A Comprehensive Introduction to Handling Date & Time in R (2024)

Date

Let us begin by looking at the current date and time. Sys.Date() and today() will return the current date.

Sys.Date()## [1] "2020-04-17"lubridate::today()## [1] "2020-04-17"

Time

Sys.time() and now() return the date, time and the timezone. In now(), we can specify the timezone using the tzone argument.

Sys.time()## [1] "2020-04-17 17:37:21 IST"lubridate::now()## [1] "2020-04-17 17:37:21 IST"lubridate::now(tzone = "UTC")## [1] "2020-04-17 12:07:21 UTC"

AM or PM?

am() and pm() allow us to check whether date/time occur in the AM or PM? They return a logical value i.e.TRUE or FALSE

lubridate::am(now())## [1] FALSElubridate::pm(now())## [1] TRUE

Leap Year

We can also check if the current year is a leap year using leap_year().

Sys.Date()## [1] "2020-04-17"lubridate::leap_year(Sys.Date())## [1] TRUE

Summary

Your Turn

• get current date
• get current time
• check whether the time occurs in am or pm?
• check whether the following years were leap years
  • 2018
  • 2016

Data

transact <- readr::read_csv('https://raw.githubusercontent.com/rsquaredacademy/datasets/master/transact.csv')## # A tibble: 2,466 x 3## Invoice Due Payment ## <date> <date> <date> ## 1 2013-01-02 2013-02-01 2013-01-15## 2 2013-01-26 2013-02-25 2013-03-03## 3 2013-07-03 2013-08-02 2013-07-08## 4 2013-02-10 2013-03-12 2013-03-17## 5 2012-10-25 2012-11-24 2012-11-28## 6 2012-01-27 2012-02-26 2012-02-22## 7 2013-08-13 2013-09-12 2013-09-09## 8 2012-12-16 2013-01-15 2013-01-12## 9 2012-05-14 2012-06-13 2012-07-01## 10 2013-07-01 2013-07-31 2013-07-26## # ... with 2,456 more rows

We will explore more about reading data sets with date/time columns after learning how toparse date/time. We have shared the code for reading the data sets used in the practicequestions both in the Learning Management System as well as in our GitHub repo.

Data Dictionary

The data set has 3 columns. All the dates are in the format (yyyy-mm-dd).

In the case study, we will try to answer a few questions we have about the transact data.

• extract date, month and year from Due
• compute the number of days to settle invoice
• compute days over due
• check if due year is a leap year
• check when due day in february is 29, whether it is a leap year
• how many invoices were settled within due date
• how many invoices are due in each quarter

Date

Sys.Date()## [1] "2020-04-17"unclass(Sys.Date())## [1] 18369

release_date <- as.Date("2019-12-12")release_date## [1] "2019-12-12"

release_date <- lubridate::as_date("2019-12-12")release_date## [1] "2019-12-12"

release_date - as.Date("1970-01-01")## Time difference of 18242 daysunclass(release_date)## [1] 18242

lubridate::origin## [1] "1970-01-01 UTC"

unclass(as.Date("1963-08-28"))## [1] -2318

as.Date(18242, origin = "1970-01-01")## [1] "2019-12-12"

as.Date(7285, origin = "2000-01-01")## [1] "2019-12-12"

ISO 8601

If you have carefully observed, the format in which we have been specifying thedates as well as of those returned by functions such as Sys.Date() orSys.time() is the same i.e.YYYY-MM-DD. It includes

• the year including the century
• the month
• the date

The month and date separated by -. This default format used in R is the ISO8601 standard for date/time. ISO 8601 is the internationally accepted way torepresent dates and times and uses the 24 hour clock system. Let us create therelease date using another function ISOdate().

ISOdate(year = 2019, month = 12, day = 12, hour = 8, min = 5, sec = 3, tz = "UTC")## [1] "2019-12-12 08:05:03 UTC"

We will look at all the different weird ways in which date/time are specified inthe real world in the Date & Time Formats section. For the time being, let uscontinue exploring date/time classes in R. The next class we are going to lookat is POSIXct/POSIXlt.

POSIX

You might be wondering what is this POSIX thing? POSIX stands for PortableOperating System Interface. It is a family of standards specified formaintaining compatibility between different operating systems. Before welearn to create POSIX objects, let us look at now() from lubridate.

class(lubridate::now())## [1] "POSIXct" "POSIXt"

now() returns current date/time as a POSIXct object. Let us look at itsinternal representation using unclass()

unclass(lubridate::now())## [1] 1587125246## attr(,"tzone")## [1] ""

The output you see is the number of seconds since January 1, 1970.

release_date <- as.POSIXct("2019-12-12 08:05:03")class(release_date)## [1] "POSIXct" "POSIXt"unclass(release_date) ## [1] 1576118103## attr(,"tzone")## [1] ""

release_date <- as.POSIXlt("2019-12-12 08:05:03")release_date## [1] "2019-12-12 08:05:03 IST"

release_date <- as.POSIXlt("2019-12-12 08:05:03")unclass(release_date)## $sec## [1] 3## ## $min## [1] 5## ## $hour## [1] 8## ## $mday## [1] 12## ## $mon## [1] 11## ## $year## [1] 119## ## $wday## [1] 4## ## $yday## [1] 345## ## $isdst## [1] 0## ## $zone## [1] "IST"## ## $gmtoff## [1] NA

release_date <- as.POSIXlt("2019-12-12 08:05:03")unlist(release_date)## sec min hour mday mon year wday yday isdst zone gmtoff ## "3" "5" "8" "12" "11" "119" "4" "345" "0" "IST" NA

release_date <- as.POSIXlt("2019-12-12 08:05:03")release_date$hour## [1] 8release_date$mon## [1] 11release_date$zone## [1] "IST"

Your Turn

R 1.0.0 was released on 2000-02-29 08:55:23 UTC. Save it as

• Date using character
• Date using origin and number
• POSIXct
• POSIXlt and extract
  • month day
  • day of year
  • month
  • zone
• ISODate

Shift Date

Time to shift the course dates. We can shift a date by days, weeks or months. Let us shift the course start date by:

• 2 days
• 3 weeks
• 1 year

course_start + days(2)## [1] "2017-04-14"course_start + weeks(3)## [1] "2017-05-03"course_start + years(1)## [1] "2018-04-12"

Case Study

transact %>% mutate( days_to_pay = Payment - Invoice )## # A tibble: 2,466 x 4## Invoice Due Payment days_to_pay## <date> <date> <date> <drtn> ## 1 2013-01-02 2013-02-01 2013-01-15 13 days ## 2 2013-01-26 2013-02-25 2013-03-03 36 days ## 3 2013-07-03 2013-08-02 2013-07-08 5 days ## 4 2013-02-10 2013-03-12 2013-03-17 35 days ## 5 2012-10-25 2012-11-24 2012-11-28 34 days ## 6 2012-01-27 2012-02-26 2012-02-22 26 days ## 7 2013-08-13 2013-09-12 2013-09-09 27 days ## 8 2012-12-16 2013-01-15 2013-01-12 27 days ## 9 2012-05-14 2012-06-13 2012-07-01 48 days ## 10 2013-07-01 2013-07-31 2013-07-26 25 days ## # ... with 2,456 more rows

transact %>% mutate( delay = Payment - Due ) %>% filter(delay > 0) %>% count(delay)## # A tibble: 36 x 2## delay n## <drtn> <int>## 1 1 days 61## 2 2 days 65## 3 3 days 51## 4 4 days 62## 5 5 days 69## 6 6 days 56## 7 7 days 55## 8 8 days 49## 9 9 days 38## 10 10 days 33## # ... with 26 more rows

Your Turn

• compute the length of a vacation which begins on 2020-04-19 and ends on 2020-04-25
• recompute the length of the vacation after shifting the vacation start and end date by 10 days and 2 weeks
• compute the days to settle invoice and days overdue from the receivables.csv data set
• compute the length of employment (only for those employees who have been terminated) from the hr-data.csv data set (use date of hire and termination)

Introduction

In the previous section, POSIXlt stored date/time components as a list. Amongthe different components it returned were

• gmtoff
• zone

gmtoff is offset in seconds from GMT i.e.difference in hours and minutes fromUTC. Wait.. What do UTC and GMT stand for?

• Coordinated Universal Time (UTC)
• Greenwich Meridian Time (GMT)

Since we are talking about UTC, GMT etc., let us spend a little time onunderstanding the basics of time zones and daylight savings.

Time Zones

Timezones exist because different parts of the Earth receive sun light atdifferent times. If there was a single timezone, noon or morning would meandifferent things in different parts of the world. The timezones are based onEarth’s rotation. The Earth moves ~15 degrees every 60 minutes i.e.360 degreesin 24 hours. The planet is divided into 24 timezones, each 15 degrees oflongitude width.

Now, you have heard of Greenwich Meridian Time (GMT) right? We just saw GMT offset in POSIXlt and you would have come across it in other time formats aswell. For example, India timezone is given as GMT +5:30. Let us explore GMT in alittle more detail. Greenwich is a suburb of London and the time at Greenwichis Greenwich Mean Time. As you move West from Greenwich, every 15degree section is one hour earlier than GMT and every 15 degree section to theEast is an hour later.

Alright! What is UTC then? Coordinated Universal Time (UTC) ,on the other hand, is the time standard commonly used across the world. Eventhough they share the same current time, GMT is a timezone whileUTC is a time standard.

So how do we check the timezone in R? When you run Sys.timezone(), you shouldbe able to see the timezone you are in.

Sys.timezone()## [1] "Asia/Calcutta"

If you do not see the timezone, use Sys.getenv() to get the value of theTZ environment variable.

Sys.getenv("TZ")## [1] ""

If nothing is returned, it means we have to set the timezone. Use Sys.setenv()to set the timezone as shown below. The author resides in India and hence thetimezone is set to Asia/Calcutta. You need to set the timezone in which youreside or work.

Sys.setenv(TZ = "Asia/Calcutta")

Another way to get the timezone is through tz() from the lubridate package.

lubridate::tz(Sys.time())## [1] "Asia/Calcutta"

If you want to view the time in a different timezone, use with_tz(). Let uslook at the current time in UTC instead of Indian Standard Time.

lubridate::with_tz(Sys.time(), "UTC")## [1] "2020-04-17 12:07:28 UTC"

Daylight Savings

Daylight savings also known as

• daylight saving time
• daylight savings time
• daylight time
• summer time

is the practice of advancing clocks during summer months so that darkness fallslater each day according to the clock. In other words

• advance clock by one hour in spring (spring forward)
• retard clocks by one hour in autumn (fall back)

In R, the dst() function is an indicator for daylight savings. It returnsTRUE if daylight saving is in force, FALSE if not and NA if unknown.

Sys.Date()## [1] "2020-04-17"dst(Sys.Date()) ## [1] FALSE

Your Turn

• check the timezone you live in
• check if daylight savings in on
• check the current time in UTC or a different time zone

Conversion Specifications

Time to work through a few examples. Let us say you are dealing with dates inthe format 19/12/12. In this format, the year comes first followed by monthand the date; each separated by a slash (/). The year consists of only 2digits i.e.it does not include the century. Let us now map each component ofthe date to the conversion specification table shown at the beginning.

Using the format argument, we will specify the conversion specification as a character vectori.e.enclosed in quotes.

as.Date("19/12/12", format = "%y/%m/%d")## [1] "2019-12-12"

Another way in which the release data can be written is 2019-Dec-12. We stillhave the year followed by the month and the date but there are a few changeshere:

• the components are separated by a - instead of /
• year has 4 digits i.e.includes the century
• the month is specified using abbreviation instead of digits

Let us map the components to the format table:

Let us specify the format for the date using the above mapping.

as.Date("2019-Dec-12", format = "%Y-%b-%d")## [1] "2019-12-12"

In both the above examples, we have not dealt with time components. Let usinclude the release time of R 3.6.2 in the next one i.e.19/12/12 08:05:03.

Since we are dealing with time, we will use as.POSIXct() instead ofas.Date().

as.POSIXct("19/12/12 08:05:03", tz = "UTC", format = "%y/%m/%d %H:%M:%S")## [1] "2019-12-12 08:05:03 UTC"

In the below table, we look at some of the most widely used conversionspecifications. You can learn more about these specifications by running?strptime or help(strptime).

We have included a lot of practice questions for you to explore the differentdate/time formats. The solutions are available in the Learning Management Systemas well as in our GitHub repo. Try them and let us know if you have any doubts.

Guess Format

guess_formats() from lubridate is a very useful function. It will guess thedate/time format if you specify the order in which year, month, date, hour,minute and second appear.

release_date_formats <- c("December 12th 2019", "Dec 12th 19", "dec 12 2019")guess_formats(release_date_formats, orders = "mdy", print_matches = TRUE)## Omdy mdy ## [1,] "December 12th 2019" "%Om %dth %Y" "%B %dth %Y"## [2,] "Dec 12th 19" "%Om %dth %y" "%b %dth %y"## [3,] "dec 12 2019" "%Om %d %Y" "%b %d %Y"## Omdy Omdy Omdy mdy mdy ## "%Om %dth %Y" "%Om %dth %y" "%Om %d %Y" "%B %dth %Y" "%b %dth %y" ## mdy ## "%b %d %Y"

Your Turn

Below, we have specified July 5th, 2019 in different ways. Create the date using as.Date() while specifying the correct format for each of them.

• 05.07.19
• 5-July 2019
• July 5th, 2019
• July 05, 2019
• 2019-July- 05
• 05/07/2019
• 07/05/2019
• 7/5/2019
• 07/5/19
• 2019-07-05

Your Turn

Below, we have specified July 5th, 2019 in different ways. Parse the dates using strptime() or parse_date_time() or any other helper function.

• July-05-19
• JUL-05-19
• 05.07.19
• 5-July 2019
• July 5th, 2019
• July 05, 2019
• 2019-July- 05
• 05/07/2019
• 07/05/2019
• 7/5/2019
• 07/5/19
• 2019-07-05

Year

release_date <- ymd_hms("2019-12-12 08:05:03")year(release_date) ## [1] 2019

Month

month() will return the month as a number i.e.12 for December.

month(release_date)## [1] 12

Instead, if you want the name of the month , use the label argument and set itto TRUE. Now it returns Dec instead of 12.

month(release_date, label = TRUE)## [1] Dec## 12 Levels: Jan < Feb < Mar < Apr < May < Jun < Jul < Aug < Sep < ... < Dec

But this is the abbreviated name and not the full name. How do we get the fullname of the month? Set the abbr argument to FALSE.

month(release_date, label = TRUE, abbr = FALSE)## [1] December## 12 Levels: January < February < March < April < May < June < ... < December

Ah! now we can see the full name of the month. months() from base R willreturn the full name of the month by default. If you want the abbreviated name,use the abbreviate argument and set it to TRUE.

months(release_date)## [1] "December"

Week

week() returns the number of complete 7 day periods between the date and 1stJanuary plus one.

week(release_date)## [1] 50

Day

Use day() to extract the date component. There are other variations such as

day(release_date)## [1] 12mday(release_date) ## [1] 12qday(release_date) ## [1] 73yday(release_date) ## [1] 346

wday can return

• a number
• abbreviation of the weekday
• full name of the weekday

wday(release_date) ## [1] 5wday(release_date, label = TRUE)## [1] Thu## Levels: Sun < Mon < Tue < Wed < Thu < Fri < Satwday(release_date, label = TRUE, abbr = FALSE) ## [1] Thursday## 7 Levels: Sunday < Monday < Tuesday < Wednesday < Thursday < ... < Saturday

weekdays() from base R also returns the day of the week (the name and notthe number). If you want the abbreviated name, use the abbreviate argument.

weekdays(release_date)## [1] "Thursday"weekdays(release_date, abbreviate = TRUE)## [1] "Thu"

Days in Month

If you want to know the number of days in the month, use days_in_month().In our example, the month is December and it has 31 days.

days_in_month(release_date)## Dec ## 31

Hour, Minute & Seconds

So far we have been looking at date components. Now, let us look at timecomponents.

hour(release_date)## [1] 8minute(release_date)## [1] 5second(release_date)## [1] 3

seconds() returns the number of seconds since 1970-01-01.

seconds(release_date)## [1] "1576137903S"

Quarter & Semester

quarter() will return the quarter from the date. December is in the 4thquarter and hence it returns 4.

quarter(release_date)## [1] 4

If you want the year along with the quarter, set the with_year argument toTRUE.

quarter(release_date, with_year = TRUE)## [1] 2019.4

In India, the fiscal starts in April and December falls in the 3rd quarter. Howcan we accommodate this change? The fiscal_start argument allows us to set themonth in which the fiscal begins. We will set it to 4 for April. Now it returns3 instead of 4.

quarter(release_date, fiscal_start = 4) ## [1] 3

quarters() from base R also returns the quarter.

quarters(release_date)## [1] "Q4"

Case Study

transact %>% mutate( due_day = day(Due), due_month = month(Due), due_year = year(Due) )## # A tibble: 2,466 x 6## Invoice Due Payment due_day due_month due_year## <date> <date> <date> <int> <dbl> <dbl>## 1 2013-01-02 2013-02-01 2013-01-15 1 2 2013## 2 2013-01-26 2013-02-25 2013-03-03 25 2 2013## 3 2013-07-03 2013-08-02 2013-07-08 2 8 2013## 4 2013-02-10 2013-03-12 2013-03-17 12 3 2013## 5 2012-10-25 2012-11-24 2012-11-28 24 11 2012## 6 2012-01-27 2012-02-26 2012-02-22 26 2 2012## 7 2013-08-13 2013-09-12 2013-09-09 12 9 2013## 8 2012-12-16 2013-01-15 2013-01-12 15 1 2013## 9 2012-05-14 2012-06-13 2012-07-01 13 6 2012## 10 2013-07-01 2013-07-31 2013-07-26 31 7 2013## # ... with 2,456 more rows

transact %>% mutate( due_day = day(Due), due_month = month(Due), due_year = year(Due), is_leap = leap_year(due_year) ) %>% filter(due_month == 2 & due_day == 29) %>% select(Due, is_leap) ## # A tibble: 4 x 2## Due is_leap## <date> <lgl> ## 1 2012-02-29 TRUE ## 2 2012-02-29 TRUE ## 3 2012-02-29 TRUE ## 4 2012-02-29 TRUE

transact %>% mutate( quarter_due = quarter(Due) ) %>% count(quarter_due)## # A tibble: 4 x 2## quarter_due n## <int> <int>## 1 1 521## 2 2 661## 3 3 618## 4 4 666

Your Turn

Get the R release dates using r_versions() from the rversions package andtabulate the following

• year
• month with label
• weekday with label
• hour
• and quarter

Create

To create date without time components, use make_date() and specify thefollowing:

• year
• month
• date

We need to specify all the components in numbers i.e.we cannot use Dec orDecember for the month. It has to be 12.

make_date(year = 2019, month = 12, day = 12)## [1] "2019-12-12"

When you need to include time components, use make_datetime().

make_datetime(year = 2019, month = 12, day = 12, hour = 08, min = 05, sec = 03, tz = "UTC")## [1] "2019-12-12 08:05:03 UTC"

Update

Let us look at another scenario. You have a date-time object and want to changeone of its components i.e.any of the following

• year
• month
• date

Instead of creating another date-time object, you can change any of thecomponents using update(). In the below example, we will start with the dateof release of R version 3.6.1 and using update(), we will change it to2019-12-12.

prev_release <- ymd("2019-07-05")prev_release %>% update(year = 2019, month = 12, mday = 12)## [1] "2019-12-12"

Date Sequence

So far we have created a single date-time instance. How about creating asequence of dates? We can do that using seq.Date(). We need to specify thefrom date as the bare minimum input. If the end date is not specified, it willcreate the sequence uptil the current date.

The interval of the sequence can be specified in any of the following units:

• day
• week
• month
• quarter
• year

We can add the following to the interval units

• integer
• + / - (increment or decrement)

Using the integer, we can specify multiples of the units mentioned and using thesign, we can specify whether to increment or decrement.

The below table displays the main arguments used in seq.Date():

In the first example, we will create a sequence of dates from 2010-01-01 to2019-12-31. The unit of increment should be a year i.e.the differencebetween the dates in the sequence should be 1 year, specified using the byargument.

seq.Date(from = as.Date("2010-01-01"), to = as.Date("2019-12-31"), by = "year")## [1] "2010-01-01" "2011-01-01" "2012-01-01" "2013-01-01" "2014-01-01"## [6] "2015-01-01" "2016-01-01" "2017-01-01" "2018-01-01" "2019-01-01"

In the next example, we change the unit of increment to a quarter i.e.thedifference between the dates in the sequence should be a quarter or 3 months.

seq.Date(from = as.Date("2009-12-12"), to = as.Date("2019-12-12"), by = "quarter")## [1] "2009-12-12" "2010-03-12" "2010-06-12" "2010-09-12" "2010-12-12"## [6] "2011-03-12" "2011-06-12" "2011-09-12" "2011-12-12" "2012-03-12"## [11] "2012-06-12" "2012-09-12" "2012-12-12" "2013-03-12" "2013-06-12"## [16] "2013-09-12" "2013-12-12" "2014-03-12" "2014-06-12" "2014-09-12"## [21] "2014-12-12" "2015-03-12" "2015-06-12" "2015-09-12" "2015-12-12"## [26] "2016-03-12" "2016-06-12" "2016-09-12" "2016-12-12" "2017-03-12"## [31] "2017-06-12" "2017-09-12" "2017-12-12" "2018-03-12" "2018-06-12"## [36] "2018-09-12" "2018-12-12" "2019-03-12" "2019-06-12" "2019-09-12"## [41] "2019-12-12"

We will now create a sequence of dates but instead of specifying the unit ofincrement, we specify the number of dates in the sequence i.e.the length of thesequence. We do this using the length.out argument which specifies the desiredlength of the sequence. We want the sequence to have 10 dates including thestart and end date, and hence we supply the value 10 for the length.outargument.

seq.Date(from = as.Date("2010-01-01"), to = as.Date("2019-12-31"), length.out = 10)## [1] "2010-01-01" "2011-02-10" "2012-03-22" "2013-05-02" "2014-06-11"## [6] "2015-07-22" "2016-08-31" "2017-10-10" "2018-11-20" "2019-12-31"

In all of the previous examples, we have specified both the start and the enddate. Let us look at a few examples where we create a sequence of dates wherewe only specify the start date. In the below example, we want to create asequence of dates starting from 2010-01-01. The unit of increment should be 1year i.e.the difference between the dates in the sequence should be 1 year andthe length of the sequence should be 10 i.e.the number of dates including thestart date should be 10.

seq.Date(from = as.Date("2010-01-01"), by = "year", length.out = 10)## [1] "2010-01-01" "2011-01-01" "2012-01-01" "2013-01-01" "2014-01-01"## [6] "2015-01-01" "2016-01-01" "2017-01-01" "2018-01-01" "2019-01-01"

The unit of increment can include multiples and +/- sign i.e.it can be anunit of increment or decrement. In the next example, we can increment the datesin the sequence by 2 i.e.the difference between the dates should be 2instead of 1. This is achieved by specifying the unit of increment (multiple)first followed by a space and then the unit. In our example, it is 2 year. Asyou can see, the sequence now goes all the way till 2028 and the gap betweenthe dates is 2 years.

seq.Date(from = as.Date("2010-01-01"), by = "2 year", length.out = 10)## [1] "2010-01-01" "2012-01-01" "2014-01-01" "2016-01-01" "2018-01-01"## [6] "2020-01-01" "2022-01-01" "2024-01-01" "2026-01-01" "2028-01-01"

Let us say instead of increment we want to decrement the dates i.e.the sequenceof dates will go backwards as shown in the next example. We achieve this byusing the - sign along with the unit of decrement. The sequence of dates innext example starts from 2010 and goes back upto 1992 and the differencebetween the dates in 2 years.

seq.Date(from = as.Date("2010-01-01"), by = "-2 year", length.out = 10)## [1] "2010-01-01" "2008-01-01" "2006-01-01" "2004-01-01" "2002-01-01"## [6] "2000-01-01" "1998-01-01" "1996-01-01" "1994-01-01" "1992-01-01"

In the last example, we will explore the along.with argument. Here we havesupplied a vector which is a sequence of numbers from 1 to 10. The length ofthis vector is 10 and the same length is used as the length of the sequence i.e.the length of value supplied to along.with is also the length of the sequence.

seq.Date(from = as.Date("2010-01-01"), by = "-2 year", along.with = 1:10)## [1] "2010-01-01" "2008-01-01" "2006-01-01" "2004-01-01" "2002-01-01"## [6] "2000-01-01" "1998-01-01" "1996-01-01" "1994-01-01" "1992-01-01"

Verify Type

How do you check if the data is a date-time object? You can do that using anyof the following from the lubridate package.

• is.Date()
• is.POSIXct()
• is.POSIXlt()

is.Date(release_date)## [1] FALSEis.POSIXct(release_date)## [1] TRUEis.POSIXlt(release_date)## [1] FALSE

Your Turn

• R 2.0.0 was released on 2004-10-04 14:24:38. Create this date using bothmake_date() and make_datetime()

• R 3.0.0 was released on 2013-04-03 07:12:36. Update the date created in theprevious step to the above using update()

Interval

An interval is a timespan defined by two date-times. Let us represent the lengthof the course using interval.

course_start <- as_date('2017-04-12')course_end <- as_date('2017-04-21')interval(course_start, course_end)## [1] 2017-04-12 UTC--2017-04-21 UTC

If you observe carefully, the interval is represented by the course start andend dates. We will learn how to use intervals in the case study.

vacation_start <- as_date('2017-04-19')vacation_end <- as_date('2017-04-25')course_interval <- interval(course_start, course_end)vacation_interval <- interval(vacation_start, vacation_end)int_overlaps(course_interval, vacation_interval)## [1] TRUE

transact %>% mutate( inv_due_interval = interval(Invoice, Due), due_next = Due + days(1), due_pay_interval = interval(due_next, Payment), overlaps = int_overlaps(inv_due_interval, due_pay_interval) ) %>% select(Invoice, Due, Payment, overlaps)## # A tibble: 2,466 x 4## Invoice Due Payment overlaps## <date> <date> <date> <lgl> ## 1 2013-01-02 2013-02-01 2013-01-15 TRUE ## 2 2013-01-26 2013-02-25 2013-03-03 FALSE ## 3 2013-07-03 2013-08-02 2013-07-08 TRUE ## 4 2013-02-10 2013-03-12 2013-03-17 FALSE ## 5 2012-10-25 2012-11-24 2012-11-28 FALSE ## 6 2012-01-27 2012-02-26 2012-02-22 TRUE ## 7 2013-08-13 2013-09-12 2013-09-09 TRUE ## 8 2012-12-16 2013-01-15 2013-01-12 TRUE ## 9 2012-05-14 2012-06-13 2012-07-01 FALSE ## 10 2013-07-01 2013-07-31 2013-07-26 TRUE ## # ... with 2,456 more rows

transact %>% mutate( inv_due_interval = interval(Invoice, Due), due_pay_interval = interval(Due, Payment), due_pay_next = int_shift(due_pay_interval, by = days(1)), overlaps = int_overlaps(inv_due_interval, due_pay_next) ) %>% select(Invoice, Due, Payment, overlaps)## # A tibble: 2,466 x 4## Invoice Due Payment overlaps## <date> <date> <date> <lgl> ## 1 2013-01-02 2013-02-01 2013-01-15 TRUE ## 2 2013-01-26 2013-02-25 2013-03-03 FALSE ## 3 2013-07-03 2013-08-02 2013-07-08 TRUE ## 4 2013-02-10 2013-03-12 2013-03-17 FALSE ## 5 2012-10-25 2012-11-24 2012-11-28 FALSE ## 6 2012-01-27 2012-02-26 2012-02-22 TRUE ## 7 2013-08-13 2013-09-12 2013-09-09 TRUE ## 8 2012-12-16 2013-01-15 2013-01-12 TRUE ## 9 2012-05-14 2012-06-13 2012-07-01 FALSE ## 10 2013-07-01 2013-07-31 2013-07-26 TRUE ## # ... with 2,456 more rows

transact %>% mutate( inv_due_interval = interval(Invoice, Due), due_pay_interval = interval(Due, Payment), overlaps = int_overlaps(inv_due_interval, due_pay_interval) ) %>% select(Invoice, Due, Payment, overlaps)## # A tibble: 2,466 x 4## Invoice Due Payment overlaps## <date> <date> <date> <lgl> ## 1 2013-01-02 2013-02-01 2013-01-15 TRUE ## 2 2013-01-26 2013-02-25 2013-03-03 TRUE ## 3 2013-07-03 2013-08-02 2013-07-08 TRUE ## 4 2013-02-10 2013-03-12 2013-03-17 TRUE ## 5 2012-10-25 2012-11-24 2012-11-28 TRUE ## 6 2012-01-27 2012-02-26 2012-02-22 TRUE ## 7 2013-08-13 2013-09-12 2013-09-09 TRUE ## 8 2012-12-16 2013-01-15 2013-01-12 TRUE ## 9 2012-05-14 2012-06-13 2012-07-01 TRUE ## 10 2013-07-01 2013-07-31 2013-07-26 TRUE ## # ... with 2,456 more rows

course_interval <- interval(course_start, course_end)# shift course_interval by 1 day int_shift(course_interval, by = days(1))## [1] 2017-04-13 UTC--2017-04-22 UTC# shift course_interval by 3 weeksint_shift(course_interval, by = weeks(3))## [1] 2017-05-03 UTC--2017-05-12 UTC# shift course_interval by 1 yearint_shift(course_interval, by = years(1))## [1] 2018-04-12 UTC--2018-04-21 UTC

Within

Let us assume that we have to attend a conference in April 2017. Does itclash with the course? We can answer this using %within% which willreturn TRUE if a date falls within an interval.

conference <- as_date('2017-04-15')conference %within% course_interval## [1] TRUE

transact %>% mutate( inv_due_interval = interval(Invoice, Due), overlaps = Payment %within% inv_due_interval ) %>% select(Due, Payment, overlaps)## # A tibble: 2,466 x 3## Due Payment overlaps## <date> <date> <lgl> ## 1 2013-02-01 2013-01-15 TRUE ## 2 2013-02-25 2013-03-03 FALSE ## 3 2013-08-02 2013-07-08 TRUE ## 4 2013-03-12 2013-03-17 FALSE ## 5 2012-11-24 2012-11-28 FALSE ## 6 2012-02-26 2012-02-22 TRUE ## 7 2013-09-12 2013-09-09 TRUE ## 8 2013-01-15 2013-01-12 TRUE ## 9 2012-06-13 2012-07-01 FALSE ## 10 2013-07-31 2013-07-26 TRUE ## # ... with 2,456 more rows

Duration

Duration is timespan measured in seconds. To create a duration object, useduration(). The timespan can be anything from seconds to years but it will berepresented as seconds. Let us begin by creating a duration object where the timespan is in seconds.

duration(50, "seconds")## [1] "50s"

Another way to specify the above timespan is shown below:

duration(second = 50)## [1] "50s"

As you can see, the output is same in both the cases. Let us increase the timespan to 60 seconds and see what happens.

duration(second = 60)## [1] "60s (~1 minutes)"

Although the timespan is primarily measured in seconds, it also shows ~1 minutes in the brackets. As the length of the timespan increases i.e.the number becomes large, it is represented using larger units such as hours and days. In the below examples, as the number of seconds increases, you can observe larger units being used to represent the timespan.

# minutesduration(minute = 50)## [1] "3000s (~50 minutes)"duration(minute = 60)## [1] "3600s (~1 hours)"# hoursduration(hour = 23)## [1] "82800s (~23 hours)"duration(hour = 24)## [1] "86400s (~1 days)"

The following helper functions can be used to create duration objects as well.

# defaultdseconds()## [1] "1s"dminutes()## [1] "60s (~1 minutes)"# secondsduration(second = 59)## [1] "59s"dseconds(59)## [1] "59s"# minutesduration(minute = 50)## [1] "3000s (~50 minutes)"dminutes(50)## [1] "3000s (~50 minutes)"# hoursduration(hour = 36)## [1] "129600s (~1.5 days)"dhours(36)## [1] "129600s (~1.5 days)"# weeksduration(week = 56)## [1] "33868800s (~1.07 years)"dweeks(56)## [1] "33868800s (~1.07 years)"

Let us use the above helper functions to get the course length in different units.

# course length in seconds course_interval / dseconds()## [1] 777600# course length in minutescourse_interval / dminutes()## [1] 12960# course length in hourscourse_interval / dhours()## [1] 216# course length in weekscourse_interval / dweeks()## [1] 1.285714# course length in yearscourse_interval / dyears()## [1] 0.02465753

Period

A period is a timespan defined in units such as years, months, and days. Inthe below examples, we use period() to represent timespan using differentunits.

# secondperiod(5, "second")## [1] "5S"period(second = 5)## [1] "5S"# minute & secondperiod(c(3, 5), c("minute", "second"))## [1] "3M 5S"period(minute = 3, second = 5)## [1] "3M 5S"# hour, minte & secondperiod(c(1, 3, 5), c("hour", "minute", "second"))## [1] "1H 3M 5S"period(hour = 1, minute = 3, second = 5)## [1] "1H 3M 5S"# day, hour, minute & secondperiod(c(3, 1, 3, 5), c("day", "hour", "minute", "second"))## [1] "3d 1H 3M 5S"period(day = 3, hour = 1, minute = 3, second = 5)## [1] "3d 1H 3M 5S"

Let us get the course length in different units using as.period().

# course length in secondas.period(course_interval, unit = "seconds")## [1] "777600S"# course length in hours and minutesas.period(course_interval, unit = "minutes")## [1] "12960M 0S"# course length in hours, minutes and secondsas.period(course_interval, unit = "hours")## [1] "216H 0M 0S"

time_length() computes the exact length of a timespan i.e.duration, interval or period. Let us use time_length() to compute the length of thecourse in different units.

# course length in secondstime_length(course_interval, unit = "seconds")## [1] 777600# course length in minutestime_length(course_interval, unit = "minutes")## [1] 12960# course length in hourstime_length(course_interval, unit = "hours")## [1] 216

Rounding Dates

We will explore functions for rounding dates

• to the nearest value using round_dates()
• down using floor_date()
• up using ceiling_date()

The unit for rounding can be any of the following:

• second
• minute
• hour
• day
• week
• month
• bimonth
• quarter
• season
• halfyear
• and year

We will look at a few examples using round_date() and you will then practiceusing the other two functions.

# minuteround_date(release_date, unit = "minute")## [1] "2019-12-12 08:05:00 UTC"round_date(release_date, unit = "mins")## [1] "2019-12-12 08:05:00 UTC"round_date(release_date, unit = "5 mins")## [1] "2019-12-12 08:05:00 UTC"# hourround_date(release_date, unit = "hour")## [1] "2019-12-12 08:00:00 UTC"# dayround_date(release_date, unit = "day")## [1] "2019-12-12 UTC"

Rollback

Use rollback() if you want to change the date to the last day of the previousmonth or the first day of the month.

rollback(release_date)## [1] "2019-11-30 08:05:03 UTC"

To change the date to the first day of the month, use the roll_to_firstargument and set it to TRUE.

rollback(release_date, roll_to_first = TRUE)## [1] "2019-12-01 08:05:03 UTC"

Your Turn

• round up R release dates to hours
• round down R release dates to minutes
• rollback R release dates to the beginning of the month