dplyr或lubridate的阅读时间

Question

在我的csv文件中，我有一个格式为“0 days 00：00：17.740000000”的字符列，代表持续时间。我怎样才能将dplyr（read_csv）作为一个持续时间读取它，或者在用lubridate读取之后进行转换？

Answer 1

假定所有的日期都有相似的结构。 模拟数据：

set.seed(12345) 

days <- sample(0:20, 10, replace = T) 
hours <- sample(0:24, 10, replace = T) 
mins <- sample(0:60, 10, replace = T) 
secs <- sample(0:60, 10, replace = T) + runif(10, 0, 1) 

DF <- data.frame(date = sprintf('%s days %s:%s:%s', days, hours, mins, secs)) 

然后我们分开的日子 “手动”，进行类型转换和重组一切都变成period：

library('tidyr') 
library('lubridate') 
library('dplyr') 

DF %>% 
    separate(date, c('days', 'time'), sep = ' days ') %>% 
    mutate(elapsed = period(days = as.numeric(days)) + hms(time)) 

    days    time     elapsed 
1 15 00:27:48.782193281 15d 0H 27M 48.782193281S 
2 18 03:19:0.429198820 18d 3H 19M 0.42919882S 
3 15 18:58:11.927273975 15d 18H 58M 11.927273975S 
4 18 00:43:41.773243225 18d 0H 43M 41.773243225S 
5  9 09:39:22.259681247 9d 9H 39M 22.259681247S 
6  3 11:23:22.321224673 3d 11H 23M 22.321224673S 
7  6 09:42:52.060195157 6d 9H 42M 52.060195157S 
8 10 10:33:55.043456454 10d 10H 33M 55.043456454S 
9 15 04:13:37.055053818 15d 4H 13M 37.055053818S 
10 20 23:29:8.625542797 20d 23H 29M 8.625542797S