Converting general time in R (without dates)

Question

I have a question on converting time in R.

1. First, I need to convert the times stored as characters into numerical. Times are stored as start times in one column, and finish times in another. For example: Work start time: 09:00 and work finish time: 17:00.
2. I then want to be able to calculate the time in between these times (i.e. the hours) for multiple rows of data by using a function. I.e. how many hours does someone work on an average day?
3. Finally, I want to compare early start times and early finish times to late start times and late finish times. For example, by assigning a category to these times. For example, someone who started work before 10:00 would be classified as "early starter" in one column, someone who started after 10:00 would be classified as "late starter" in another, and then someone who finished work before 17:00 would be classified as "early finisher" in one column, and then someone who finished work after 17:00 would be classified as "late finisher" in another column. Is there a way for R to recognise times in this way, when you don't have a date to assign it to?

All the advice I have read so far seems to be geared towards a particular time within a date. E.g. DD/MM/YY HH:MM. I am only concerned with a daily time.

Thanks in advance.

Answer 1

You can try with lubridate:

library(lubridate)
set.seed(4)
df <- data.frame(start=paste0(sample(7:11,10,replace = T),".00"),finish=paste0(sample(16:19,10,replace = T),".00"))
df$duration <- hm(df$finish)-hm(df$start)
df$start_cat <- ifelse(hm(df$start)<hm("10.00"),"early_starter","late_starter")
df$finish_cat <- ifelse(hm(df$finish)<hm("17.00"),"early_finisher","late_finishe")

output:

   start finish  duration     start_cat     finish_cat
1   9.00  19.00 10H 0M 0S early_starter   late_finishe
2   9.00  18.00  9H 0M 0S early_starter   late_finishe
3   9.00  18.00  9H 0M 0S early_starter   late_finishe
4  10.00  18.00  8H 0M 0S  late_starter   late_finishe
5   9.00  18.00  9H 0M 0S early_starter   late_finishe
6  11.00  16.00  5H 0M 0S  late_starter early_finisher
7   8.00  19.00 11H 0M 0S early_starter   late_finishe
8   9.00  19.00 10H 0M 0S early_starter   late_finishe
9   8.00  16.00  8H 0M 0S early_starter early_finisher
10  7.00  16.00  9H 0M 0S early_starter early_finisher

Answer 2

The easiest way to go about this is to use the hms package.

Suppose your data is something like this:

df <- data.frame(employee = LETTERS[1:4],
                 starts = c("08:00", "09:00", "10:00", "11:00"),
                 finishes = c("16:00", "17:00", "17:30", "18:00"))

df
#>   employee starts finishes
#> 1        A  08:00    16:00
#> 2        B  09:00    17:00
#> 3        C  10:00    17:30
#> 4        D  11:00    18:00

Then you can do:

library(hms)
library(dplyr)

df %>% mutate(early_start = as_hms(paste0(starts, ":00")) < as_hms("10:00:00"),
              late_finish = as_hms(paste0(finishes, ":00")) > as_hms("17:00:00"))
#>   employee starts finishes early_start late_finish
#> 1        A  08:00    16:00        TRUE       FALSE
#> 2        B  09:00    17:00        TRUE       FALSE
#> 3        C  10:00    17:30       FALSE        TRUE
#> 4        D  11:00    18:00       FALSE        TRUE

^{Created on 2021-10-16 by the reprex package (v2.0.0)}