I have data below for work hours which I need to compare - start and stop with date and time. I first extract the time portion of each as start and stop variables, then use the chron package to change them from factor data to something I can compare more easily.
require(chron)
eg_data3 <- data.frame(
id = c('42', '42', '42', '42', '42'),
time_in = as.factor(c('11/5/2017 13:52', '11/4/2017 14:25', '11/5/2017 15:30', '11/5/2017 17:10', '11/6/2017 18:20')),
time_out = as.factor(c('11/5/2017 13:59', '11/4/2017 14:59', '11/5/2017 16:00', '11/5/2017 17:45', '11/6/2017 18:50')))
eg_data3$start_time <- substring(strptime(eg_data3$time_in, format = "%m/%d/%Y %H:%M"),12,19)
eg_data3$end_time <- substring(strptime(eg_data3$time_out, format = "%m/%d/%Y %H:%M"),12,19)
eg_data3$end_time <- chron(times = eg_data3$end_time)
eg_data3$start_time <- chron(times = eg_data3$start_time)
Next, I generate another variable which compares the difference between stop time 1, and start time 2, IE stop time in row 1 with start time in row 2, to see the gap between them.
require(dplyr)
eg_data3 <- eg_data3 %>% group_by(id) %>% mutate(diff_outX0_inX1 = start_time - lag(end_time))
When I do this, the variable is formatted as a decimal. I cannot for the life of me get it to display as hh:mm:ss. I have tried specifying out.format as hh:mm:ss in chron, changing time_in / time_out to numeric and character before and after extraction and applying chron(times), changing the format of the diff_ variable after, etc.
What seems like a very simple question -
How do I get the result comparison (diff_outX0_inX1) variable to display as time, either hh:mm or hh:mm:ss ?? I know the formula to convert fractional days into minutes in Excel, but I'd prefer to not write out a two step function, I assume it's a simple formatting issue.
Any help is appreciated.
EDIT - got flagged as a duplicate...OK. I asked if there was a way to do this that did not involve writing a function. The answer that was linked involves a function. First comment provided a clean simple answer. I can reproduce the answer in the comment, I could not reproduce the function myself, not nearly as helpful. I also added another solution that does not requre dplyr. No where I looked online showed me something as simple as "just format the result with chron."