'UTC')) * 1  
                as.numeric(strftime(start_time, format = '%M', tz = 'UTC'))/60 + 
                as.numeric(strftime(start_time, format = '%H', tz = 'UTC'))/3600)
# A tibble: 10 x 4
   trip_id start_time          end_time              dec
     <int> <dttm>              <dttm>              <dbl>
 1       1 2020-10-05 11:11:36 2020-10-05 12:12:54 11.2 
 2       2 2020-10-05 16:09:16 2020-10-05 20:00:42 16.2 
 3       3 2020-10-05 09:16:33 2020-10-05 11:16:27  9.27
 4       4 2020-10-05 14:16:38 2020-10-05 14:37:38 14.3 
 5       5 2020-10-05 13:08:16 2020-10-05 13:13:16 13.1 
 6       6 2020-10-05 11:02:23 2020-10-05 13:04:16 11.0 
 7       7 2020-10-05 13:15:19 2020-10-05 15:54:19 13.3 
 8   56562 2020-10-09 11:05:25 2020-10-09 13:37:44 11.1 
 9   56563 2020-10-09 14:11:30 2020-10-09 14:12:30 14.2 
10   56564 2020-10-09 16:00:40 2020-10-09 16:46:58 16.0

Answer 2

使用 R 基础。

df <- read.csv(text ="start_time    end_time
05/10/2020 11:11:36 05/10/2020 12:12:54
05/10/2020 16:09:16 05/10/2020 20:00:42
05/10/2020 09:16:33 05/10/2020 11:16:27
05/10/2020 14:16:38 05/10/2020 14:37:38
05/10/2020 13:08:16 05/10/2020 13:13:16
05/10/2020 11:02:23 05/10/2020 13:04:16
05/10/2020 13:15:19 05/10/2020 15:54:19
09/10/2020 11:05:25 09/10/2020 13:37:44
09/10/2020 14:11:30 09/10/2020 14:12:30
09/10/2020 16:00:40 09/10/2020 16:46:58
", sep="\t" )


df$test <- lapply(str_split(str_sub(df$start_time,12), ":"),  
                  function(x) sum(as.numeric(unlist(x)) * c(1,1/60, 1/3600)))
df


            start_time            end_time     test
1  05/10/2020 11:11:36 05/10/2020 12:12:54 11.19333
2  05/10/2020 16:09:16 05/10/2020 20:00:42 16.15444
3  05/10/2020 09:16:33 05/10/2020 11:16:27 9.275833
4  05/10/2020 14:16:38 05/10/2020 14:37:38 14.27722
5  05/10/2020 13:08:16 05/10/2020 13:13:16 13.13778
6  05/10/2020 11:02:23 05/10/2020 13:04:16 11.03972
7  05/10/2020 13:15:19 05/10/2020 15:54:19 13.25528
8  09/10/2020 11:05:25 09/10/2020 13:37:44 11.09028
9  09/10/2020 14:11:30 09/10/2020 14:12:30 14.19167
10 09/10/2020 16:00:40 09/10/2020 16:46:58 16.01111

Answer 3

您可以通过从日期时间对象中减去日期对象来使用 difftime() 函数。

例如：

difftime(data$start_time, as.Date(data$start_time))
#or to specify the units
difftime(data$start_time, as.Date(data$start_time), units= "day")

如何将小时：分钟：秒转换为R中的十进制数

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