tidyr Exercises in R: 30 Real-World Practice Problems
Thirty practice problems on the tidyr verbs that move data between shapes: pivot_longer, pivot_wider, separate, unite, fill, complete, drop_na, replace_na, nest, and unnest. Each exercise embeds a realistic data task, with the expected output stated up front and the solution hidden until you click to reveal.
Section 1. Pivot longer: wide-to-long reshaping (5 problems)
Exercise 1.1: Pivot three numeric columns of mtcars into long format
Task: A reporting analyst wants a tidy long-format slice of the mtcars dataset showing model, metric, and value for the columns mpg, cyl, and hp. Take the first five rows of mtcars, push the row names into a model column, then pivot the three metrics into key-value pairs and save to ex_1_1.
Expected result:
#> # A tibble: 15 x 3
#> model metric value
#> <chr> <chr> <dbl>
#> 1 Mazda RX4 mpg 21
#> 2 Mazda RX4 cyl 6
#> 3 Mazda RX4 hp 110
#> 4 Mazda RX4 Wag mpg 21
#> 5 Mazda RX4 Wag cyl 6
#> 6 Mazda RX4 Wag hp 110
#> 7 Datsun 710 mpg 22.8
#> 8 Datsun 710 cyl 4
#> 9 Datsun 710 hp 93
#> 10 Hornet 4 Drive mpg 21.4
#> # ... 5 more rowsDifficulty: Beginner
The row names are not a column yet, so promote them first; then collapse the three metric columns into one name column and one value column.
Chain rownames_to_column("model"), slice_head(n = 5), select(), then pivot_longer() with names_to and values_to.
Click to reveal solution
Explanation: pivot_longer() collapses several columns into two: a names column (metric) holding the original column names and a values column (value) holding the cells. Use it whenever each row should represent one observation of one variable. The cols argument accepts tidy-select helpers like starts_with() or where(is.numeric) when the column list is large.
Exercise 1.2: Reshape all iris measurements into a single value column
Task: A statistics tutor wants iris in long form so they can facet a ggplot by measurement. Pivot every numeric column (Sepal.Length, Sepal.Width, Petal.Length, Petal.Width) into a long table keeping Species intact, and save the result to ex_1_2.
Expected result:
#> # A tibble: 600 x 3
#> Species measurement value
#> <fct> <chr> <dbl>
#> 1 setosa Sepal.Length 5.1
#> 2 setosa Sepal.Width 3.5
#> 3 setosa Petal.Length 1.4
#> 4 setosa Petal.Width 0.2
#> 5 setosa Sepal.Length 4.9
#> 6 setosa Sepal.Width 3
#> ...
#> # 594 more rows hiddenDifficulty: Intermediate
Every measurement column should stack into a single value column while the species label stays untouched on each row.
Call pivot_longer() with cols = -Species, names_to = "measurement", and values_to = "value".
Click to reveal solution
Explanation: Using cols = -Species is a negative tidy-select: pivot every column except Species. It is more maintainable than spelling out the four numeric columns because adding a new metric to the source no longer breaks the pipeline. where(is.numeric) is an equivalent guard that excludes the factor automatically.
Exercise 1.3: Pivot wide exam scores into a per-student long table
Task: A junior analyst has a wide exam tibble where each student row holds math, science, and english scores. Build the tibble inline, then pivot it longer so each row is one student-subject-score combination, ordered by student. Save the result to ex_1_3.
Expected result:
#> # A tibble: 9 x 3
#> student subject score
#> <chr> <chr> <dbl>
#> 1 Alice math 88
#> 2 Alice science 72
#> 3 Alice english 91
#> 4 Bob math 64
#> 5 Bob science 80
#> 6 Bob english 77
#> 7 Carla math 95
#> 8 Carla science 89
#> 9 Carla english 84Difficulty: Intermediate
Each subject column should become a row, with the student identifier carried along into every new row.
Call pivot_longer() with cols = -student, names_to = "subject", and values_to = "score".
Click to reveal solution
Explanation: This is the classic gradebook reshape. The wide layout is good for entering data but bad for analysis: you cannot easily filter on subject or join in a per-subject pass mark. The long layout has one observation per row, which is what dplyr group_by(subject) and ggplot facet_wrap(~subject) expect.
Exercise 1.4: Use names_sep to split a compound header during pivot
Task: A marketing analyst pulled a campaign report where columns are named like clicks_jan, clicks_feb, signups_jan, signups_feb. Pivot longer in a single call, splitting each header on the underscore into two new columns metric and month, and save to ex_1_4.
Expected result:
#> # A tibble: 8 x 4
#> campaign metric month value
#> <chr> <chr> <chr> <dbl>
#> 1 A clicks jan 120
#> 2 A clicks feb 150
#> 3 A signups jan 18
#> 4 A signups feb 24
#> 5 B clicks jan 90
#> 6 B clicks feb 110
#> 7 B signups jan 9
#> 8 B signups feb 14Difficulty: Intermediate
Each header carries two facts joined by a delimiter, so the reshape can split them apart as it pivots.
Pass a two-element vector to names_to and set names_sep = "_" inside pivot_longer().
Click to reveal solution
Explanation: When column names encode two pieces of information separated by a delimiter, pass a character vector to names_to and a delimiter to names_sep. tidyr applies the split during the pivot so you skip a follow-up separate() step. For more complex headers, use names_pattern with a capture-group regex.
Exercise 1.5: Parse multi-variable column names with names_pattern
Task: A public-health analyst has aggregated TB counts coded in the WHO-style header new_sp_m65 (new cases, smear-positive, male, age 65 plus). Given a small inline tibble with four such columns, pivot longer and split the header into diagnosis, gender, and age columns using names_pattern. Save to ex_1_5.
Expected result:
#> # A tibble: 8 x 5
#> country diagnosis gender age cases
#> <chr> <chr> <chr> <chr> <int>
#> 1 IND sp m 65 12
#> 2 IND sp f 65 7
#> 3 IND ep m 65 3
#> 4 IND ep f 65 2
#> 5 PAK sp m 65 9
#> 6 PAK sp f 65 5
#> 7 PAK ep m 65 1
#> 8 PAK ep f 65 4Difficulty: Advanced
When a header is not a clean delimiter join, a regex with one capture group per output column can pull each piece apart.
Call pivot_longer() with a three-element names_to and a names_pattern regex containing three capture groups.
Click to reveal solution
Explanation: names_pattern accepts a regex with one capture group per output column. The three groups (\\w+), (m|f), and (\\d+) map onto diagnosis, gender, and age in order. Use this pattern when the header is not a clean delimiter join: a regex captures arbitrary structure. The age column comes back as character; cast with as.integer() if you need numeric.
Section 2. Pivot wider: long-to-wide reshaping (5 problems)
Exercise 2.1: Average ChickWeight by Diet across time
Task: A nutrition team needs a wide summary of ChickWeight where each column is a diet and each row is a time point holding the mean chick weight. Group by Time and Diet, average the weight, then pivot the four diets into columns. Save to ex_2_1.
Expected result:
#> # A tibble: 12 x 5
#> Time `1` `2` `3` `4`
#> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 0 41.4 40.7 40.8 41.0
#> 2 2 47.2 49.4 50.4 51.8
#> 3 4 56.5 59.8 62.2 64.5
#> 4 6 66.8 75.4 77.9 83.9
#> 5 8 79.7 91.7 98.4 105.6
#> 6 10 93.1 108.5 123.4 126.0
#> 7 12 108.5 131.3 150.6 151.4
#> 8 14 123.4 141.2 174.9 166.7
#> ...Difficulty: Beginner
Average the weights for each time-and-diet pair first, then lift the diet labels out into their own columns.
After group_by() and summarise(mean(...)), call pivot_wider() with names_from = Diet and values_from.
Click to reveal solution
Explanation: pivot_wider() is the inverse of pivot_longer(): it lifts the values in one column out into new columns. Aggregate first, then pivot, so each (Time, Diet) cell has a single value. Without summarising you would get a list-column or a values_fn warning because some combinations have multiple chicks.
Exercise 2.2: Build a quarterly sales matrix from a long ledger
Task: A finance team keeps a long-format sales tibble with region, quarter, and sales. Pivot it to a wide matrix where regions are rows and quarters Q1 to Q4 are columns, with each cell holding the recorded sale. Save to ex_2_2.
Expected result:
#> # A tibble: 3 x 5
#> region Q1 Q2 Q3 Q4
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 East 120 135 140 160
#> 2 North 100 110 130 125
#> 3 West 80 95 105 120Difficulty: Intermediate
Each quarter label should become a column heading, with the recorded sale dropped into the cell.
Call pivot_wider() with names_from = quarter and values_from = sales.
Click to reveal solution
Explanation: The wide layout is the right shape for a stakeholder report because each row is one region and each column is one period. Notice tidyr preserves the order of quarter values as it encountered them in the data. To force a specific column order, set quarter to a factor with the right levels before pivoting.
Exercise 2.3: Pivot ToothGrowth mean length by dose within supplement
Task: A pharmacology team is comparing dose-response curves for two supplements in ToothGrowth. Group by supp and dose, take the mean tooth length, then pivot doses into columns prefixed dose_ so each row is one supplement with three dose readings. Save to ex_2_3.
Expected result:
#> # A tibble: 2 x 4
#> supp dose_0.5 dose_1 dose_2
#> <fct> <dbl> <dbl> <dbl>
#> 1 OJ 13.2 22.7 26.1
#> 2 VC 7.98 16.8 26.1Difficulty: Intermediate
Summarise the mean length per supplement-and-dose pair, then turn the doses into prefixed columns.
Use pivot_wider() with names_from = dose, values_from, and names_prefix = "dose_".
Click to reveal solution
Explanation: names_prefix is useful when the source values are numeric like 0.5, 1, 2; without a prefix the new columns would be named with backtick-needing tokens. Two equivalent customisations: names_glue = "dose_{dose}" for templated output, or convert dose to a string before pivot for full control.
Exercise 2.4: Pivot two value columns simultaneously into wide format
Task: A trading desk wants daily price AND volume side by side for two tickers across three dates. Starting from a long tibble with one row per ticker-date, pivot both price and volume so each date becomes a pair of columns named like price_2026-04-01 and volume_2026-04-01. Save to ex_2_4.
Expected result:
#> # A tibble: 2 x 7
#> ticker `price_2026-04-01` `price_2026-04-02` `price_2026-04-03` `volume_2026-04-01` `volume_2026-04-02` `volume_2026-04-03`
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 AAPL 185 187 190 1.2 1.4 1.1
#> 2 MSFT 410 412 415 0.8 0.9 0.7Difficulty: Advanced
Both metrics should spread across the dates at once, producing one column group per metric.
Call pivot_wider() with names_from = date and a vector values_from = c(price, volume).
Click to reveal solution
Explanation: Passing a vector to values_from tells pivot_wider() to spread each value column into its own group of new columns. The default column names take the form {values_from}_{names_from}, which is configurable via names_glue. This is the right move when reports need parallel metrics per period.
Exercise 2.5: Control wide column names with names_glue
Task: A growth team wants a polished campaign report where the spread columns are named like Q1_2026_revenue rather than the default revenue_Q1_2026. Reshape the long quarterly tibble using names_glue to assemble that order, sorted ticker, and save to ex_2_5.
Expected result:
#> # A tibble: 2 x 5
#> product Q1_2026_revenue Q2_2026_revenue Q3_2026_revenue Q4_2026_revenue
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 Basic 200 250 300 280
#> 2 Pro 450 520 600 650Difficulty: Advanced
A glue-style template lets you dictate the exact order of the pieces in each new column name.
Use pivot_wider() with names_from = quarter, values_from = revenue, and names_glue = "{quarter}_revenue".
Click to reveal solution
Explanation: names_glue accepts a glue-style template that references variables from names_from and the literal .value placeholder when multiple value columns exist. It is the cleanest way to enforce a strict naming convention for downstream consumers like BI tools, where stable column names are part of the contract.
Section 3. Split, unite, and extract columns (5 problems)
Exercise 3.1: Split a "Last, First" name field into two columns
Task: An HR analyst received a roster with a single full_name column formatted as "Doe, Jane". Use separate_wider_delim() to split on the comma plus space and produce clean last_name and first_name columns. Save the resulting tibble to ex_3_1.
Expected result:
#> # A tibble: 3 x 2
#> last_name first_name
#> <chr> <chr>
#> 1 Doe Jane
#> 2 Smith Aaron
#> 3 Patel RiyaDifficulty: Beginner
The single name field holds two pieces joined by a comma, so a delimiter split gives two clean columns.
Call separate_wider_delim() with delim = ", " and names = c("last_name", "first_name").
Click to reveal solution
Explanation: separate_wider_delim() is the modern replacement for separate(). The "wider" suffix signals that the result grows columns rather than rows. Use the too_few and too_many arguments to control what happens when a row has an unexpected number of pieces. The retired separate() still works but emits a deprecation note.
Exercise 3.2: Combine year, month, day columns into an ISO date string
Task: A reporting analyst has a tibble with three integer columns yr, mo, dy and needs a single date column formatted as YYYY-MM-DD for export to a CSV. Use unite() with a hyphen separator, drop the source columns, and save to ex_3_2.
Expected result:
#> # A tibble: 4 x 1
#> date
#> <chr>
#> 1 2026-01-15
#> 2 2026-02-03
#> 3 2026-02-28
#> 4 2026-03-10Difficulty: Intermediate
Three columns need to be glued into one date string with a separator between the parts.
Use unite() with col = "date", the three source columns, sep = "-", and remove = TRUE.
Click to reveal solution
Explanation: unite() is the inverse of separate(): it concatenates columns with a separator. Setting remove = TRUE discards the source columns once united. If month and day are stored as integers, format them with sprintf("%02d", ...) first to keep the leading zero. For real date conversion, pipe through as.Date().
Exercise 3.3: Parse a SKU code with separate_wider_regex
Task: A retail analyst inherits a sales table with an opaque sku column like "AB-1234-XL" coding a brand, a numeric ID, and a size. Use separate_wider_regex() to extract the three components into brand, sku_id, and size columns and save to ex_3_3.
Expected result:
#> # A tibble: 3 x 3
#> brand sku_id size
#> <chr> <chr> <chr>
#> 1 AB 1234 XL
#> 2 CD 9876 M
#> 3 EF 0042 SDifficulty: Intermediate
The code has a fixed brand-id-size structure, so a pattern with one named piece per component extracts it cleanly.
Call separate_wider_regex() with a patterns vector naming brand, sku_id, and size, with unnamed strings for the hyphens.
Click to reveal solution
Explanation: separate_wider_regex() is more reliable than a hand-rolled regex with capture groups when the field has fixed structure. Each named element becomes a new column; unnamed strings are matched and discarded (the hyphens here). The function fails loudly on rows that do not match, which catches bad inputs early in a pipeline.
Exercise 3.4: Expand a semicolon-separated tags column with separate_longer_delim
Task: A product team stores article tags in a single string column like "r;tidyverse;data". Use separate_longer_delim() on a sample tibble of three articles so each tag becomes its own row, preserving the article ID. Save to ex_3_4.
Expected result:
#> # A tibble: 8 x 2
#> article_id tag
#> <int> <chr>
#> 1 1 r
#> 2 1 tidyverse
#> 3 1 data
#> 4 2 r
#> 5 2 ggplot2
#> 6 3 sql
#> 7 3 etl
#> 8 3 cloudDifficulty: Intermediate
Each delimited token in the tags string should expand into its own row, with the article id recycled.
Use separate_longer_delim() on the tag column with delim = ";".
Click to reveal solution
Explanation: separate_longer_delim() is the "growing rows" counterpart to separate_wider_delim(). Each delimited token becomes its own row; non-delimited columns are recycled. This is the right tool for normalising many-to-one fields like tags, authors, or roles before joining to a lookup table.
Exercise 3.5: Parse server log lines with named regex groups
Task: An ops engineer needs to break a raw log column into timestamp, level, and message columns. Each line has the shape "2026-05-12T08:30:00 [INFO] Service started". Use separate_wider_regex() with three named patterns and save to ex_3_5.
Expected result:
#> # A tibble: 3 x 3
#> timestamp level message
#> <chr> <chr> <chr>
#> 1 2026-05-12T08:30:00 INFO Service started
#> 2 2026-05-12T08:32:14 WARN Slow query
#> 3 2026-05-12T08:35:01 ERROR Connection refusedDifficulty: Advanced
Break each line into timestamp, level, and message by matching the fixed shape and dropping the brackets.
Call separate_wider_regex() with a patterns vector naming timestamp, level, and message, plus unnamed literals for the brackets and spaces.
Click to reveal solution
Explanation: The named patterns become columns; the literal strings between them ("[", "]") are matched and dropped. Anchoring with \\S+ for the timestamp and .* for the trailing message lets each row vary in length while still parsing reliably. For very large log files, prefer readr::read_log() which is optimised for the common Apache or syslog formats.
Section 4. Handle missing data: fill, complete, drop_na, replace_na (5 problems)
Exercise 4.1: Drop rows of airquality with any NA in the measurement columns
Task: A climatologist wants a clean subset of airquality where every numeric measurement is observed. Use drop_na() across the Ozone, Solar.R, Wind, and Temp columns and save the result to ex_4_1. Print the resulting row count and head.
Expected result:
#> [1] 111
#> # A tibble: 6 x 6
#> Ozone Solar.R Wind Temp Month Day
#> <int> <int> <dbl> <int> <int> <int>
#> 1 41 190 7.4 67 5 1
#> 2 36 118 8 72 5 2
#> 3 12 149 12.6 74 5 3
#> 4 18 313 11.5 62 5 4
#> 5 23 299 8.6 65 5 7
#> 6 19 99 13.8 59 5 8Difficulty: Beginner
Keep only the rows where every named measurement is actually observed.
Use drop_na() and list the columns Ozone, Solar.R, Wind, and Temp.
Click to reveal solution
Explanation: drop_na() removes rows that have a missing value in any of the named columns; calling it with no arguments drops rows with NA anywhere. Be conservative: a careless drop_na() can wipe out entire studies. Always inspect nrow() before and after, and consider whether imputation is more honest than deletion for the analysis at hand.
Exercise 4.2: Replace NA with column-specific defaults
Task: A junior analyst has a tibble of customer responses with missing ratings and comments. Use replace_na() to fill rating with 0 and comment with "no comment" in one call, then save to ex_4_2.
Expected result:
#> # A tibble: 4 x 3
#> customer_id rating comment
#> <int> <dbl> <chr>
#> 1 1 4 great service
#> 2 2 0 no comment
#> 3 3 5 fast
#> 4 4 0 no commentDifficulty: Intermediate
Each missing cell should get a default that matches its column's type, all in one call.
Call replace_na() with a named list mapping rating to 0 and comment to "no comment".
Click to reveal solution
Explanation: replace_na() accepts a named list mapping columns to their replacement values. The replacement must be the same type as the column: numeric for numerics, character for character. For type-aware imputation across many columns, look at mutate(across(where(is.numeric), \\(x) tidyr::replace_na(x, 0))) to keep the pipeline declarative.
Exercise 4.3: Forward-fill grouping headers with fill()
Task: A reporting analyst has a sales tibble where the region column is only populated on the first row of each block, like a printed report. Use fill() with .direction = "down" to propagate the region down the empty rows and save to ex_4_3.
Expected result:
#> # A tibble: 6 x 3
#> region store_id sales
#> <chr> <int> <dbl>
#> 1 East 1 1200
#> 2 East 2 1400
#> 3 East 3 1100
#> 4 West 4 1700
#> 5 West 5 1500
#> 6 West 6 1600Difficulty: Intermediate
The blank rows below each header should inherit the last value that was actually present.
Use fill() on region with .direction = "down".
Click to reveal solution
Explanation: fill() copies the last non-NA value into subsequent NA cells. .direction = "down" (the default) is what you want for header-style fills; "up", "downup", and "updown" handle other patterns. It is most often used after reading Excel sheets with merged cells, where a header value only appears in the first row of each group.
Exercise 4.4: Expand all combinations with complete() and fill missing counts with zero
Task: A retail analyst has a sales tibble with only the (region, product) pairs that actually transacted. They need a balanced panel showing every region-product combination, with zeros where there were no sales. Use complete() with fill = list(units = 0) and save to ex_4_4.
Expected result:
#> # A tibble: 6 x 3
#> region product units
#> <chr> <chr> <dbl>
#> 1 East A 30
#> 2 East B 0
#> 3 East C 12
#> 4 West A 0
#> 5 West B 25
#> 6 West C 18Difficulty: Intermediate
Build the full cross of every region and product, then put zero where no sale happened.
Call complete() with region, product, and fill = list(units = 0).
Click to reveal solution
Explanation: complete() cross-joins the unique values of the named columns and fills any newly created rows with the values in fill. This is the right tool when downstream code expects a balanced panel: a missing zero is silent, a present zero is countable. For larger grids, expand() returns just the skeleton and you can left-join your data onto it.
Exercise 4.5: Build an irregular-frequency monthly panel from gappy sensor data
Task: A field engineer logs sensor readings on irregular months and you need a continuous panel with no gaps. Use complete() to insert any missing months between the observed minimum and maximum, then fill() forward the sensor_id and replace_na() the reading with -999 as a sentinel. Save to ex_4_5.
Expected result:
#> # A tibble: 6 x 3
#> month sensor_id reading
#> <dbl> <chr> <dbl>
#> 1 1 S1 5.2
#> 2 2 S1 -999
#> 3 3 S1 6.1
#> 4 4 S1 -999
#> 5 5 S1 -999
#> 6 6 S1 4.8Difficulty: Advanced
Insert every missing month between the first and last, carry the sensor id down, then mark unseen readings with a sentinel.
Chain complete(month = full_seq(month, 1)), fill(sensor_id, .direction = "down"), and replace_na(list(reading = -999)).
Click to reveal solution
Explanation: full_seq(month, 1) builds a regular monthly skeleton from min to max. complete() injects rows for the missing months with NA values; fill() then propagates the sensor ID and replace_na() substitutes the sentinel for the missing reading. A real pipeline would more likely impute with zoo::na.approx() for a numeric series, but the pattern is the same.
Section 5. List columns: nest, unnest, hoist (5 problems)
Exercise 5.1: Nest iris rows by Species into a list-column
Task: A statistics tutor wants iris collapsed into three rows, one per Species, with a data list-column holding the matching subset of measurements. Use nest() with the .by argument so the result is one row per species and save to ex_5_1.
Expected result:
#> # A tibble: 3 x 2
#> Species data
#> <fct> <list>
#> 1 setosa <tibble [50 x 4]>
#> 2 versicolor <tibble [50 x 4]>
#> 3 virginica <tibble [50 x 4]>Difficulty: Beginner
Collapse the table to one row per species, tucking the matching measurements into a packed column.
Use nest() with data = -Species.
Click to reveal solution
Explanation: nest(data = -Species) says "pack every column except Species into a list-column called data." The result is one row per species with the per-species table tucked inside, which is the foundation of the many-models pattern. You can also write iris |> group_by(Species) |> nest() for the same outcome with a grouping prefix.
Exercise 5.2: Fit one regression per Species and pull out the coefficients
Task: A regression tutor wants to compare slopes of Petal.Length on Sepal.Length across the three iris species. Nest by Species, fit lm() inside each nested tibble using purrr::map(), tidy the model with broom::tidy(), then unnest the tidied output. Save the coefficient table to ex_5_2.
Expected result:
#> # A tibble: 6 x 6
#> Species term estimate std.error statistic p.value
#> <fct> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 setosa (Intercept) 0.803 0.344 2.34 0.0238
#> 2 setosa Sepal.Length 0.132 0.0685 1.92 0.0607
#> 3 versicolor (Intercept) 0.185 0.514 0.36 0.720
#> 4 versicolor Sepal.Length 0.686 0.0863 7.95 0.0000000272
#> 5 virginica (Intercept) 0.610 0.417 1.46 0.149
#> 6 virginica Sepal.Length 0.750 0.0628 11.9 0.0000000000114Difficulty: Intermediate
Pack each species into its own table, fit a model inside every row, tidy it, then flatten the coefficients back out.
After nest(), use mutate() with map() to call lm(), map() again for broom::tidy(), then unnest() the tidied column.
Click to reveal solution
Explanation: The many-models pattern (nest, then map() model fits, then tidy(), then unnest()) is one of the most powerful idioms in the tidyverse. Each model lives in its own row and remains addressable by its grouping keys. Compare slopes by filtering term == "Sepal.Length" and inspecting the estimate column.
Exercise 5.3: Explode a list-column of skills with unnest_longer
Task: A recruiter has a tibble where each candidate row carries a skills list-column of variable length. Use unnest_longer() to expand the table so each skill becomes its own row, preserving the candidate ID. Save to ex_5_3.
Expected result:
#> # A tibble: 6 x 2
#> candidate skill
#> <int> <chr>
#> 1 1 R
#> 2 1 SQL
#> 3 1 Python
#> 4 2 R
#> 5 3 Python
#> 6 3 JavaDifficulty: Intermediate
Each element of the variable-length list should be stacked into its own row.
Call unnest_longer() on the skills column with values_to = "skill".
Click to reveal solution
Explanation: unnest_longer() is the right verb when each element of the list-column is a vector you want stacked into rows. If the elements were named, you would also get an index column with the names. Compare with unnest_wider(), which spreads named elements into columns instead. Both are descendants of the deprecated unnest().
Exercise 5.4: Spread a list-column of named lists into typed columns with unnest_wider
Task: An API client returned a tibble with a details list-column where each element is a named list of weight, colour, and in_stock. Use unnest_wider() to turn each named element into its own typed column and save to ex_5_4.
Expected result:
#> # A tibble: 3 x 4
#> item_id weight colour in_stock
#> <int> <dbl> <chr> <lgl>
#> 1 1 1.2 red TRUE
#> 2 2 0.9 blue FALSE
#> 3 3 2.1 green TRUEDifficulty: Intermediate
Each named entry inside the list element should become its own typed column, keeping the row count unchanged.
Use unnest_wider() on the details column.
Click to reveal solution
Explanation: unnest_wider() keeps the row count constant and grows columns. It is the right verb for parsing JSON-style responses where each row has the same schema. When elements are missing or unnamed, use hoist() for fine-grained control over which paths to extract and what to name them.
Exercise 5.5: Per-group regressions on mtcars with predictions joined back
Task: An ML engineer wants per-cylinder linear models predicting mpg from wt. Nest mtcars by cyl, fit one lm() per group, augment each model with predictions on its own data, then unnest a tibble containing cyl, wt, mpg, and the fitted value .fitted. Save to ex_5_5.
Expected result:
#> # A tibble: 32 x 4
#> cyl wt mpg .fitted
#> <dbl> <dbl> <dbl> <dbl>
#> 1 6 2.62 21 20.7
#> 2 6 2.88 21 19.4
#> 3 4 2.32 22.8 25.6
#> 4 6 3.21 21.4 17.6
#> 5 8 3.44 18.7 17.3
#> 6 6 3.46 18.1 16.4
#> 7 8 3.57 14.3 16.6
#> ...Difficulty: Advanced
Fit one model per cylinder group, attach each model's predictions to its own data, then flatten the result.
After nest(data = -cyl), map() an lm(), map2() broom::augment() over model and data, then unnest() and select() .fitted.
Click to reveal solution
Explanation: map2() walks both the model and its original data in lockstep, calling broom::augment() which appends .fitted, .resid, and friends. This is how you generate predictions while keeping group identity attached. The pattern scales naturally to time series partitions, A/B experiments, or any per-group modelling problem.
Section 6. End-to-end tidying workflows (5 problems)
Exercise 6.1: Turn a wide survey response sheet into a tidy long answer log
Task: A market researcher exported a survey as a wide tibble where each column is a question (q1, q2, q3) and rows are respondents. Pivot longer so each row holds one respondent-question-answer triple, drop rows where the answer is NA, and save to ex_6_1.
Expected result:
#> # A tibble: 7 x 3
#> respondent question answer
#> <int> <chr> <chr>
#> 1 1 q1 yes
#> 2 1 q2 no
#> 3 1 q3 yes
#> 4 2 q1 no
#> 5 2 q3 no
#> 6 3 q1 yes
#> 7 3 q2 yesDifficulty: Intermediate
Stack every question column into one answer column, then discard the rows where no answer was given.
Use pivot_longer() with cols = starts_with("q"), then drop_na() on answer.
Click to reveal solution
Explanation: This two-step combo is the workhorse for survey ETL: pivot_longer() gets the data into one-observation-per-row shape, then drop_na() removes the skipped questions. The result feeds straight into count(question, answer) for a response distribution by question.
Exercise 6.2: Build a region by quarter report with row and column totals
Task: A finance team wants a wide quarterly sales report by region with a total column summing across quarters. Starting from the long sales ledger from earlier, pivot wider, then mutate() a row total using rowSums(). Save to ex_6_2.
Expected result:
#> # A tibble: 3 x 6
#> region Q1 Q2 Q3 Q4 total
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 East 120 135 140 160 555
#> 2 North 100 110 130 125 465
#> 3 West 80 95 105 120 400Difficulty: Intermediate
Reshape the ledger into one row per region, then add a column that sums across the quarter columns.
After pivot_wider(), mutate() a total using rowSums(across(starts_with("Q"))).
Click to reveal solution
Explanation: rowSums(across(starts_with("Q"))) is the modern tidyverse way to sum a tidy-selected set of columns into a new column. The across() helper inside dplyr is the right tool here because the column names are known by pattern after the pivot. This idiom replaces older mutate(total = Q1 + Q2 + Q3 + Q4), which is brittle if you add a Q5 next year.
Exercise 6.3: Reconstruct a full daily panel from gappy weather readings
Task: A climatologist has temperature readings for two stations on irregular days in May and wants a complete daily panel for the month. For each station, complete() the date range from 2026-05-01 to 2026-05-07, fill station down, and impute missing temp_c with the station's mean using coalesce() and mean() inside group_by(). Save to ex_6_3.
Expected result:
#> # A tibble: 14 x 3
#> station date temp_c
#> <chr> <date> <dbl>
#> 1 A 2026-05-01 18
#> 2 A 2026-05-02 19
#> 3 A 2026-05-03 19.5
#> 4 A 2026-05-04 19.5
#> 5 A 2026-05-05 21
#> 6 A 2026-05-06 19.5
#> 7 A 2026-05-07 20
#> 8 B 2026-05-01 22
#> 9 B 2026-05-02 23.5
#> 10 B 2026-05-03 23.5
#> 11 B 2026-05-04 25
#> ...Difficulty: Advanced
Fill in every day of the range for each station, then replace gaps with that station's average reading.
Use complete(station, date = seq(...)), then group_by(station) and mutate(temp_c = coalesce(temp_c, mean(temp_c, na.rm = TRUE))).
Click to reveal solution
Explanation: Passing both grouping columns to complete() cross-joins the unique stations against the explicit date sequence. The mean-imputation step uses coalesce() which returns the first non-NA value, so observed readings remain untouched. For modelling work a more principled imputation such as mice::mice() or recipes::step_impute_* is usually preferable.
Exercise 6.4: Reshape a wide sensor log for a faceted ggplot
Task: A platform engineer has wide sensor data with cpu, mem, and disk columns sampled across timestamps and needs it in long format for a faceted ggplot. Pivot the three metric columns longer, then arrange by timestamp and metric so plotting code can use facet_wrap(~metric). Save to ex_6_4.
Expected result:
#> # A tibble: 12 x 3
#> timestamp metric value
#> <chr> <chr> <dbl>
#> 1 t1 cpu 12.5
#> 2 t1 disk 50
#> 3 t1 mem 30.1
#> 4 t2 cpu 14.7
#> 5 t2 disk 52
#> 6 t2 mem 31
#> 7 t3 cpu 20.2
#> 8 t3 disk 53
#> 9 t3 mem 33.8
#> 10 t4 cpu 18
#> ...Difficulty: Advanced
Stack the three metric columns into long form, then sort so the facet variable follows a predictable order.
Use pivot_longer() on c(cpu, mem, disk), then arrange() by timestamp and metric.
Click to reveal solution
Explanation: Plotting libraries like ggplot2 expect long data because they map a single variable to position and another to colour or facet. Pivoting before plotting is so common that some teams build a small helper that pipes long form straight into the plot. Sort by the facet variable so legends and small multiples follow a predictable order.
Exercise 6.5: Aggregate a raw event log into a stakeholder rollup
Task: A growth analyst has a raw event log where each row is one user action with user_id, event_type, and value. The product manager asks for a wide rollup by event type showing total value and event count, with zeros for absent events. Pivot wider with two value functions, then replace_na() zero into the missing cells. Save to ex_6_5.
Expected result:
#> # A tibble: 3 x 7
#> user_id total_click count_click total_view count_view total_purchase count_purchase
#> <int> <dbl> <int> <dbl> <int> <dbl> <int>
#> 1 1 5 2 0 0 20 1
#> 2 2 0 0 4 2 0 0
#> 3 3 3 1 2 1 50 1Difficulty: Advanced
Summarise total value and event count per user-and-type, then spread both metrics into columns with zero where an event is absent.
After group_by() and summarise(total, count), call pivot_wider() with values_from = c(total, count), names_glue, and values_fill = 0.
Click to reveal solution
Explanation: The values_from = c(total, count) shape creates two parallel column families per event type. names_glue enforces the metric-first column-name convention stakeholders expect. values_fill = 0 saves a follow-up replace_na() call. This single pipeline is what a reporting analyst would otherwise hand-build with a half-dozen joins.
What to do next
- Build on these patterns with the focused tidyr Pivot Exercises for deeper pivot_longer and pivot_wider drills.
- Move to list-column workflows in tidyr Nest and Unnest Exercises.
- Combine tidying with row-wise transformations in dplyr Exercises in R.
- Practice the cleaning side of the pipeline in Data Cleaning Exercises in R.
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