How to manipulate the time part of a date column? - r

Question

How do I write this code (hour is from lubridate package)?
Objective: if hour part of PICK_DATE is later than 16:00, the ADJ_PICK_DATE should be next day 03:00. If the hour part of PICK_DATE is earlier than 03:00, then ADJ_PICK_DATE is to be same day 03:00. Problem is, when there is no change needed, the code still adds 3 hours to the PICK_DATE i.e. when the hour part of PICK_DATE is within 03:00 and 16:00.
x$PICK_TIME <- cut(hour(x$PICK_DATE), c(-1, 2, 15, 24), c("EARLY", "OKAY", "LATE"))
x$ADJ_PICK_DATE <- ifelse(x$PICK_TIME=="EARLY",
as.POSIXct(paste(format(x$PICK_DATE, "%d-%b-%Y"), "03:00"),
format="%d-%b-%Y %H:%M"), x$PICK_DATE)
x$ADJ_PICK_DATE <- ifelse(x$PICK_TIME=="LATE",
as.POSIXct(paste(format(x$PICK_DATE+86400, "%d-%b-%Y"),
"03:00"), format="%d-%b-%Y %H:%M"),
x$ADJ_PICK_DATE)
x$ADJ_PICK_DATE <- as.POSIXct(x$ADJ_PICK_DATE, origin = "1970-01-01")
Help please.
Sample data:
PICK_DATE SHIP_DATE
01-APR-2017 00:51 02-APR-2017 06:55 AM
01-APR-2017 00:51 02-APR-2017 12:11 PM
01-APR-2017 00:51 02-APR-2017 12:11 PM
01-APR-2017 00:51 02-APR-2017 09:39 AM

Answer 1

Here is a simple, reproducible example. I had to make up some sample data, based on an earlier question you asked. I suggest reading into dplyr and lubridate as they will help you with your work on manipulating dates.
EDIT: Updated to work with end-of-month dates.
library(lubridate)
library(dplyr)
df <- data.frame(pick_date = c("01-APR-2017 00:51", "02-APR-2017 08:53", "15-APR-2017 16:12", "23-APR-2017 02:04", "30-APR-2017 20:08"), ship_date = c("05-APR-2017 06:55", "09-APR-2017 12:11", "30-APR-2017 13:11", "02-MAY-2017 15:16", "05-MAY-2017 09:57"))
df %>%
mutate(pick_date = dmy_hm(pick_date)) %>%
mutate(ship_date = dmy_hm(ship_date)) %>%
mutate(pick_time = case_when(
hour(pick_date) <= 3 ~ "early",
hour(pick_date) >= 16 ~ "late",
TRUE ~ "okay")
) %>%
mutate(new_pick_time = case_when(
pick_time == "early" ~ hms(hours(3)),
pick_time == "late" ~ hms(hours(3)),
TRUE ~ hms(paste0(hour(pick_date), "H ", minute(pick_date), "M ", second(pick_date), "S")))
) %>%
mutate(temp_pick_date = case_when(
pick_time == "early" ~ pick_date,
pick_time == "late" ~ pick_date + days(1),
TRUE ~ pick_date)
) %>%
mutate(new_pick_date = make_datetime(year(temp_pick_date), month(temp_pick_date), day(temp_pick_date), hour(new_pick_time), minute(new_pick_time), second(new_pick_time))) %>%
select(-new_pick_time, -temp_pick_date)
This returns
pick_date ship_date pick_time new_pick_date
1 2017-04-01 00:51:00 2017-04-05 06:55:00 early 2017-04-01 03:00:00
2 2017-04-02 08:53:00 2017-04-09 12:11:00 okay 2017-04-02 08:53:00
3 2017-04-15 16:12:00 2017-04-30 13:11:00 late 2017-04-16 03:00:00
4 2017-04-23 02:04:00 2017-05-02 15:16:00 early 2017-04-23 03:00:00
5 2017-04-30 20:08:00 2017-05-05 09:57:00 late 2017-05-01 03:00:00

Answer 2

So it sounds like you just need to do two different arithmetic operations, conditional on the hour of a date time?
The simplest way I can think to access the hour component is to store the time in a POSIXlt. I believe the "l" stands or "list", and this lets you treat a timestamp like a list with the different time measurements being accessible attributes accordingly.
Like this:
> time <- as.POSIXlt('2017-07-29 15:12:01')
> time
[1] "2017-07-29 15:12:01 EDT"
> time$hour
[1] 15
So you could write a function that does the operation you desire, and feed it your date column. Hard for me to take it further because I don't quite understand the question, but here's a skeleton:
ComputeDifference <- function(time) {
if (time$hour < 3) {
# code to count orders between 0 and 3 "from same day 3:00"
}
if (time$hour > 16) {
# code to consider late orders
}
}
If you throw in sample data and refine the question, maybe I can take a more thorough crack at this.