当字符串改变c时，从R中的字符串中间提取时间

Question

我有一个数据集，包含多个运动员在不同日期/时间的训练数据。其中一列包含会话的日期和开始时间。我只想在本栏中保留开始时间，即我想删除"2020/01/05"one_answers"UTC"。如何删除时间前后的所有内容(有400万行，日期/时间各不相同(。

start.time
1 2020/01/05 21:30:04 UTC 
2 2020/01/05 21:30:04 UTC 
3 2020/01/05 21:30:04 UTC 
4 2020/01/05 21:30:04 UTC 
5 2020/01/05 21:30:04 UTC 
6 2020/01/05 21:30:04 UTC 

抱歉，这个问题可能已经在某个地方得到了回答。

感谢

Answer 1

实现这一点的多种方法：

1( 使用regex

df$time <- sub('.*\s+(.*) UTC', '\1', df$start.time)
df
#               start.time     time
#1 2020/01/05 21:30:04 UTC 21:30:04
#2 2020/01/05 21:30:04 UTC 21:30:04
#3 2020/01/05 21:30:04 UTC 21:30:04
#4 2020/01/05 21:30:04 UTC 21:30:04
#5 2020/01/05 21:30:04 UTC 21:30:04
#6 2020/01/05 21:30:04 UTC 21:30:04

在这里，我们捕获空白和"UTC"之间的所有内容。CCD_ 2被用作后参考以捕获提取的值。

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2(转换为POSIXct，然后转换为format

这可以在基本R:中完成

format(as.POSIXct(df$start.time, format = "%Y/%m/%d %T"), "%T")

或使用lubridate

format(lubridate::ymd_hms(df$start.time), "%T")

数据

df <- structure(list(start.time = structure(c(1L, 1L, 1L, 1L, 1L, 1L
), .Label = "2020/01/05 21:30:04 UTC", class = "factor")), 
class = "data.frame", row.names = c(NA,-6L))

Answer 2

我们可以从anytime使用anytime

library(anytime)
format(anytime(df$start.time), "%T")

---

或使用as.ITime

library(data.table)
as.ITime(df$start.time)
#[1] "21:30:04" "21:30:04" "21:30:04" "21:30:04" "21:30:04" "21:30:04"

数据

df <- structure(list(start.time = structure(c(1L, 1L, 1L, 1L, 1L, 1L
), .Label = "2020/01/05 21:30:04 UTC", class = "factor")), 
class = "data.frame", row.names = c(NA,-6L))