A.10 Solutions (10)

Here are the solutions to the exercises on creating and computing with dates and times of Chapter 10 (Section 10.6).

Please note

The lubridate package is not part of the core tidyverse packages. Hence, do not forget loading this package if you want to use its commands:

library(tidyverse)      # core tidyverse
library(lubridate)
library(ds4psy, unikn)  # other packages

Note that some key tasks (e.g., computing someone’s age, determining the weekday of some date) occur repeatedly throughout these exercises. If this gets boring, use different solution paths for solving them.

A.10.1 Exercise 1

Reading dates and times

1. Use the appropriate lubridate function to parse each of the following dates:

d1 <- "January 24, 2020"
d2 <- "2024-Apr-01"
d3 <- "11-Nov-2024"
d4 <- c("July 13 (1969)", "August 23 (1972)", "July 1 (1975)")

                 # Date: 
d5 <- "08/12/10" # Oct 12, 2008
d6 <- d5         # Aug 12, 2010
d7 <- d5         # Oct 08, 2012

Solution

mdy(d1)
#> [1] "2020-01-20"
ymd(d2)
#> [1] "2020-04-01"
dmy(d3)
#> [1] "2020-11-11"
mdy(d4)
#> [1] "1969-07-13" "1972-08-23" "1975-07-01"

ydm(d5)
#> [1] "2008-10-12"
mdy(d6)
#> [1] "2010-08-12"
dym(d7)
#> [1] "2012-10-08"

2. Use the appropriate lubridate function to parse each of the following date-times:

t1 <- "2024-11-11 11:11:01"
t2 <- "2024/12/24 07:30"
t3 <- "31:12:24 12:45:59"

t4 <- c("8:05 01/01/2024", "9:20 29/02/2024", "12:30 24/12/2024", "23:58 30/12/2024")

Hint: Note that t4 contains the time component before the date component. To handle this vector, consider creating a tibble and then using dplyr commands for separating its time and date components, and pasting them in reversed order (date before time).

Solution

ymd_hms(t1)
#> [1] "2020-11-11 11:11:01 UTC"
ymd_hm(t2)
#> [1] "2020-12-24 07:30:00 UTC"
dmy_hms(t3)
#> [1] "2020-12-31 12:45:59 UTC"

# t4: 
tb <- tibble(t4 = t4)  # as tibble

tb <- tb %>% 
  separate(t4, into = c("t", "d"), sep = " ", remove = FALSE) %>%
  mutate(ds = paste(d, t),  # as text string
         dt = dmy_hm(ds)    # parse text 
         )

# Print tibble:
knitr::kable(tb, caption = "A tibble with `t4` separated and mutated into `dt`.")

Table A.27: A tibble with t4 separated and mutated into dt.

t4	t	d	ds	dt
8:05 01/01/2020	8:05	01/01/2020	01/01/2020 8:05	2020-01-01 08:05:00
9:20 29/02/2020	9:20	29/02/2020	29/02/2020 9:20	2020-02-29 09:20:00
12:30 24/12/2020	12:30	24/12/2020	24/12/2020 12:30	2020-12-24 12:30:00
23:58 30/12/2020	23:58	30/12/2020	30/12/2020 23:58	2020-12-30 23:58:00

# Print vector: 
tb$dt
#> [1] "2020-01-01 08:05:00 UTC" "2020-02-29 09:20:00 UTC"
#> [3] "2020-12-24 12:30:00 UTC" "2020-12-30 23:58:00 UTC"

3. Determine the weekdays of the 7 dates in d4 and t4.

Hint: First combine the seven dates into a vector. Then choose from an abundance of options — the base R function weekdays(), the lubridate function wday(), or the ds4psy function what_wday() — to solve the task.

Solution

# (a) with base R:
dates_1 <- c(as.Date(d4, format = "%B %d (%Y)"), as.Date(tb$d))
# dates_1
base::weekdays(dates_1)
#> [1] "Sunday"    "Wednesday" "Tuesday"   "Saturday"  "Tuesday"   "Friday"   
#> [7] "Friday"
# base::weekdays(as.Date(dates_1, format = "%d/%m/%y"))

# (b) with lubridate: 
dates_2 <- c(lubridate::mdy(d4), lubridate::as_date(tb$dt))
# dates_2
lubridate::wday(dates_2, label = TRUE, week_start = 1, abbr = FALSE)
#> [1] Sunday    Wednesday Tuesday   Wednesday Saturday  Thursday  Wednesday
#> 7 Levels: Monday < Tuesday < Wednesday < Thursday < Friday < ... < Sunday

# (c) with ds4psy:
ds4psy::what_wday(dates_1)
#> [1] "Sunday"    "Wednesday" "Tuesday"   "Saturday"  "Tuesday"   "Friday"   
#> [7] "Friday"

A.10.2 Exercise 2

Birth dates and times

The table dt_10 (available from ds4psy or rpository.com) contains the birth dates and times of ten non-existent people. Read the data into a tibble dt_10:

# dt_10 <- readr::read_csv("./data/dt_10.csv")  # from local file
# dt_10 <- readr::read_csv("http://rpository.com/ds4psy/data/dt_10.csv")  # online
dt_10 <- ds4psy::dt_10  # from ds4psy

# Show data:
knitr::kable(dt_10, caption = "Data of table `dt_10`.")

Table A.28: Data of table dt_10.

name	day	month	year	hour	min	sec
Anna	8	8	1994	11	47	57
Beowulf	1	6	1994	5	35	43
Cassandra	14	11	2000	5	58	6
David	17	1	1991	13	3	12
Eva	21	1	2001	21	33	55
Frederic	19	7	2000	13	47	12
Gwendoline	20	9	1996	8	28	37
Hamlet	5	5	1996	17	7	8
Ian	18	8	1996	8	27	17
Joy	18	12	1990	14	44	35

1. Use base R commands (with appropriate “POSIX” specifications) or the corresponding lubridate functions to parse the data of birth dob and time of birth tob as two new columns of dt_10.

Hint: When using base R commands, consider using paste() for creating a character string with appropriate separators from the date- and time-related variables contained in dt_10.

Solution

# (a) base R:
dt_10 <- dt_10 %>%
  mutate(dob = as.Date(paste(year, month, day, sep = "-"), format = "%Y-%m-%d"), 
         tob = as.POSIXct(paste0(year, "-", month, "-", day, " ", hour, ":", min, ":", sec), 
                          format = "%Y-%m-%d %H:%M:%S", 
                          tz = "UTC")   # using standard time zone
  )

# (b) lubridate: 
dt_11 <- dt_10 %>%
  mutate(dob = make_date(year, month, day),
         tob = make_datetime(year, month, day, hour, min, sec, 
                             tz = "UTC")  # using standard time zone
  )

# Verify equality:
all.equal(dt_10$dob, dt_11$dob)
#> [1] TRUE
all.equal(dt_10$tob, dt_11$tob)
#> [1] TRUE

# Show data:
knitr::kable(dt_10, caption = "Data of table `dt_10` with `dob` and `tob` variables.")

Table A.29: Data of table dt_10 with dob and tob variables.

name	day	month	year	hour	min	sec	dob	tob
Anna	8	8	1994	11	47	57	1994-08-08	1994-08-08 11:47:57
Beowulf	1	6	1994	5	35	43	1994-06-01	1994-06-01 05:35:43
Cassandra	14	11	2000	5	58	6	2000-11-14	2000-11-14 05:58:06
David	17	1	1991	13	3	12	1991-01-17	1991-01-17 13:03:12
Eva	21	1	2001	21	33	55	2001-01-21	2001-01-21 21:33:55
Frederic	19	7	2000	13	47	12	2000-07-19	2000-07-19 13:47:12
Gwendoline	20	9	1996	8	28	37	1996-09-20	1996-09-20 08:28:37
Hamlet	5	5	1996	17	7	8	1996-05-05	1996-05-05 17:07:08
Ian	18	8	1996	8	27	17	1996-08-18	1996-08-18 08:27:17
Joy	18	12	1990	14	44	35	1990-12-18	1990-12-18 14:44:35

2. As it turns out, all the people of dt_10 were born in Denmark. Create a second tibble dt_10_2 that considers this fact for the tob variable (e.g., when using the make_datetime() function) and quantify and explain any discrepancies between dt_10$tob and the corresponding variable in dt_10_2.

dt_10$tob  # default time zone is UTC (Universal Time, Coordinated)
#>  [1] "1994-08-08 11:47:57 UTC" "1994-06-01 05:35:43 UTC"
#>  [3] "2000-11-14 05:58:06 UTC" "1991-01-17 13:03:12 UTC"
#>  [5] "2001-01-21 21:33:55 UTC" "2000-07-19 13:47:12 UTC"
#>  [7] "1996-09-20 08:28:37 UTC" "1996-05-05 17:07:08 UTC"
#>  [9] "1996-08-18 08:27:17 UTC" "1990-12-18 14:44:35 UTC"

# Using tz of Denmark (see ?OlsonNames() for options): 
dt_10_2 <- dt_10 %>%
  mutate(tob_2 = make_datetime(year, month, day, hour, min, sec, 
                               tz = "Europe/Copenhagen")
  )

dt_10_2$tob_2  # are CET/CEST (Central European Time)
#>  [1] "1994-08-08 11:47:57 CEST" "1994-06-01 05:35:43 CEST"
#>  [3] "2000-11-14 05:58:06 CET"  "1991-01-17 13:03:12 CET" 
#>  [5] "2001-01-21 21:33:55 CET"  "2000-07-19 13:47:12 CEST"
#>  [7] "1996-09-20 08:28:37 CEST" "1996-05-05 17:07:08 CEST"
#>  [9] "1996-08-18 08:27:17 CEST" "1990-12-18 14:44:35 CET"

# Time differences: 
dt_10$tob - dt_10_2$tob_2
#> Time differences in hours
#>  [1] 2 2 1 1 1 2 2 2 2 1

Answer: The make_datetime() function used tz = "UTC" (Coordinated Universal Time) by default. By contrast, Denmark lies in the “CET” (Central European Time) time zone and switches to daylight saving time (indicated by “CEST”) during the summer months. Hence, the variables in dt_10$tob were lagging 1 or 2 hours behind the actual times in dt_10_2$tob_2.

3. Use the appropriate lubridate functions to add 2 columns that specify – given each person’s DOB – the weekday dob_wd (from Monday to Sunday) of their birthday and their current age age_fy in full years (i.e., the numeric value of their age, as an integer).

Hint: Their current age can be computed by subtracting their DOB from today’s date today(). One way of computing their age in full years is by dividing the interval() of their current age by a duration() in the unit of “years”. (Alternatively, rounding can also work.)

Solution

# Today's date:
today <- lubridate::today()
today
#> [1] "2022-09-10"

# Redo and simplify dt_10 (from above):
dt_10 <- dt_10 %>%
  mutate(dob = make_date(year, month, day),
         tob = make_datetime(year, month, day, hour, min, sec,
                             tz = "Europe/Copenhagen")) %>%
  select(name, dob, tob)
# dt_10

# Compute age (in different ways):
today - dt_10$dob  # age (in days)
#> Time differences in days
#>  [1] 10260 10328  7970 11559  7902  8088  9486  9624  9519 11589
lubridate::as.duration(today - dt_10$dob)  # as duration
#>  [1] "886464000s (~28.09 years)"  "892339200s (~28.28 years)" 
#>  [3] "688608000s (~21.82 years)"  "998697600s (~31.65 years)" 
#>  [5] "682732800s (~21.63 years)"  "698803200s (~22.14 years)" 
#>  [7] "819590400s (~25.97 years)"  "831513600s (~26.35 years)" 
#>  [9] "822441600s (~26.06 years)"  "1001289600s (~31.73 years)"
interval(dt_10$dob, today) / duration(num = 1, units = "years")  # interval in years
#>  [1] 28.09035 28.27652 21.82067 31.64682 21.63450 22.14374 25.97125 26.34908
#>  [9] 26.06160 31.72895

dt_10 <- dt_10 %>%
  select(-tob) %>% 
  mutate(dob_wd = wday(dob, label = TRUE, week_start = 1, abbr = FALSE),
         age_yr = interval(dob, today) / duration(num = 1, units = "years"),
         age_fy = floor(age_yr))
knitr::kable(dt_10, caption = "Danish people with DOB weekday and current age.") 

Table A.30: Danish people with DOB weekday and current age.

name	dob	dob_wd	age_yr	age_fy
Anna	1994-08-08	Monday	28.09035	28
Beowulf	1994-06-01	Wednesday	28.27652	28
Cassandra	2000-11-14	Tuesday	21.82067	21
David	1991-01-17	Thursday	31.64682	31
Eva	2001-01-21	Sunday	21.63450	21
Frederic	2000-07-19	Wednesday	22.14374	22
Gwendoline	1996-09-20	Friday	25.97125	25
Hamlet	1996-05-05	Sunday	26.34908	26
Ian	1996-08-18	Sunday	26.06160	26
Joy	1990-12-18	Tuesday	31.72895	31

A.10.3 Exercise 3

This exercise uses the fame dataset included in the ds4psy package. Actually, the entries of the dataset were populated by the submissions of previous students. So think carefully about your entries — they might end up in the dataset studied by future generations of students.

Add to fame

1. Pick at least 4 famous people — some of which are still alive, some of which have already died — and enter their name, area of occupation, date of birth (DOB), and date of death (DOD, if deceased) in a tibble fame, in analogy to the following:

fame <- tibble(name = c("Napoleon Bonaparte", "Jimi Hendrix", "Michael Jackson", "Frida Kahlo", 
                        "Angela Merkel", "Kobe Bryant", "Lionel Messi", "Zinedine Zidane"),
               area = c("politics", "guitarist/music", "singer/music", "arts/painter", 
                        "politics", "basketball/sports", "football/sports", "football/sports"), 
               DOB = c("August 15, 1769", "November 27, 1942", "August 29, 1958", "July 06, 1907", 
                       "July 17, 1954", "August 23, 1978", "June 24, 1987", "June 23, 1972"),
               DOD = c("May 05, 1821", "September 18, 1970", "June 25, 2009", "July 13, 1954", 
                       NA, "January 26, 2020", NA, NA))

knitr::kable(fame, caption = "Basic info on some famous people.")

Table A.31: Basic info on some famous people.

name	area	DOB	DOD
Napoleon Bonaparte	politics	August 15, 1769	May 05, 1821
Jimi Hendrix	guitarist/music	November 27, 1942	September 18, 1970
Michael Jackson	singer/music	August 29, 1958	June 25, 2009
Frida Kahlo	arts/painter	July 06, 1907	July 13, 1954
Angela Merkel	politics	July 17, 1954	NA
Kobe Bryant	basketball/sports	August 23, 1978	January 26, 2020
Lionel Messi	football/sports	June 24, 1987	NA
Zinedine Zidane	football/sports	June 23, 1972	NA

Note: Please remember to enter any rare and unusual symbols as Unicode characters (see Section 9.2.2).

2. Use the appropriate lubridate functions to replace the DOB and DOD variables in fame by corresponding dob and dod variables of type “Date”.

Solution

fame <- fame %>% 
  mutate(dob = lubridate::mdy(DOB),
         dod = lubridate::mdy(DOD)) %>%
  select(name, area, dob, dod)
# knitr::kable(fame, caption = "Info on some famous people.")

3. Add two variables to fame that specify the weekday (from “Monday” to “Sunday”) of their birth (dob_wd) and — if applicable — of their death (dob_wd).

Solution

fame %>%
  mutate(dob_wd = lubridate::wday(dob, label = TRUE, week_start = 1, abbr = FALSE),
         dod_wd = lubridate::wday(dod, label = TRUE, week_start = 1, abbr = FALSE)
  )
#> # A tibble: 8 × 6
#>   name               area              dob        dod        dob_wd    dod_wd  
#>   <chr>              <chr>             <date>     <date>     <ord>     <ord>   
#> 1 Napoleon Bonaparte politics          1769-08-15 1821-05-05 Tuesday   Saturday
#> 2 Jimi Hendrix       guitarist/music   1942-11-27 1970-09-18 Friday    Friday  
#> 3 Michael Jackson    singer/music      1958-08-29 2009-06-25 Friday    Thursday
#> 4 Frida Kahlo        arts/painter      1907-07-06 1954-07-13 Saturday  Tuesday 
#> 5 Angela Merkel      politics          1954-07-17 NA         Saturday  <NA>    
#> 6 Kobe Bryant        basketball/sports 1978-08-23 2020-01-26 Wednesday Sunday  
#> 7 Lionel Messi       football/sports   1987-06-24 NA         Wednesday <NA>    
#> 8 Zinedine Zidane    football/sports   1972-06-23 NA         Friday    <NA>

4. Add a variable age_days that computes their age in days (relative to today’s date). Then compute two more variables age_yr1 and age_yr2 that determines their age in years (as a decimal number) in two different ways. Finally, add a variable age_fy that specifies their current age (in full years) as an integer (i.e., what they would say if they truthfully responded to the question “How old are you today?”).

Solution

The answer to “How old are you today?” can be computed in many different ways. The following dplyr pipe implements four different solutions:

# Determine today's date:
today <- Sys.Date()
# today <- lubridate::today()

fame %>%
  mutate(# 1. time difference (in days/days-in-average-year):  
         age_days = (today - dob),
         age_yr1 = as.numeric(age_days)/365.25,
         # 2. interval (in duration of years):
         # age_yr2 = interval(dob, today) / duration(num = 1, units = "years"),
         age_yr2 = interval(dob, today) / dyears(1),         
         # 3. interval (in period of years): 
         # age_yr3 = interval(dob, today) / period(num = 1, units = "years"),
         age_yr3 = interval(dob, today) / years(1),
         # Round down year values:
         age_fy1 = floor(age_yr1),          
         age_fy2 = floor(age_yr2),
         age_fy3 = floor(age_yr3),
         # 4. interval and periods with integer division:
         age_fy4 = interval(dob, today) %/% years(1),  
         # 5. If dead people do no longer age: 
         age_fy5 = ifelse(is.na(dod), age_fy4, interval(dob, dod) %/% years(1))
         ) %>%
  select(-area, -age_days, -age_yr1, -age_yr2, -age_yr3)
#> # A tibble: 8 × 8
#>   name             dob        dod        age_fy1 age_fy2 age_fy3 age_fy4 age_fy5
#>   <chr>            <date>     <date>       <dbl>   <dbl>   <dbl>   <dbl>   <dbl>
#> 1 Napoleon Bonapa… 1769-08-15 1821-05-05     253     253     253     253      51
#> 2 Jimi Hendrix     1942-11-27 1970-09-18      79      79      79      79      27
#> 3 Michael Jackson  1958-08-29 2009-06-25      64      64      64      64      50
#> 4 Frida Kahlo      1907-07-06 1954-07-13     115     115     115     115      47
#> 5 Angela Merkel    1954-07-17 NA              68      68      68      68      68
#> 6 Kobe Bryant      1978-08-23 2020-01-26      44      44      44      44      41
#> 7 Lionel Messi     1987-06-24 NA              35      35      35      35      35
#> 8 Zinedine Zidane  1972-06-23 NA              50      50      50      50      50

Note: The four solutions shown here will mostly yield the same results, but may still vary for some cases. This is quite a common situation when solving problems in R (and the same problem will re-occur below in Exercise 6). To find out which solutions are reliable, we would need to check critical cases (e.g., people whose birthday was yesterday, today, or tomorrow, born in different years).

The computation of age_fy5 assumes the premise that deceased people do not age any further, which suggests limiting their maximum age at their date of death. (See Section 11.3 of Chapter 11 on Functions for the ifelse() statement.)

5. Correct your previous age_fyr variable so that — for those people who have already died — it should remain at the age at which they died (i.e., dead people do not age further).

Solution

fame %>%
  filter(!is.na(dod)) %>%
  mutate(age_days = (dod - dob),
         age_yr1 = as.numeric(age_days)/365, 
         age_yr2 = interval(dob, dod) / duration(num = 1, units = "years"),
         age_fyr = floor(age_yr2)
         ) %>%
  select(-area)
#> # A tibble: 5 × 7
#>   name               dob        dod        age_days   age_yr1 age_yr2 age_fyr
#>   <chr>              <date>     <date>     <drtn>       <dbl>   <dbl>   <dbl>
#> 1 Napoleon Bonaparte 1769-08-15 1821-05-05 18890 days    51.8    51.7      51
#> 2 Jimi Hendrix       1942-11-27 1970-09-18 10157 days    27.8    27.8      27
#> 3 Michael Jackson    1958-08-29 2009-06-25 18563 days    50.9    50.8      50
#> 4 Frida Kahlo        1907-07-06 1954-07-13 17174 days    47.1    47.0      47
#> 5 Kobe Bryant        1978-08-23 2020-01-26 15131 days    41.5    41.4      41

A.10.4 Exercise 4

Time conversions

1. Define a time point of the New Year fireworks in Sydney, Australia, as “2021-01-01 00:00:01” (including time zone information).

Solution

# Time of Sydney NY fireworks:
(t_fw <- ymd_hms("2021-01-01 00:00:01", tz = "Australia/Sydney"))
#> [1] "2021-01-01 00:00:01 AEDT"

2. Predict and explain the results of the following commands in your own words.

with_tz(t_fw,  tz = "Europe/Berlin")
#> [1] "2020-12-31 14:00:01 CET"
force_tz(t_fw, tz = "Europe/Berlin")
#> [1] "2021-01-01 00:00:01 CET"

Solution

# (a) Convert a fixed time point into a different time zone: 
with_tz(t_fw,  tz = "Europe/Berlin")
#> [1] "2020-12-31 14:00:01 CET"
# => looks like a 10 hour time difference.

# in base R: 
format(t_fw, "%F %T %Z (UTC %z)")
#> [1] "2021-01-01 00:00:01 AEDT (UTC +1100)"
format(as.POSIXlt(t_fw, tz = "Europe/Berlin"), "%F %T %Z (UTC %z)")
#> [1] "2020-12-31 14:00:01 CET (UTC +0100)"

# (b) Same time display (but different time) in a different time zone: 
force_tz(t_fw, tz = "Europe/Berlin")
#> [1] "2021-01-01 00:00:01 CET"

3. Predict and explain the outcome of the following commands.

t_fw -  with_tz(t_fw, tz = "Europe/Berlin")
t_fw - force_tz(t_fw, tz = "Europe/Berlin")

Hint: This is possible without actually running them (after having done 2.).

Solution

t_fw -  with_tz(t_fw, tz = "Europe/Berlin")
#> Time difference of 0 secs
t_fw - force_tz(t_fw, tz = "Europe/Berlin")
#> Time difference of -10 hours

Answer:

• As with_tz() does not change the actual time represented (only its display), the time difference must be zero.

• As force_tz() changes the time (but not the time dislayed) and we have seen in 2. that the time in Sydney is 10 hours ahead of Berlin, the time difference must be \(-10\) hours.

A.10.5 Exercise 5

Hoop times

This exercise uses the lakers dataset included in lubridate (originally from http://www.basketballgeek.com/data/), which contains play-by-play statistics of each Los Angeles Lakers (LAL) basketball game in the 2008/2009 season of the NBA. (See ?lakers for details.)

1. Select only those games against the Dallas Mavericks (abbreviated as “DAL”) and save the corresponding data as a tibble LAL_DAL.

Solution

LAL_DAL <- as_tibble(lubridate::lakers) %>% 
  filter(opponent == "DAL")

2. Use your tidyverse knowledge acquired so far to answer some basic questions about those games:

   • How many such (home vs. away) games exist?
     
   • On which dates were they played?
     
   • What were their scores? Who won the game?

Hint: All these questions can be answered with a single dplyr pipe.

Solution

# How many (home vs. away) games?
# Game scores?  Who won?
as_tibble(LAL_DAL) %>% 
  filter(team != "OFF") %>%
  group_by(date, game_type, team) %>% 
  summarise(point_sum = sum(points)) %>%
  spread(key = team, val = point_sum)
#> # A tibble: 3 × 4
#> # Groups:   date, game_type [3]
#>       date game_type   DAL   LAL
#>      <int> <chr>     <int> <int>
#> 1 20081111 away         99   106
#> 2 20081128 home        107   114
#> 3 20090315 home        100   107

3. Create and add the following date and time variables to LAL_DAL:

   • date should be a variable of type “Date” (rather than a character string)
     
   • t_clock should represent the time shown on the clock (as a period)
     
   • t_psec should represent the time elapsed in the current period (a duration in seconds)
     
   • t_game should represent the time elapsed in the game overall (as a duration).

Hint: An NBA game consists of 4 periods, each of which lasts 12 minutes (i.e., each game’s time should add up to a total of 48 minutes).

Solution

# Data: 
# LAL_DAL
all(ds4psy::is_wholenumber(LAL_DAL$date))
#> [1] TRUE

# Define a constant:
# 4 periods of 12 minutes each (48 minutes in total)
t_period <- dminutes(12)

lada_1 <- LAL_DAL %>%
  select(-opponent, -game_type, -x, -y, -player, -result) %>% 
  mutate(date = ymd(date),  # convert integer into date
         t_clock = ms(LAL_DAL$time),
         t_psec = t_period - as.duration(t_clock),
         t_game = t_period * (period - 1) + t_psec
         )
tail(lada_1)
#> # A tibble: 6 × 10
#>   date       time  period etype team  points type  t_clock t_psec               
#>   <date>     <chr>  <int> <chr> <chr>  <int> <chr> <Perio> <Duration>           
#> 1 2009-03-15 00:07      4 free… LAL        1 ""    7S      713s (~11.88 minutes)
#> 2 2009-03-15 00:07      4 free… LAL        1 ""    7S      713s (~11.88 minutes)
#> 3 2009-03-15 00:07      4 time… DAL        0 "reg… 7S      713s (~11.88 minutes)
#> 4 2009-03-15 00:07      4 sub   LAL        0 ""    7S      713s (~11.88 minutes)
#> 5 2009-03-15 00:00      4 shot  DAL        0 "3pt" 0S      720s (~12 minutes)   
#> 6 2009-03-15 00:00      4 rebo… LAL        0 "def" 0S      720s (~12 minutes)   
#> # … with 1 more variable: t_game <Duration>

4. Prominent players:

   • For which individual player on each team do the data record the highest number of events?
     
   • How many points did each of these two players score (over all games)?
     
   • What would it take to compute the time difference between all recorded events for these two players as lubridate intervals?
     
   • Bonus task: Compute these intervals for each of these two players.

Solution

# Player with most events on each team?  
# How many points did they score?
LAL_DAL %>% 
  group_by(team, player) %>% 
  summarise(n_events = n(), 
            point_sum = sum(points)) %>%
  arrange(desc(n_events)) %>%
  head(2)
#> # A tibble: 2 × 4
#> # Groups:   team [2]
#>   team  player        n_events point_sum
#>   <chr> <chr>            <int>     <int>
#> 1 DAL   Dirk Nowitzki      114        53
#> 2 LAL   Kobe Bryant        113        90

- What would it take to compute the time difference between all recorded events for these two players as **lubridate** intervals?  

Answer:

All measures of time in the data so far are provided in terms of clock-time. Thus, they only denote the relative time elapsed within each game (i.e., ranging from 0 to a maximum of 48 minutes).

As lubridate intervals are anchored in actual calendar time, computing them would require looking up the starting time of each game (in the correct time zone) and adding t_game to it. This would yield a new date-time (or “POSIXct”) variable denoting actual calendar time. We could then use the lubridate function int_diff() on this variable to compute the time spans between events as intervals.

5. Cumulative points per game:

   • Compute and add a variable for the cumulative point_total of each game and team.
     
   • Compute the final score f_score of each game and team (and compare your result to the one obtained to answer 2. above).
     
   • Plot the (cumulative) point_total for each game per team as a function of t_game.

Solution

# lada_1  # data (from above)

# Add cumulative points per team:
lada_2 <- lada_1 %>%
  filter(team != "OFF") %>% 
  group_by(date, team) %>%
  mutate(point_total = cumsum(points))
# lada_2

# Compute and check final scores: 
lada_2 %>% 
  group_by(as.character(date), team) %>%
  summarise(f_score = max(point_total)
  ) %>%
  spread(key = team, value = f_score)
#> # A tibble: 3 × 3
#> # Groups:   as.character(date) [3]
#>   `as.character(date)`   DAL   LAL
#>   <chr>                <int> <int>
#> 1 2008-11-11              99   106
#> 2 2008-11-28             107   114
#> 3 2009-03-15             100   107

# Plot point_total as a function of t_game for each game and team: 
ggplot(lada_2, aes(x = t_game, y = point_total, group = team)) +
  geom_line(aes(col = team), size = 1, alpha = .8) + 
  facet_wrap(~date) +
  scale_color_manual(values = unikn::usecol(c(unikn::Seeblau, unikn::Pinky))) + 
  labs(title = "Cumulative points in NBA games between LAL and DAL (season 2008/2009)", 
       color = "Team:", x = "Time (sec)", y = "Cumulative points", 
       caption = "Based on lubridate's `lakers` data.") + 
  theme_ds4psy()

Please note: This dataset and questions like the ones asked here are a good illustration of a possible Data science project. At this point, you should be starting to think about datasets and questions for your own project. (See Appendix C for some guidelines for and the scope of a successful data science projects.)

A.10.6 Exercise 6

DOB and study times

The dataset exp_num_dt (available in the ds4psy package or as a CSV-file from rpository.com) contains the birth dates and study participation times of 1000 ficticious, but surprisingly friendly people.

We read the data file into a tibble dt and select only its date-related variables:

# dt <- readr::read_csv("http://rpository.com/ds4psy/data/dt.csv")  # online 
dt <- ds4psy::exp_num_dt  # ds4psy package
# dt

# Select only its date-time related variables:
dt_t <- dt %>% select(name:byear, t_1, t_2)

# Check:
# dt  # 1000 x 7
knitr::kable(head(dt_t), caption = "Time-related variables of table `dt`.")

Table A.32: Time-related variables of table dt.

name	gender	bday	bmonth	byear	t_1	t_2
I.G.	male	14	12	1968	2020-01-16 11:00:58	2020-01-16 11:32:21
O.B.	male	10	4	1974	2020-01-17 14:11:07	2020-01-17 15:05:14
M.M.	male	28	9	1987	2020-01-16 10:06:06	2020-01-16 10:51:47
V.J.	female	15	2	1978	2020-01-10 10:06:04	2020-01-10 10:39:48
O.E.	male	18	5	1985	2020-01-20 09:23:51	2020-01-20 10:11:36
Q.W.	male	1	3	1968	2020-01-13 11:10:09	2020-01-13 11:54:07

1. The variables bday, bmonth, and byear contain each participant’s date of birth.

   • Compute a variable DOB that summarizes bday, bmonth, and byear (in a “Date” variable) and a variable bweekday that shows the weekday of each participant’s DOB (as a chacter variable).

Hint: A base R solution is about as long as the dplyr/lubridate solution.

Solution

# (a) base R solution:
dt_1 <- dt_t  # copy file
DOB_strings <- paste(dt_1$byear, dt_1$bmonth, dt_1$bday, sep = "-")  # paste DOB string
dt_1$DOB <- as.Date(DOB_strings, format = "%Y-%m-%d")                # parse DOB
dt_1$bweekday <- format(dt_1$DOB, "%a")                              # retrieve weekday
dt_1
#> # A tibble: 1,000 × 9
#>    name  gender  bday bmonth byear t_1                 t_2                
#>    <chr> <chr>  <dbl>  <dbl> <dbl> <dttm>              <dttm>             
#>  1 I.G.  male      14     12  1968 2020-01-16 11:00:58 2020-01-16 11:32:21
#>  2 O.B.  male      10      4  1974 2020-01-17 14:11:07 2020-01-17 15:05:14
#>  3 M.M.  male      28      9  1987 2020-01-16 10:06:06 2020-01-16 10:51:47
#>  4 V.J.  female    15      2  1978 2020-01-10 10:06:04 2020-01-10 10:39:48
#>  5 O.E.  male      18      5  1985 2020-01-20 09:23:51 2020-01-20 10:11:36
#>  6 Q.W.  male       1      3  1968 2020-01-13 11:10:09 2020-01-13 11:54:07
#>  7 H.K.  male      27      4  1994 2020-01-19 13:54:15 2020-01-19 14:17:26
#>  8 T.R.  female     5      6  1961 2020-01-19 09:38:54 2020-01-19 10:33:33
#>  9 F.J.  male       1     10  1983 2020-01-15 08:24:11 2020-01-15 09:08:13
#> 10 J.R.  female    29     12  1941 2020-01-18 08:54:27 2020-01-18 09:35:21
#> # … with 990 more rows, and 2 more variables: DOB <date>, bweekday <chr>

# (b) lubridate solution:
dt_2 <- dt_t %>% 
  mutate(DOB = lubridate::make_date(day = bday, month = bmonth, year = byear),
         bweekday = lubridate::wday(DOB, label = TRUE, abbr = TRUE)) %>% 
  select(name:byear, DOB, bweekday, everything()) %>%
  mutate(bweekday = as.character(bweekday))
dt_2  
#> # A tibble: 1,000 × 9
#>    name  gender  bday bmonth byear DOB        bweekday t_1                
#>    <chr> <chr>  <dbl>  <dbl> <dbl> <date>     <chr>    <dttm>             
#>  1 I.G.  male      14     12  1968 1968-12-14 Sat      2020-01-16 11:00:58
#>  2 O.B.  male      10      4  1974 1974-04-10 Wed      2020-01-17 14:11:07
#>  3 M.M.  male      28      9  1987 1987-09-28 Mon      2020-01-16 10:06:06
#>  4 V.J.  female    15      2  1978 1978-02-15 Wed      2020-01-10 10:06:04
#>  5 O.E.  male      18      5  1985 1985-05-18 Sat      2020-01-20 09:23:51
#>  6 Q.W.  male       1      3  1968 1968-03-01 Fri      2020-01-13 11:10:09
#>  7 H.K.  male      27      4  1994 1994-04-27 Wed      2020-01-19 13:54:15
#>  8 T.R.  female     5      6  1961 1961-06-05 Mon      2020-01-19 09:38:54
#>  9 F.J.  male       1     10  1983 1983-10-01 Sat      2020-01-15 08:24:11
#> 10 J.R.  female    29     12  1941 1941-12-29 Mon      2020-01-18 08:54:27
#> # … with 990 more rows, and 1 more variable: t_2 <dttm>

# Verify equality:
all.equal(dt_1$DOB, dt_2$DOB)
#> [1] TRUE
all.equal(dt_1$bweekday, dt_2$bweekday)
#> [1] TRUE

Note: We could also parse DOB as calendar times/date-times (using the as.POSIXct() and make_datetime() functions). However, to obtain identical results in base R and lubridate, we need to specify the same time zone in both solutions (e.g., by setting tz = "").

2. What would each participant respond to the question

   • “How old are you?”

(i.e., what was each person’s age in completed years, when starting the study in January 2020)? Verify your result for those participants who took part in the study on their birthday.

Hint: This task requires considering both DOB and t_1 (to check whether the person already celebrated his or her birthday in the current year when starting the study at the time t_1).

Solution

dt_2 <- dt_t %>% 
  mutate(DOB = lubridate::make_date(day = bday, month = bmonth, year = byear),
         study_date = as.Date(t_1),         # time as date
         year_diff = lubridate::year(t_1) - lubridate::year(DOB),  # difference (in date years)
         life_time = DOB %--% study_date,   # a time interval (between dates)         
         life_time_2 = DOB %--% t_1,        # a time interval (between times)
         age = life_time_2 %/% years(1)) %>%  # completed years
  select(name, DOB, t_1, age, year_diff)
dt_2
#> # A tibble: 1,000 × 5
#>    name  DOB        t_1                   age year_diff
#>    <chr> <date>     <dttm>              <dbl>     <dbl>
#>  1 I.G.  1968-12-14 2020-01-16 11:00:58    51        52
#>  2 O.B.  1974-04-10 2020-01-17 14:11:07    45        46
#>  3 M.M.  1987-09-28 2020-01-16 10:06:06    32        33
#>  4 V.J.  1978-02-15 2020-01-10 10:06:04    41        42
#>  5 O.E.  1985-05-18 2020-01-20 09:23:51    34        35
#>  6 Q.W.  1968-03-01 2020-01-13 11:10:09    51        52
#>  7 H.K.  1994-04-27 2020-01-19 13:54:15    25        26
#>  8 T.R.  1961-06-05 2020-01-19 09:38:54    58        59
#>  9 F.J.  1983-10-01 2020-01-15 08:24:11    36        37
#> 10 J.R.  1941-12-29 2020-01-18 08:54:27    78        79
#> # … with 990 more rows

# Check: Participants with bmonth of 1 (January), who may 
#        already have celebrated their birthday in 2020: 
dt_2 %>% filter(lubridate::month(DOB) == 1)
#> # A tibble: 79 × 5
#>    name  DOB        t_1                   age year_diff
#>    <chr> <date>     <dttm>              <dbl>     <dbl>
#>  1 U.W.  1996-01-12 2020-01-13 10:33:52    24        24
#>  2 U.V.  1990-01-13 2020-01-20 13:00:44    30        30
#>  3 G.H.  1948-01-17 2020-01-17 15:29:00    72        72
#>  4 V.U.  1952-01-22 2020-01-17 11:09:41    67        68
#>  5 T.M.  1994-01-14 2020-01-12 14:45:03    25        26
#>  6 Y.B.  1956-01-10 2020-01-18 15:38:54    64        64
#>  7 H.V.  1973-01-07 2020-01-13 14:28:10    47        47
#>  8 F.H.  1947-01-21 2020-01-19 10:34:42    72        73
#>  9 H.R.  1974-01-14 2020-01-15 13:57:58    46        46
#> 10 R.S.  1972-01-12 2020-01-12 14:20:19    48        48
#> # … with 69 more rows

# Check: Participants starting the study on their birthday:
dt_2 %>% 
  filter(lubridate::month(DOB) == lubridate::month(t_1)) %>%
  filter(lubridate::day(DOB) == lubridate::day(t_1))
#> # A tibble: 4 × 5
#>   name  DOB        t_1                   age year_diff
#>   <chr> <date>     <dttm>              <dbl>     <dbl>
#> 1 G.H.  1948-01-17 2020-01-17 15:29:00    72        72
#> 2 R.S.  1972-01-12 2020-01-12 14:20:19    48        48
#> 3 Z.Q.  1992-01-20 2020-01-20 09:08:33    28        28
#> 4 N.Z.  1994-01-13 2020-01-13 10:08:16    26        26

3. The time variables t_1 and t_2 indicate the start and end times of each person’s participation in this study.

   • Compute the duration of each person’s participation (in minutes and seconds) and plot the distribution of the resulting durations (e.g., as a histogram).

Solution

dt_3 <- dt %>%
  mutate(t_diff = (t_2 - t_1), 
         dur = as.duration(t_2 - t_1))
# dt_3

# Get means:
dur_mn <- mean(dt_3$dur)    # mean
dur_md <- median(dt_3$dur)  # median

# Plot histograms:

## base R:
# hist(as.numeric(dt$dur), breaks = 20, col = unikn::Seeblau)  

# ggplot: 
ggplot(dt_3, aes(x = as.numeric(dur))) +
  geom_histogram(col = "black", binwidth = 200, fill = unikn::Seeblau) +
  geom_vline(xintercept = dur_mn, col = "gold", linetype = 1, size = 1) +
  geom_vline(xintercept = dur_md, col = unikn::Pinky, linetype = 2, size = 1) +
  labs(title = "Distribution of durations", x = "Duration (in seconds)") + 
  ds4psy::theme_ds4psy()

4. The study officially only ran for 5 days — from “2020-01-13” to “2020-01-18” — and should only include participants that responded in up to one hour (60 minutes).

   • Add a filter variable valid that enforces these criteria (i.e., allows filtering out participants with other dates and durations longer than 60 minutes).

Solution

dt_4 <- dt_3 %>%
  mutate(date = as_date(t_1),
         valid_date = (date >= "2020-01-13") & (date <= "2020-01-18"),
         valid_dur  = (dur <= as.duration(60 * 60)),
         valid      = valid_date & valid_dur) %>%
  filter(valid)

# Filtered data: 
dt_4
#> # A tibble: 519 × 21
#>    name  gender  bday bmonth byear height blood_…¹ bnt_1 bnt_2 bnt_3 bnt_4  g_iq
#>    <chr> <chr>  <dbl>  <dbl> <dbl>  <dbl> <chr>    <dbl> <dbl> <dbl> <dbl> <dbl>
#>  1 I.G.  male      14     12  1968    169 O+           1     0     0     1   113
#>  2 O.B.  male      10      4  1974    181 O+           1     1     1    NA   114
#>  3 M.M.  male      28      9  1987    183 A-           0     1     0     0   108
#>  4 Q.W.  male       1      3  1968    172 A+           1     1     1     0   103
#>  5 F.J.  male       1     10  1983    158 O+           0     0     0     0   107
#>  6 J.R.  female    29     12  1941    157 O+           1     1     0     1   107
#>  7 K.E.  male      10     12  1951    161 A+           0     0     1     1   104
#>  8 U.W.  female    12      1  1996    161 O+           0     1     0     0   104
#>  9 J.Y.  female    20      5  1987    155 O-           0     1     1    NA   106
#> 10 S.X.  female     5      3  1986    169 O+           1     0     0     1    95
#> # … with 509 more rows, 9 more variables: s_iq <dbl>, t_1 <dttm>, t_2 <dttm>,
#> #   t_diff <drtn>, dur <Duration>, date <date>, valid_date <lgl>,
#> #   valid_dur <lgl>, valid <lgl>, and abbreviated variable name ¹​blood_type

# Check: Does the filter work as intended?
min(dt_4$t_1)
#> [1] "2020-01-13 08:09:30 UTC"
max(dt_4$t_1)
#> [1] "2020-01-18 17:50:53 UTC"
max(dt_4$dur)
#> [1] 3600

5. Finally, we can compute some basic descriptives of the participants considered to be valid:

   • How many participants remain in the sample of valid data?
     
   • What is their average height and g_iq score?

Solution

# Get descriptives (by hand):
nrow(dt_4)  # N of valid participants
#> [1] 519
mean(dt_4$height)
#> [1] 166.2852
mean(dt_4$g_iq, na.rm = TRUE)  # There are NA values!
#> [1] 101.5207
sum(!is.na(dt_4$g_iq))         # N of non-NA values?
#> [1] 507

# All in one dplyr pipe:
dt_4 %>% 
  summarise(N = n(),
            mn_height = mean(height),
            N_hg_nonNA = sum(!is.na(height)), 
            mn_iq = mean(g_iq, na.rm = TRUE),
            N_iq_nonNA = sum(!is.na(g_iq)))
#> # A tibble: 1 × 5
#>       N mn_height N_hg_nonNA mn_iq N_iq_nonNA
#>   <int>     <dbl>      <int> <dbl>      <int>
#> 1   519      166.        519  102.        507

A.10.7 Exercise 7

Bonus task: Evaluating time differences

This exercise creates random time differences and compares the results of computing them in two different ways.

1. Use the sample_time() function of ds4psy to generate vectors of N random starting times and N random end times.

2. Compute and compare the time difference between both vectors for various units of time. Specifically, compare the solutions of the diff_times() function of ds4psy with the corresponding lubridate solution (using time intervals and periods).

3. Continue comparing the results of both solution methods until you find some examples with different solutions for the same time difference. Can you explain the discrepancies?

Hint: Here is a possible setup for an investigation of this type:

# Parameters:
N <- 10
t1 <- "2020-01-01 00:00:00"
t2 <- Sys.time()

# Random time vectors:
t_start <- ds4psy::sample_time(from = t1, to = t2, size = N)
t_end   <- ds4psy::sample_time(from = t1, to = t2, size = N)

# in months:
ds4psy::diff_times(t_start, t_end, unit = "months", as_character = FALSE)
lubridate::as.period(lubridate::interval(t_start, t_end), unit = "months")

# in days:
ds4psy::diff_times(t_start, t_end, unit = "days", as_character = FALSE)
lubridate::as.period(lubridate::interval(t_start, t_end), unit = "days")

Solution

Here are some examples with discrepancies between solutions (different day counts):

# (1) 
t1 <- "2020-04-14 10:00:00"
t2 <- "2020-02-25 05:00:00"

ds4psy::diff_times(t1, t2, unit = "months", as_character = TRUE)
#> [1] "-1m 20d 5H 0M 0S"
lubridate::as.period(lubridate::interval(t1, t2), unit = "months")
#> [1] "-1m -18d -5H 0M 0S"

# (2)
t1 <- "2020-05-11 12:00:00" 
t2 <- "2020-02-15 10:00:00"

ds4psy::diff_times(t1, t2, unit = "months", as_character = TRUE)
#> [1] "-2m 26d 2H 0M 0S"
lubridate::as.period(lubridate::interval(t1, t2), unit = "months")
#> [1] "-2m -25d -2H 0M 0S"

# (3)
t1 <- "2020-03-15 15:00:00" 
t2 <- "2020-01-28 16:00:00"

ds4psy::diff_times(t1, t2, unit = "months", as_character = TRUE)
#> [1] "-1m 15d 23H 0M 0S"
lubridate::as.period(lubridate::interval(t1, t2), unit = "months")
#> [1] "-1m -17d -23H 0M 0S"

Answer: Negative time intervals are occasionally handled differently by both packages. In ds4psy, the solution of diff_times(t1, t2) is the negation of diff_times(t2, t1), which corresponds to our understanding of subtraction.

This concludes our exercises on creating and computing with dates and times.