Extracting time from middle of string in R when string changes c

user_12752087 0
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I have a dataset containing training data for multiple athletes on different days/times. One column contains the date and start time of the session. I only wish to keep the start time in this column i.e. i want to remove "2020/01/05" and "UTC". How to I remove everything before and after the time (there are 4 million rows with varying dates/times). 

 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

Sorry this has probably already been answered somewhere.

Thanks

strsplit extract r
2 Answers

Multiple ways to do this :

1) Using 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

Here, we capture everything between whitespace and "UTC". \\1 is used as a backreference to capture the extracted value.

2) Convert to POSIXct and then format

This can be done in base R : 

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

Or using lubridate

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

data

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))
user_39629140

We can use anytime from anytime

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

Or with 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"

data

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))
user_37322710
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