R Date & Time Exercises: 20 lubridate Practice Problems
Twenty practice problems covering date parsing, component extraction, period and interval arithmetic, timezone handling, and date-driven analytics workflows in R. Each problem has a hidden, fully worked solution with an explanation, runnable directly in your browser.
Section 1. Parsing dates and times (4 problems)
Exercise 1.1: Parse a year-month-day string with ymd
Task: Parse the character string "2024-03-15" into a proper R Date object using ymd() from lubridate. The result should print as a single date and have class "Date". Save the parsed value to ex_1_1 and inspect both the value and its class.
Expected result:
#> [1] "2024-03-15"
#> [1] "Date"Difficulty: Beginner
Identify which calendar component the string lists first, second, and third, then pick the parser keyed to that exact order.
Hand the raw string straight to ymd() - no format string is required.
Click to reveal solution
Explanation: ymd() is a strict, fast parser keyed by component ORDER rather than separators, so "2024/03/15", "2024-03-15", and "20240315" all work. Compared to base R's as.Date() you do not have to pass a format string. The result is a Date (1 day precision); use ymd_hms() when you need sub-day precision.
Exercise 1.2: Parse a CSV column with mixed date formats
Task: A marketing analyst received conversion logs from three vendors and concatenated them. The raw vector below contains mixed formats: ISO, US slash, and European dash. Use parse_date_time() with all three format hints so every entry parses correctly. Save the cleaned Date vector to ex_1_2 and confirm none are NA.
Expected result:
#> [1] "2024-03-15 UTC" "2024-03-15 UTC" "2024-03-15 UTC" "2024-04-01 UTC" "2024-04-02 UTC"
#> [1] 0Difficulty: Intermediate
The vector mixes three different component orderings, so you need a parser you can give several format hints to at once.
Call parse_date_time() with orders = c("ymd", "mdy", "dmy"), listing the most specific order first.
Click to reveal solution
Explanation: parse_date_time() tries each order in turn and picks the first that matches without ambiguity, so messy real-world feeds parse without writing a custom regex. Order matters: put the most specific format first; if mdy came before ymd, the string "2024-03-15" would still parse as ymd because ymd is unambiguous, but for ambiguous strings the first listed order wins.
Exercise 1.3: Parse a datetime with seconds and treat it as POSIXct
Task: Convert the string "2024-03-15 14:30:45" into a POSIXct value using ymd_hms(). Save the result to ex_1_3. Print it and confirm via class() that it carries both POSIXct and POSIXt classes, which is what date-aware functions look for.
Expected result:
#> [1] "2024-03-15 14:30:45 UTC"
#> [1] "POSIXct" "POSIXt"Difficulty: Intermediate
The string carries hours, minutes, and seconds, so reach for the parser variant that captures sub-day precision.
Use ymd_hms() on the string; it returns a POSIXct value in UTC by default.
Click to reveal solution
Explanation: ymd_hms() returns POSIXct, an integer number of seconds since 1970-01-01 UTC. The default timezone is UTC when no tz argument is given. Use POSIXct (not POSIXlt) for data frames and vectorized math; POSIXlt is a list-of-components structure that breaks dplyr columns. If your input lacks seconds, use ymd_hm() so missing seconds do not generate NA.
Exercise 1.4: Truncate noisy timestamps before parsing
Task: An audit team received timestamps with extra text appended: "2024-03-15 14:30:00 EST (audit log)". The trailing parenthetical breaks ymd_hms(). Use parse_date_time() with truncated = 3 and an orders hint so the parser tolerates trailing noise. Parse the full vector and save the cleaned datetimes to ex_1_4.
Expected result:
#> [1] "2024-03-15 14:30:00 UTC" "2024-03-16 09:00:15 UTC"
#> [3] "2024-03-17 23:59:59 UTC"Difficulty: Intermediate
The trailing parenthetical is noise the parser can be told to tolerate rather than something you must scrub out first.
Call parse_date_time() with orders = "Ymd HMS" and truncated = 3.
Click to reveal solution
Explanation: parse_date_time() ignores trailing characters once it has consumed the requested format. truncated = N allows up to N missing trailing components. An alternative is to pre-clean with sub("\\s*\\(.*\\)$", "", raw) then call ymd_hms(); that is more explicit when the noise pattern is regular. Prefer regex pre-cleaning when the same noise appears on every row; use truncated when noise is varied.
Section 2. Extracting date components (3 problems)
Exercise 2.1: Pull year, month, day, and hour from a POSIXct vector
Task: Given the POSIXct vector stamps, extract the year, month, day, and hour into four separate integer vectors using year(), month(), day(), and hour(). Combine them into a tibble with columns year, month, day, hour and save it to ex_2_1.
Expected result:
#> # A tibble: 3 x 4
#> year month day hour
#> <dbl> <dbl> <int> <int>
#> 1 2024 3 15 9
#> 2 2024 7 4 14
#> 3 2024 12 31 23Difficulty: Beginner
Each calendar component can be pulled out with its own dedicated accessor, one accessor per column.
Wrap year(), month(), day(), and hour() calls inside a tibble().
Click to reveal solution
Explanation: Each accessor returns a numeric vector the same length as its input, so they slot directly into a tibble column. For month names instead of numbers, use month(stamps, label = TRUE, abbr = FALSE). The lubridate::wday() companion gives weekday with a label flag. These accessors are vectorized over POSIXct, Date, and character (after coercion), making them the workhorse for feature engineering on datetime columns.
Exercise 2.2: Tag a retailer's orders with weekday name and weekend flag
Task: A retailer wants to see whether orders skew toward weekends. Given the inline orders tibble of order timestamps, add two columns: dow (weekday name like "Mon") and is_weekend (logical TRUE for Saturday or Sunday). Save the augmented tibble to ex_2_2.
Expected result:
#> # A tibble: 6 x 4
#> order_id placed_at dow is_weekend
#> <int> <dttm> <ord> <lgl>
#> 1 1 2024-03-15 10:14:00 Fri FALSE
#> 2 2 2024-03-16 11:30:00 Sat TRUE
#> 3 3 2024-03-17 19:02:00 Sun TRUE
#> 4 4 2024-03-18 08:45:00 Mon FALSE
#> 5 5 2024-03-19 14:20:00 Tue FALSE
#> 6 6 2024-03-23 16:00:00 Sat TRUEDifficulty: Intermediate
You need two derived facts about each timestamp: the weekday name and whether that weekday lands on a weekend.
In mutate(), use wday(..., label = TRUE) for the name and wday(..., week_start = 1) >= 6 for the flag.
Click to reveal solution
Explanation: wday(label = TRUE) returns an ordered factor with weekday names, which is great for axis labels and arrange(). The numeric form depends on week_start: the default (1 for Monday) makes Saturday 6 and Sunday 7, so >= 6 cleanly tags weekends. The default in some locales is Sunday-start, which would flip the comparison; always set week_start explicitly for reproducible reports.
Exercise 2.3: Compute ISO week and quarter from economics dates
Task: Use the built-in economics tibble. Take the first ten rows and add quarter (e.g. "2024 Q1") and iso_week (1 to 53) columns derived from the date column. Save the augmented ten-row tibble to ex_2_3 and print it.
Expected result:
#> # A tibble: 10 x 8
#> date pce pop psavert uempmed unemploy quarter iso_week
#> <date> <dbl> <int> <dbl> <dbl> <int> <chr> <dbl>
#> 1 1967-07-01 507. 198712 12.6 4.5 2944 1967 Q3 26
#> 2 1967-08-01 510. 198911 12.6 4.7 2945 1967 Q3 31
#> 3 1967-09-01 516. 199113 11.9 4.6 2958 1967 Q3 35
#> 4 1967-10-01 513. 199311 12.9 4.9 3143 1967 Q4 39
#> ...
#> # 6 more rows hiddenDifficulty: Intermediate
One column needs the calendar quarter, the other needs the international standard week number, both derived from the date column.
In mutate(), combine year() with quarter() for the label and use isoweek() for the week.
Click to reveal solution
Explanation: quarter() returns the calendar quarter as an integer 1 to 4, while isoweek() returns the ISO 8601 week number, where week 1 is the week containing the first Thursday of the year. Use week() for the simpler "day-of-year / 7" definition; ISO weeks matter when aligning to international reporting standards, payroll, or sports seasons that span year boundaries.
Section 3. Period, duration, and interval arithmetic (4 problems)
Exercise 3.1: Add thirty days to a date
Task: Take the date "2024-01-15" and add 30 days to it using days(30) from lubridate. Save the resulting date to ex_3_1 and print it. The output should be a Date that is exactly thirty calendar days later than the input, with no time component attached.
Expected result:
#> [1] "2024-02-14"Difficulty: Beginner
Adding a calendar span to a date leaves you with a date - no clock component appears.
Add days(30) to the value returned by ymd("2024-01-15").
Click to reveal solution
Explanation: days(30) builds a Period of 30 days, which respects calendar boundaries. Compared with ddays(30), which is a fixed Duration of 30 * 86400 seconds, periods are usually what humans expect for "30 days later". The two diverge only across DST transitions or leap seconds. For business arithmetic ("30 days from today"), prefer days() over ddays().
Exercise 3.2: Compute trial expiration using months vs dmonths
Task: A SaaS subscription team needs to mark every trial as expiring exactly one calendar month after signup_date. The team has tried signup_date + dmonths(1) and seen off-by-one bugs near month ends. Using months(1), build the inline tibble's expires_on column and save the result to ex_3_2.
Expected result:
#> # A tibble: 3 x 3
#> user signup_date expires_on
#> <chr> <date> <date>
#> 1 alice 2024-01-31 2024-02-29
#> 2 bob 2024-02-15 2024-03-15
#> 3 carol 2024-03-30 2024-04-30Difficulty: Intermediate
Plain addition of a one-month span fails at month ends because the target day may not exist, so you need the operator that rolls back to the last valid day.
Combine months(1) with the %m+% operator inside mutate().
Click to reveal solution
Explanation: signup_date + months(1) returns NA for 2024-01-31 because there is no Feb 31. The %m+% operator rolls back to the last valid day of the target month, which is what billing systems want. dmonths(1) is a fixed 30.4375 days and drifts; never use it for calendar-month math. Pair %m+% with months() whenever you need to add or subtract calendar months safely.
Exercise 3.3: Days between two dates as an integer
Task: Given the start date "2024-01-01" and the end date "2024-03-15", compute the number of full days between them as an integer scalar. Use interval() and division by days(1), then coerce to integer. Save the integer count to ex_3_3 and print it along with its class.
Expected result:
#> [1] 74
#> [1] "integer"Difficulty: Intermediate
Anchor the span to a real start and end, then express its length as a count of whole days.
Divide interval(start_d, end_d) by days(1) and wrap the result in as.integer().
Click to reveal solution
Explanation: interval(start, end) builds a time span anchored to a real start/end, and dividing by days(1) returns the number of days as a Period ratio. This survives DST and month-length variation. The shorter as.integer(difftime(end_d, start_d, units = "days")) works for plain Dates but can produce non-integer results across DST when applied to POSIXct, so the interval/period idiom is safer for mixed types.
Exercise 3.4: Detect overlapping booking windows
Task: An operations team manages meeting-room reservations and needs to flag any pair of bookings whose intervals overlap. From the inline bookings tibble, build a tibble of all i < j index pairs where int_overlaps() returns TRUE on their intervals. Save the result to ex_3_4 with columns i, j, and room.
Expected result:
#> # A tibble: 2 x 3
#> i j room
#> <int> <int> <chr>
#> 1 1 2 A
#> 2 4 5 BDifficulty: Advanced
Compare every distinct pair of bookings in the same room to see whether their time spans intersect.
Build spans with interval(), generate the i < j index pairs, and test each pair with int_overlaps().
Click to reveal solution
Explanation: int_overlaps() treats intervals as closed-open so a booking that ends exactly when another starts is NOT flagged as overlapping, which matches calendar app behavior. Restricting to i < j avoids double counting and self-pairs. For large booking tables, sort by start time and sweep once instead of comparing all pairs; the all-pairs approach here is the cleanest correct baseline, fine up to a few thousand rows.
Section 4. Timezones and formatting (3 problems)
Exercise 4.1: Convert a UTC log timestamp to New York exchange time
Task: A trading desk records all order events in UTC. Convert the UTC timestamp vector events to America/New_York local time without changing the underlying moment. Use with_tz(). Save the converted POSIXct vector to ex_4_1 and verify the printed timezone abbreviation matches EST or EDT.
Expected result:
#> [1] "2024-01-15 09:30:00 EST" "2024-07-15 10:30:00 EDT"Difficulty: Intermediate
Re-express the same absolute instant under a different regional clock, without shifting the moment itself.
Call with_tz() with tzone = "America/New_York".
Click to reveal solution
Explanation: with_tz() keeps the absolute moment fixed and re-labels it in a new zone, so the wall-clock display changes. Its sibling force_tz() does the opposite: it keeps the wall-clock unchanged and reinterprets it in a new zone, which shifts the absolute moment. Use with_tz() for display conversions, force_tz() to fix mislabeled raw data. Notice how DST flips the offset between January (EST, UTC-5) and July (EDT, UTC-4).
Exercise 4.2: Format a date as Mon, Jan 15 2024
Task: Format the date "2024-01-15" as the human-friendly string "Mon, Jan 15 2024" using format() with the right format codes. Save the formatted character scalar to ex_4_2 and print it. The day-of-week abbreviation comes first, then comma, then month-day-year.
Expected result:
#> [1] "Mon, Jan 15 2024"Difficulty: Intermediate
Turn the date into a display string by spelling out which pieces appear and in what order.
Pass the format string "%a, %b %d %Y" to format().
Click to reveal solution
Explanation: format.Date uses C-style strftime codes: %a weekday abbreviated, %A weekday full, %b month abbreviated, %B month full, %d zero-padded day, %Y four-digit year. For locale-independent output set Sys.setlocale("LC_TIME", "C") first, otherwise weekday and month names follow the user's locale. lubridate has a stamp() function that infers the format from an example string and can be handy for irregular outputs.
Exercise 4.3: Detect a DST spring-forward in a timestamp sequence
Task: An ops engineer wants to flag the spring-forward day in a daily timestamp sequence. Build a daily sequence of POSIXct values at 02:30 local time for 2024-03-09 through 2024-03-11 in America/New_York. Compute the dst() flag for each. Save a tibble with columns ts and is_dst as ex_4_3.
Expected result:
#> # A tibble: 3 x 2
#> ts is_dst
#> <dttm> <lgl>
#> 1 2024-03-09 02:30:00 FALSE
#> 2 2024-03-10 03:30:00 TRUE
#> 3 2024-03-11 02:30:00 TRUEDifficulty: Advanced
Build the regular daily series first, then test each timestamp for whether it sits inside Daylight Saving.
Generate the sequence with seq(..., by = "1 day") in the New York zone, then apply dst() inside a tibble().
Click to reveal solution
Explanation: Notice that the printed clock time on March 10 jumps from 02:30 to 03:30: the hour 02:00 to 03:00 simply does not exist in America/New_York that day. dst() returns TRUE for moments inside Daylight Saving. This is why scheduling jobs with naive local timestamps is fragile; mission-critical schedules should run in UTC and convert to local for display. For non-DST math use force_tz(ts, "UTC") first.
Section 5. Sequences, filtering, and aggregation (3 problems)
Exercise 5.1: Generate the next seven daily dates starting today
Task: Build a length-7 vector of consecutive daily Dates starting on "2024-03-15" using seq.Date() or seq() with by = "1 day". Save the vector to ex_5_1 and print it. The result should be a clean sequence with class Date and length seven.
Expected result:
#> [1] "2024-03-15" "2024-03-16" "2024-03-17" "2024-03-18"
#> [5] "2024-03-19" "2024-03-20" "2024-03-21"Difficulty: Beginner
Ask for a regular run of consecutive days with a fixed start and a known count.
Call seq() on ymd("2024-03-15") with by = "1 day" and length.out = 7.
Click to reveal solution
Explanation: seq() dispatches to seq.Date() when the start is a Date. You can use by = "1 day", "1 week", "1 month", "1 year", or any duration string. Replacing length.out = 7 with to = ymd("2024-03-21") gives the same result. To pull only weekdays, generate seven dates then filter with wday(x, week_start = 1) <= 5. For business-day calendars across holidays, the bizdays package is the cleaner tool.
Exercise 5.2: Filter rows to the last 90 days relative to a fixed today
Task: Given the inline events tibble and a fixed reference date "2024-04-01", return rows whose ts is strictly within the last 90 days, that is ts > today - 90 days. Use difftime() or interval division to keep the comparison safe. Save the filtered tibble to ex_5_2.
Expected result:
#> # A tibble: 4 x 2
#> id ts
#> <int> <date>
#> 1 2 2024-01-05
#> 2 3 2024-01-15
#> 3 4 2024-02-10
#> 4 5 2024-03-20Difficulty: Intermediate
Keep only rows inside a trailing window measured back from the reference date and stopping before it.
In filter(), compare ts against today - days(90) and against today.
Click to reveal solution
Explanation: Subtracting days(90) from a Date returns another Date, so the comparison stays type-stable. The strict inequalities exclude rows exactly 90 days old and rows AT today, matching most "trailing window, excluding today" reports. Replacing today - days(90) with today - 90 works for plain Dates but breaks if the column is POSIXct; using days(90) is the idiom-safe form for either type.
Exercise 5.3: Aggregate the economics series by year
Task: Use the built-in economics tibble. Group rows by calendar year of the date column and compute the mean unemploy (in thousands) per year. Sort descending by mean unemployment. Save the resulting tibble of year and mean_unemploy to ex_5_3 and print the top six rows.
Expected result:
#> # A tibble: 6 x 2
#> year mean_unemploy
#> <dbl> <dbl>
#> 1 2010 14825.
#> 2 2011 13747.
#> 3 2012 12506.
#> 4 2009 14265.
#> 5 2013 11460.
#> 6 2014 9617.Difficulty: Intermediate
Roll the monthly series up to one row per calendar year, then order rows by the summarised value.
Add a year column with year(), then group_by(), summarise(mean(unemploy)), and arrange(desc(...)).
Click to reveal solution
Explanation: Extracting year() then grouping is the most readable pattern for yearly rollups. The newer dplyr 1.1+ inline form summarise(.by = year(date), ...) also works and avoids creating a permanent column. For multi-grain aggregations (year-quarter, year-month), build all grain columns once with mutate() then call summarise() separately; recomputing inside summarise makes the SQL plan harder to read for collaborators.
Section 6. Real-world workflows (3 problems)
Exercise 6.1: Compute patient age in years at encounter date
Task: A healthcare analyst needs the patient's age in completed years at the time of each clinic encounter. From the inline patients tibble, compute age using interval division by years(1) and integer truncation so a birthday not yet reached this year rounds down. Save the augmented tibble to ex_6_1.
Expected result:
#> # A tibble: 4 x 4
#> patient_id date_of_birth encounter_dt age
#> <chr> <date> <date> <int>
#> 1 P001 1980-06-15 2024-03-15 43
#> 2 P002 1995-12-31 2024-03-15 28
#> 3 P003 2010-03-15 2024-03-15 14
#> 4 P004 1955-07-04 2024-03-15 68Difficulty: Advanced
Measure the span as a fractional age that knows whether this year's birthday has passed, then floor it to completed years.
Divide interval(date_of_birth, encounter_dt) by years(1), then apply trunc() and as.integer().
Click to reveal solution
Explanation: interval(dob, enc) / years(1) returns a fractional age (e.g. 43.75) that correctly accounts for whether the birthday has occurred this year. trunc() then floors toward zero, giving completed years, the standard clinical and legal definition of age. Using (enc - dob) / 365.25 is wrong: leap years and calendar drift make it disagree with the legal age near birthdays. Always prefer the interval-period idiom for age computations.
Exercise 6.2: Tag invoices as on-time, due-soon, or overdue
Task: A finance team scans an invoice ledger and needs each row tagged. Given the inline invoices tibble and today = ymd("2024-04-01"), add a column status with three values: "overdue" if due_date < today, "due_soon" if due within 7 days, otherwise "on_time". Save the tagged tibble to ex_6_2.
Expected result:
#> # A tibble: 5 x 4
#> invoice_id amount due_date status
#> <chr> <dbl> <date> <chr>
#> 1 INV001 1200 2024-03-15 overdue
#> 2 INV002 850 2024-04-04 due_soon
#> 3 INV003 4500 2024-04-15 on_time
#> 4 INV004 300 2024-03-30 overdue
#> 5 INV005 2200 2024-04-30 on_timeDifficulty: Intermediate
Assign each row one of three labels by testing the due date against the reference date in priority order, most urgent first.
Use case_when() with due_date < today as the first branch and due_date <= today + days(7) as the second.
Click to reveal solution
Explanation: case_when() reads top-to-bottom and the first matching branch wins, so listing overdue first prevents an overdue row from being relabeled due_soon. Adding days(7) to a Date returns a Date, so the comparison stays type-stable; mixing in dweeks(1) (a Duration) would coerce the Date to POSIXct and surprise downstream joins. Use Periods for calendar windows in business logic.
Exercise 6.3: Build a calendar tibble with month, quarter, and weekend flag
Task: Build a daily calendar tibble for every day in Q1 2024 (Jan 1 through Mar 31) with columns date, month_name, quarter, iso_week, and is_weekend. This is the kind of dimension table a BI team prebuilds so dashboards do not have to recompute date features. Save the result to ex_6_3 and inspect the first six rows.
Expected result:
#> # A tibble: 91 x 5
#> date month_name quarter iso_week is_weekend
#> <date> <ord> <int> <dbl> <lgl>
#> 1 2024-01-01 January 1 1 FALSE
#> 2 2024-01-02 January 1 1 FALSE
#> 3 2024-01-03 January 1 1 FALSE
#> 4 2024-01-04 January 1 1 FALSE
#> 5 2024-01-05 January 1 1 FALSE
#> 6 2024-01-06 January 1 1 TRUE
#> ...
#> # 85 more rows hiddenDifficulty: Advanced
Lay down one row per day across the quarter first, then attach each date feature as its own column.
Build the date series with seq(..., by = "1 day"), then mutate() with month(), quarter(), isoweek(), and wday().
Click to reveal solution
Explanation: A pre-built daily calendar table is the single best optimization for date-driven BI: every dashboard joins to it instead of recomputing weekday/quarter/holiday flags. In production extend this with is_holiday, fiscal_year, and a working_day_index column. The tidyquant::tk_make_timeseries_signature() helper produces a thirty-plus-column version automatically; rolling your own gives you control over the columns you actually need.
What to do next
Now that you have practiced date and time work, here are the next stops:
- lubridate Exercises in R for a deeper drill on lubridate's full API.
- R Syntax 101 to revisit the foundations that the date system builds on.
- dplyr Exercises in R to pair date features with grouped summaries.
- Apply Family Exercises in R for the vectorization patterns that complement date math.
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