datetime object to minutes: I need 3 packages-CodePudding

I am wondering if I am missing something?!

I would like to know: is there a better/shorter way to get minutes from a datetime object:

Lessons studied so far:

Extract time (HMS) from lubridate date time object?

converting from hms to hour:minute in r, or rounding in minutes from hms

My tibble:

df <- structure(list(dttm = structure(c(-2209068000, -2209069200, -2209061520, 
-2209064100, -2209065240), tzone = "UTC", class = c("POSIXct", 
"POSIXt"))), row.names = c(NA, -5L), class = c("tbl_df", "tbl", 
"data.frame"))

# A tibble: 5 x 1
  dttm               
  <dttm>             
1 1899-12-31 02:00:00
2 1899-12-31 01:40:00
3 1899-12-31 03:48:00
4 1899-12-31 03:05:00
5 1899-12-31 02:46:00

I would like to add a new column with minutes as integer:

My approach so far:

library(dplyr)
library(lubridate)  # ymd_hms
library(hms)        # as_hms
library(chron)      # times

test %>% 
    mutate(dttm_min = as_hms(ymd_hms(dttm)),
           dttm_min = 60*24*as.numeric(times(dttm_min)))

# A tibble: 5 x 2
  dttm                dttm_min
  <dttm>                 <dbl>
1 1899-12-31 02:00:00      120
2 1899-12-31 01:40:00      100
3 1899-12-31 03:48:00      228
4 1899-12-31 03:05:00      185
5 1899-12-31 02:46:00      166

This gives me the result I want, but I need for this operation 3 packages. Is there a more direct way?

CodePudding user response：

Here is a base R way -

You can extract the time using format, change the date to '1970-01-01' (Since R datetime starts with '1970-01-01'), convert to numeric and divide the time by 60 to get the duration in minutes.

as.numeric(as.POSIXct(paste('1970-01-01', format(df$dttm, '%T')), tz = 'UTC'))/60
#[1] 120 100 228 185 166

CodePudding user response：

Here are two ways.

Base R

with(df, as.integer(format(dttm, "%M"))   60*as.integer(format(dttm, "%H")))
#[1] 120 100 228 185 166

Another base R option, using class "POSIXlt" as proposed here.

minute_of_day <- function(x){
  y <- as.POSIXlt(x)
  60*y$hour   y$min
}

minute_of_day(df$dttm)
#[1] 120 100 228 185 166

Package `lubridate`

lubridate::minute(df$dttm)   60*lubridate::hour(df$dttm)
#[1] 120 100 228 185 166

If the package is loaded, this can be simplified, with the same output, to

library(lubridate)
minute(df$dttm)   60*hour(df$dttm)

CodePudding user response：

We can use

library(data.table)
 as.numeric(as.ITime(format(df$dttm, '%T')))/60
[1] 120 100 228 185 166

CodePudding user response：

For sake of completeness, the time of day (in minutes) can be calculated by taking the difftime() between the POSIXct datetime object and the beginning of the day, e.g.,

difftime(df$dttm, lubridate::floor_date(df$dttm, "day"), units = "min")

Time differences in mins
[1] 120 100 228 185 166

Besides base R only one other package is required.

According to help("difftime"), difftime() returns an object of class "difftime" with an attribute indicating the units.

Base R

Package lubridate

Package `lubridate`