【发布时间】:2019-04-25 22:58:54
【问题描述】:
我正在尝试结合日期和时间。这些是从文件中导入的,如下所示:
library(tidyverse)
library(lubridate)
bookings <- structure(list(booking_date = structure(c(1549670400, 1550275200,
1550880000, 1551484800, 1552089600, 1552694400), class = c("POSIXct",
"POSIXt"), tzone = "UTC"), start_time = structure(c(-2209043700,
-2209043700, -2209043700, -2209043700, -2209043700, -2209043700
), class = c("POSIXct", "POSIXt"), tzone = "UTC")), row.names = c(NA,
-6L), class = c("tbl_df", "tbl", "data.frame"))
看起来像这样:
# A tibble: 6 x 2
booking_date start_time
<dttm> <dttm>
1 2019-02-09 00:00:00 1899-12-31 08:45:00
2 2019-02-16 00:00:00 1899-12-31 08:45:00
3 2019-02-23 00:00:00 1899-12-31 08:45:00
4 2019-03-02 00:00:00 1899-12-31 08:45:00
5 2019-03-09 00:00:00 1899-12-31 08:45:00
6 2019-03-16 00:00:00 1899-12-31 08:45:00
显然start_time 列中的日期是错误的。它应该与预订日期相结合,因此第一行应为2019-02-09 08:45:00。
最好的方法是什么?我已经尝试过这个(based on this other answer),这在我的情况下并没有真正起作用。
bookings %>%
select(booking_date, start_time) %>%
mutate(time_2 = as.character(start_time)) %>%
mutate(time_3 = str_sub(time_2, -8, -1)) %>%
mutate(booking_start = dmy(paste(booking_date, time_3)))
谢谢。
【问题讨论】:
标签: r dplyr tidyverse lubridate stringr