dplyr filter() and select() Exercises: 20 Practice Problems

Solve 20 short, focused R problems that drill the most important filter() and select() patterns. Every exercise lists the task, the expected output, and a difficulty marker. Click any solution panel to reveal the worked answer and a short explanation when you are ready to compare notes.

By Selva Prabhakaran · Published May 13, 2026 · Last updated May 13, 2026

RRun this once before any exercise

library(dplyr) library(ggplot2)

Section 1. Single-condition row filtering (3 problems)

Exercise 1.1: Keep the fuel-efficient cars from mtcars

Task: Use filter() to keep only the rows of the built-in mtcars dataset where mpg is greater than 25 miles per gallon. The result is a shortlist of the most fuel-efficient cars in the dataset. Save the result to ex_1_1.

Expected result:

#>                 mpg cyl  disp  hp drat    wt  qsec vs am gear carb
#> Fiat 128       32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1
#> Honda Civic    30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2
#> Toyota Corolla 33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1
#> Fiat X1-9      27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
#> Porsche 914-2  26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2
#> Lotus Europa   30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2

Difficulty: Beginner

RYour turn

ex_1_1 <- # your code here ex_1_1

Click to reveal solution

RSolution

ex_1_1 <- mtcars |> filter(mpg > 25) ex_1_1

Explanation: filter() keeps rows where the supplied condition evaluates to TRUE. Six cars clear the 25-mpg bar and every one of them is a four-cylinder economy compact. A common slip for newcomers is writing = instead of > when typing fast: only the comparison operators (>, <, >=, <=, ==, !=) belong inside filter(). Assignment with = or <- would silently set a column and then drop nothing.

Exercise 1.2: Pull a budget tier from the diamonds catalog

Task: A jeweller preparing a seasonal sale wants the budget tier of the diamonds catalog, defined as stones priced strictly below $1,000. Filter the diamonds dataset to that subset and save the result to ex_1_2 so it can be exported to the discount flyer.

Expected result:

#> # A tibble: 14,524 x 10
#>    carat cut       color clarity depth table price     x     y     z
#>    <dbl> <ord>     <ord> <ord>   <dbl> <dbl> <int> <dbl> <dbl> <dbl>
#> 1   0.23 Ideal     E     SI2      61.5    55   326  3.95  3.98  2.43
#> 2   0.21 Premium   E     SI1      59.8    61   326  3.89  3.84  2.31
#> 3   0.23 Good      E     VS1      56.9    65   327  4.05  4.07  2.31
#> 4   0.29 Premium   I     VS2      62.4    58   334  4.20  4.23  2.63
#> 5   0.31 Good      J     SI2      63.3    58   335  4.34  4.35  2.75
#> # 14,519 more rows hidden

Difficulty: Beginner

RYour turn

ex_1_2 <- # your code here ex_1_2

Click to reveal solution

RSolution

ex_1_2 <- diamonds |> filter(price < 1000) ex_1_2

Explanation: Strictly less than (<) is correct because the brief says "strictly below $1,000". Switch to <= if the boundary should be inclusive. About 27% of the catalog falls in this tier, which often surprises people because the dataset's price range stretches up to $18,823. The long right tail of the price distribution hides how many entry-level stones the catalog actually contains.

Exercise 1.3: Keep only compact cars from the mpg dataset

Task: Use filter() to keep only the mpg rows where the class column is exactly "compact". The mpg dataset ships with the ggplot2 package and catalogues 234 car models with their EPA fuel economy ratings. Save the compact-only subset to ex_1_3.

Expected result:

#> # A tibble: 47 x 11
#>    manufacturer model    displ  year   cyl trans drv     cty   hwy fl    class
#>    <chr>        <chr>    <dbl> <int> <int> <chr> <chr> <int> <int> <chr> <chr>
#> 1  audi         a4         1.8  1999     4 auto~ f        18    29 p     compact
#> 2  audi         a4         1.8  1999     4 manu~ f        21    29 p     compact
#> 3  audi         a4         2.0  2008     4 manu~ f        20    31 p     compact
#> # 44 more rows hidden

Difficulty: Beginner

RYour turn

ex_1_3 <- # your code here ex_1_3

Click to reveal solution

RSolution

ex_1_3 <- mpg |> filter(class == "compact") ex_1_3

Explanation: class == "compact" is element-wise equality on a character column, so each row's class value is compared to the literal string. Quotes matter here: writing class == compact without quotes would search for a variable named compact and throw object 'compact' not found. There are 47 compact cars across seven manufacturers, making it one of the most common classes in the dataset.

Section 2. Combining multiple conditions (4 problems)

Exercise 2.1: Pick setosa flowers with wide petals

Task: Filter the iris dataset to rows where Species equals "setosa" AND Petal.Width is at least 0.4 centimetres. Combining two conditions inside filter() with a comma (implicit AND) is the single most common dplyr filter pattern in real analysis code. Save the result to ex_2_1.

Expected result:

#>   Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 1          5.4         3.9          1.7         0.4  setosa
#> 2          5.7         4.4          1.5         0.4  setosa
#> 3          5.1         3.5          1.4         0.3  setosa
#> ...
#> nrow(ex_2_1):
#> [1] 9

Difficulty: Intermediate

RYour turn

ex_2_1 <- # your code here ex_2_1

Click to reveal solution

RSolution

ex_2_1 <- iris |> filter(Species == "setosa", Petal.Width >= 0.4) ex_2_1

Explanation: Inside filter(), commas separate conditions and combine them with logical AND. You could also write Species == "setosa" & Petal.Width >= 0.4 and get an identical result. Use | for OR. A common pitfall: && and || (double form) short-circuit and only return one value, which silently breaks vectorised row-wise filtering. Stick to single & and | inside filter() calls.

Exercise 2.2: Climatologist flags hot and broken-sensor days

Task: A climatologist studying poor air quality wants the airquality rows where Ozone is greater than 100 OR Solar.R is missing entirely. The first condition flags pollution hotspots and the second flags sensor outages. Combine the two with the OR operator | and save the result to ex_2_2.

Expected result:

#>    Ozone Solar.R Wind Temp Month Day
#> 1     NA      NA 14.3   56     5   5
#> 2    115     223  5.7   79     5  30
#> 3    135     269  4.1   84     7   1
#> ...
#> nrow(ex_2_2):
#> [1] 14

Difficulty: Intermediate

RYour turn

ex_2_2 <- # your code here ex_2_2

Click to reveal solution

RSolution

ex_2_2 <- airquality |> filter(Ozone > 100 | is.na(Solar.R)) ex_2_2

Explanation: Combining conditions with | (OR) keeps rows that satisfy either side. Note that the Ozone > 100 clause alone would silently drop rows where Ozone is NA, because any comparison against NA returns NA, which filter() treats as FALSE. To explicitly include or exclude NAs, wrap with is.na() and chain it with | or &. There are seven rows where Solar.R is missing.

Exercise 2.3: Use %in% for four or six-cylinder cars

Task: Use filter() together with the %in% operator to keep only mtcars rows whose cyl column is either 4 or 6. The %in% operator reads cleaner than chaining == with | once you have three or more allowed values. Save the result to ex_2_3.

Expected result:

#>                 mpg cyl  disp  hp drat    wt  qsec vs am gear carb
#> Mazda RX4      21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
#> Mazda RX4 Wag  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
#> Datsun 710     22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1
#> Hornet 4 Drive 21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
#> ...
#> nrow(ex_2_3):
#> [1] 18

Difficulty: Intermediate

RYour turn

ex_2_3 <- # your code here ex_2_3

Click to reveal solution

RSolution

ex_2_3 <- mtcars |> filter(cyl %in% c(4, 6)) ex_2_3

Explanation: %in% returns TRUE for each element of the left side that appears anywhere in the right-hand vector. Compared to cyl == 4 | cyl == 6, it scales better: just add another value to the vector for an extra allowed level. There are 11 four-cylinder cars and 7 six-cylinder cars in mtcars, totalling 18. The complement !(cyl %in% c(4, 6)) catches the V8s.

Exercise 2.4: Filter mid-carat diamonds with between()

Task: A jeweller's pricing model focuses on stones between 0.5 and 1.0 carats inclusive, the sweet spot for engagement rings. Use between() inside filter() to keep only the diamonds rows whose carat falls in that inclusive range, and save the result to ex_2_4 for the pricing model input.

Expected result:

#> # A tibble: 24,624 x 10
#>    carat cut       color clarity depth table price     x     y     z
#>    <dbl> <ord>     <ord> <ord>   <dbl> <dbl> <int> <dbl> <dbl> <dbl>
#> 1   0.5  Premium   F     SI2      60.7    58  1414  5.13  5.16  3.12
#> 2   0.51 Premium   F     VS2      59.4    62  1746  5.27  5.30  3.13
#> 3   0.52 Ideal     F     VS1      61.5    56  1624  5.18  5.21  3.19
#> ...
#> # 24,619 more rows hidden

Difficulty: Intermediate

RYour turn

ex_2_4 <- # your code here ex_2_4

Click to reveal solution

RSolution

ex_2_4 <- diamonds |> filter(between(carat, 0.5, 1.0)) ex_2_4

Explanation: between(x, left, right) is shorthand for x >= left & x <= right. Both bounds are inclusive, matching the jeweller's brief. It is purely a readability convenience: the underlying comparison runs at the same speed as the long form. If you need one bound exclusive, fall back to the explicit carat >= 0.5 & carat < 1.0 form. About 46% of the catalog fits the engagement-ring range.

Section 3. Missing values and slicing helpers (2 problems)

Exercise 3.1: Drop rows where Ozone is missing

Task: An analyst preparing the airquality dataset for a regression model needs to drop every row where Ozone is missing, because the model response variable cannot be NA. Use filter() together with !is.na() to keep only the rows that have a recorded ozone value, and save the result to ex_3_1.

Expected result:

#>    Ozone Solar.R Wind Temp Month Day
#> 1     41     190  7.4   67     5   1
#> 2     36     118  8.0   72     5   2
#> 3     12     149 12.6   74     5   3
#> ...
#> nrow(ex_3_1):
#> [1] 116

Difficulty: Intermediate

RYour turn

ex_3_1 <- # your code here ex_3_1

Click to reveal solution

RSolution

ex_3_1 <- airquality |> filter(!is.na(Ozone)) ex_3_1

Explanation: is.na() returns TRUE for missing values, so !is.na() inverts the test and keeps only the present ones. airquality has 37 missing ozone readings out of 153, leaving 116 complete rows. A frequent trap is writing Ozone != NA, which returns NA for every row because you cannot compare a value to NA with == or !=, and filter() then silently drops every row.

Exercise 3.2: Top five highest horsepower cars with slice_max

Task: Use slice_max() to extract the five mtcars rows with the highest hp. slice_max() is the modern dplyr replacement for sorting by descending value and taking head(n). Save the result to ex_3_2 and confirm five rows are returned for the top-power leaderboard.

Expected result:

#>                      mpg cyl  disp  hp drat    wt  qsec vs am gear carb
#> Maserati Bora       15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8
#> Ford Pantera L      15.8   8 351.0 264 3.22 3.170 14.50  0  1    5    4
#> Duster 360          14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4
#> Camaro Z28          13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4
#> Chrysler Imperial   14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4

Difficulty: Intermediate

RYour turn

ex_3_2 <- # your code here ex_3_2

Click to reveal solution

RSolution

ex_3_2 <- mtcars |> slice_max(hp, n = 5) ex_3_2

Explanation: slice_max(hp, n = 5) orders rows in descending hp and returns the top five. Use slice_min() for the opposite end. Ties at the boundary are kept by default, so you may receive more than n rows when multiple rows share the cutoff value: pass with_ties = FALSE to cap at exactly five. The helper is more readable and faster than the older arrange(desc(hp)) |> head(5) idiom for this job.

Section 4. Selecting and renaming columns (3 problems)

Exercise 4.1: Keep three core columns from mtcars

Task: Use select() to keep only the mpg, cyl, and hp columns from mtcars and save the result to ex_4_1. This is the most basic select() pattern: name the columns you want, in the order you want them, and dplyr drops everything you do not name from the result.

Expected result:

#>                    mpg cyl  hp
#> Mazda RX4         21.0   6 110
#> Mazda RX4 Wag     21.0   6 110
#> Datsun 710        22.8   4  93
#> Hornet 4 Drive    21.4   6 110
#> Hornet Sportabout 18.7   8 175
#> Valiant           18.1   6 105
#> ...
#> # 26 more rows hidden

Difficulty: Beginner

RYour turn

ex_4_1 <- # your code here ex_4_1

Click to reveal solution

RSolution

ex_4_1 <- mtcars |> select(mpg, cyl, hp) head(ex_4_1)

Explanation: select() returns a data frame with only the named columns, in the order you list them. The output order matches the call order: select(hp, cyl, mpg) would yield the same three columns but rearranged. Unlike filter(), which operates on rows, select() works on columns, so its arguments are unquoted column names rather than row-level conditions. This is the first verb most analysts reach for after loading data.

Exercise 4.2: Drop one column with negative selection

Task: From the mpg dataset, drop only the fl column (fuel type) while keeping all other columns. Negative selection with a minus sign is the cleanest way to remove one or two columns without listing every survivor by hand. Save the trimmed tibble to ex_4_2.

Expected result:

#> # A tibble: 234 x 10
#>   manufacturer model  displ  year   cyl trans drv     cty   hwy class
#>   <chr>        <chr>  <dbl> <int> <int> <chr> <chr> <int> <int> <chr>
#> 1 audi         a4       1.8  1999     4 auto~ f        18    29 compact
#> 2 audi         a4       1.8  1999     4 manu~ f        21    29 compact
#> 3 audi         a4       2.0  2008     4 manu~ f        20    31 compact
#> ...
#> # 231 more rows hidden

Difficulty: Intermediate

RYour turn

ex_4_2 <- # your code here ex_4_2

Click to reveal solution

RSolution

ex_4_2 <- mpg |> select(-fl) ex_4_2

Explanation: Putting a minus sign before a column name removes it from the result; every other column passes through. To drop multiple columns at once, wrap them in c(): select(-c(fl, trans)). Negative selection composes with helpers too, so select(-starts_with("c")) removes every column whose name starts with the letter c. A common confusion is mixing positive and negative selections in one call: keep the call all-positive or all-negative for clarity.

Exercise 4.3: Rename columns inline while selecting

Task: A reporting analyst preparing a stakeholder summary wants only three columns from mtcars, renamed so they are self-explanatory: mpg becomes fuel_economy, hp becomes horsepower, and wt becomes weight_tons. Do both the selection and the rename in a single select() call and save the result to ex_4_3.

Expected result:

#>                    fuel_economy horsepower weight_tons
#> Mazda RX4                  21.0        110       2.620
#> Mazda RX4 Wag              21.0        110       2.875
#> Datsun 710                 22.8         93       2.320
#> Hornet 4 Drive             21.4        110       3.215
#> ...
#> # 28 more rows hidden

Difficulty: Intermediate

RYour turn

ex_4_3 <- # your code here ex_4_3

Click to reveal solution

RSolution

ex_4_3 <- mtcars |> select(fuel_economy = mpg, horsepower = hp, weight_tons = wt) head(ex_4_3)

Explanation: Inside select(), the syntax new_name = old_name renames a column while selecting it. The new name goes on the left and the original on the right, which is the opposite of base R's names()<- idiom. If you only want to rename without dropping any other columns, reach for rename() instead. Doing both in one select() call keeps the pipeline shorter for ad-hoc reports that go straight into a slide deck.

Section 5. Selection helpers (4 problems)

Exercise 5.1: Pull Sepal columns with starts_with

Task: Use the starts_with() selection helper inside select() to keep only the columns of iris whose name begins with "Sepal". This pattern is essential when your tables follow a naming convention like prefix_*, which is common in survey data and feature engineering output. Save the result to ex_5_1.

Expected result:

#>   Sepal.Length Sepal.Width
#> 1          5.1         3.5
#> 2          4.9         3.0
#> 3          4.7         3.2
#> 4          4.6         3.1
#> 5          5.0         3.6
#> ...
#> # 145 more rows hidden

Difficulty: Intermediate

RYour turn

ex_5_1 <- # your code here ex_5_1

Click to reveal solution

RSolution

ex_5_1 <- iris |> select(starts_with("Sepal")) head(ex_5_1)

Explanation: starts_with() matches column names that begin with the supplied string and returns them in original order. Companion helpers include ends_with(), contains(), and matches() (full regex). The match is case-insensitive by default; pass ignore.case = FALSE if you need exact case. For wide tables (think survey data with q1_a, q1_b, ...), helpers like these turn a 50-column select into a one-liner.

Exercise 5.2: Pull dimension columns with any_of

Task: A jewellery analyst auditing physical measurements needs only the three dimension columns of the diamonds catalog: x, y, and z. Use select() with the any_of() helper and a name vector to grab them in one call, then save the result to ex_5_2. Using any_of() makes the call robust when the column list is built dynamically.

Expected result:

#> # A tibble: 53,940 x 3
#>       x     y     z
#>   <dbl> <dbl> <dbl>
#> 1  3.95  3.98  2.43
#> 2  3.89  3.84  2.31
#> 3  4.05  4.07  2.31
#> 4  4.20  4.23  2.63
#> ...
#> # 53,936 more rows hidden

Difficulty: Intermediate

RYour turn

ex_5_2 <- # your code here ex_5_2

Click to reveal solution

RSolution

dims <- c("x", "y", "z") ex_5_2 <- diamonds |> select(any_of(dims)) ex_5_2

Explanation: any_of(dims) selects every column in the vector that actually exists in the data and silently ignores any that don't. The strict counterpart all_of() errors when a name is missing, which is useful when you want a hard contract. Either is far safer than the bare unquoted form select(x, y, z) for code that builds column names dynamically (from config files, function arguments, or user input).

Exercise 5.3: Extract numeric columns with where(is.numeric)

Task: Use select() together with where() and the predicate is.numeric to keep only the numeric columns of the mpg dataset. The where() helper applies a predicate function to each column and returns the columns where the function evaluates to TRUE. Save the numeric-only tibble to ex_5_3.

Expected result:

#> # A tibble: 234 x 5
#>   displ  year   cyl   cty   hwy
#>   <dbl> <int> <int> <int> <int>
#> 1   1.8  1999     4    18    29
#> 2   1.8  1999     4    21    29
#> 3   2.0  2008     4    20    31
#> ...
#> # 231 more rows hidden

Difficulty: Advanced

RYour turn

ex_5_3 <- # your code here ex_5_3

Click to reveal solution

RSolution

ex_5_3 <- mpg |> select(where(is.numeric)) ex_5_3

Explanation: where(predicate) evaluates the predicate against each column and keeps the columns where the result is TRUE. is.numeric matches both integer and double columns. Common variants include where(is.character), where(is.factor), and custom anonymous functions like where(~ mean(.x, na.rm = TRUE) > 0) for value-based selection. It is the dplyr equivalent of Filter(is.numeric, df) in base R, but composes naturally inside the tidyverse grammar.

Exercise 5.4: Move the last column to the front with last_col

Task: A data engineer publishing the mpg dataset wants the very last column of the table moved to the front and every other column to follow in its original order. Use select() with last_col() and everything() to reorder without retyping any column names, then save the result to ex_5_4.

Expected result:

#> # A tibble: 234 x 11
#>   class   manufacturer model displ  year   cyl trans drv     cty   hwy fl
#>   <chr>   <chr>        <chr> <dbl> <int> <int> <chr> <chr> <int> <int> <chr>
#> 1 compact audi         a4      1.8  1999     4 auto~ f        18    29 p
#> 2 compact audi         a4      1.8  1999     4 manu~ f        21    29 p
#> 3 compact audi         a4      2.0  2008     4 manu~ f        20    31 p
#> ...
#> # 231 more rows hidden

Difficulty: Intermediate

RYour turn

ex_5_4 <- # your code here ex_5_4

Click to reveal solution

RSolution

ex_5_4 <- mpg |> select(last_col(), everything()) head(ex_5_4)

Explanation: last_col() returns the index of the rightmost column, and dplyr keeps each named column only once even when it appears in multiple helpers. The result: class jumps to position 1, then everything() adds the rest in original order without duplicating class. The same trick reorders the second-to-last column with last_col(offset = 1). Useful for surfacing identifier columns that exporters tend to bury at the end of a wide table.

Section 6. Filter and select together (4 problems)

Exercise 6.1: Houston-only housing trend feed

Task: A real estate analyst building a Houston-only price trend dashboard needs from txhousing only the columns date, median, and sales, and only the rows where city equals "Houston". Chain filter() and select() together and save the result to ex_6_1 for the dashboard data feed.

Expected result:

#> # A tibble: 187 x 3
#>     date median sales
#>    <dbl>  <dbl> <int>
#> 1  2000  129200  3041
#> 2  2000. 131800  4154
#> 3  2000. 133000  4202
#> 4  2000. 134300  4061
#> ...
#> # 183 more rows hidden

Difficulty: Intermediate

RYour turn

ex_6_1 <- # your code here ex_6_1

Click to reveal solution

RSolution

ex_6_1 <- txhousing |> filter(city == "Houston") |> select(date, median, sales) ex_6_1

Explanation: Order matters here only for efficiency, not correctness: filtering first means select() operates on far fewer rows. Reversing the order would still give an identical result. The pipe |> reads top to bottom as "take txhousing, filter to Houston, then select the three columns" - much clearer than the nested call select(filter(txhousing, ...), ...). The Houston subset spans roughly 16 years of monthly observations.

Exercise 6.2: Premium-cut diamond pricing slice

Task: A jeweller pricing premium-cut stones wants only the rows of diamonds where cut equals "Premium", and only the columns carat, color, clarity, and price. Combine filter() and select() into one pipeline, then save the slice to ex_6_2 so it can be handed off to the pricing model.

Expected result:

#> # A tibble: 13,791 x 4
#>    carat color clarity price
#>    <dbl> <ord> <ord>   <int>
#> 1   0.21 E     SI1       326
#> 2   0.29 I     VS2       334
#> 3   0.22 F     SI1       342
#> 4   0.20 E     SI2       345
#> ...
#> # 13,787 more rows hidden

Difficulty: Intermediate

RYour turn

ex_6_2 <- # your code here ex_6_2

Click to reveal solution

RSolution

ex_6_2 <- diamonds |> filter(cut == "Premium") |> select(carat, color, clarity, price) ex_6_2

Explanation: This is the canonical analyst pattern: narrow rows first, then narrow columns. The cut column in diamonds is an ordered factor, but "Premium" still works as a character literal because R compares factor levels by their label string. There are 13,791 Premium-cut diamonds in the catalog, just over 25% of the total. Pricing teams typically pair this with a group_by(color) summary downstream.

Exercise 6.3: Summer pollution audit

Task: A climatologist auditing summer pollution wants from airquality only the months of June, July, and August, only the rows that have a recorded Ozone value, and only the columns Ozone, Temp, and Month. Combine two filter() conditions with a select() step and save the result to ex_6_3.

Expected result:

#>    Ozone Temp Month
#> 1     NA   NA    NA
#> 2     71   81     6
#> 3     49   69     6
#> 4     39   84     6
#> ...
#> nrow(ex_6_3):
#> [1] 83

Difficulty: Advanced

RYour turn

ex_6_3 <- # your code here ex_6_3

Click to reveal solution

RSolution

ex_6_3 <- airquality |> filter(Month %in% c(6, 7, 8), !is.na(Ozone)) |> select(Ozone, Temp, Month) ex_6_3

Explanation: Two comma-separated filter conditions combine with implicit AND, the same as writing & between them. airquality$Month is numeric, not a factor, so %in% c(6, 7, 8) works without quotes around the months. Of the 92 summer rows, 83 have an ozone reading. The 10% missingness rate climbs late in August. Always filter NAs before downstream summary statistics so the row counts in your report match what users see.

Exercise 6.4: Fuel-efficient small cars shortlist

Task: A consumer-reports analyst wants from mpg only the rows where class is either "compact" or "subcompact" and hwy (highway mpg) is at least 30. From those rows, keep just manufacturer, model, class, and hwy. Chain filter() and select() and save the result to ex_6_4 for the shortlist.

Expected result:

#> # A tibble: 39 x 4
#>   manufacturer model     class      hwy
#>   <chr>        <chr>     <chr>    <int>
#> 1 audi         a4        compact     30
#> 2 audi         a4        compact     31
#> 3 audi         a4 quattro compact     31
#> ...
#> # 36 more rows hidden

Difficulty: Advanced

RYour turn

ex_6_4 <- # your code here ex_6_4

Click to reveal solution

RSolution

ex_6_4 <- mpg |> filter(class %in% c("compact", "subcompact"), hwy >= 30) |> select(manufacturer, model, class, hwy) ex_6_4

Explanation: %in% plus a numeric threshold is the typical "small fuel-efficient car" shopping query. The pipeline order (filter then select) is conventional: it narrows the data before exposing only the columns the consumer report cares about. Thirty-nine rows survive both filters. Watch out for duplicate manufacturer-model pairs across model years: here that's expected, because each row is one model-year combination rather than a single car.

What to do next

You've drilled the full grammar of filter() and select() across 20 problems. To go deeper, work through these next:

dplyr filter() in R: the long-form reference for filter() patterns including row-wise comparisons, if_any(), and if_all().
dplyr select() in R: every selection helper (starts_with, ends_with, matches, num_range, where) with worked examples.
dplyr arrange and slice: combine row filters with ordering, slice_head, slice_min, and slice_sample to build leaderboards.
dplyr Exercises in R: the umbrella exercise hub covering all six core dplyr verbs end to end.

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dplyr filter() and select() Exercises: 20 Practice Problems

Section 1. Single-condition row filtering (3 problems)

Exercise 1.1: Keep the fuel-efficient cars from mtcars

Exercise 1.2: Pull a budget tier from the diamonds catalog

Exercise 1.3: Keep only compact cars from the mpg dataset

Section 2. Combining multiple conditions (4 problems)

Exercise 2.1: Pick setosa flowers with wide petals

Exercise 2.2: Climatologist flags hot and broken-sensor days

Exercise 2.3: Use %in% for four or six-cylinder cars

Exercise 2.4: Filter mid-carat diamonds with between()

Section 3. Missing values and slicing helpers (2 problems)

Exercise 3.1: Drop rows where Ozone is missing

Exercise 3.2: Top five highest horsepower cars with slice_max

Section 4. Selecting and renaming columns (3 problems)

Exercise 4.1: Keep three core columns from mtcars

Exercise 4.2: Drop one column with negative selection

Exercise 4.3: Rename columns inline while selecting

Section 5. Selection helpers (4 problems)

Exercise 5.1: Pull Sepal columns with starts_with

Exercise 5.2: Pull dimension columns with any_of

Exercise 5.3: Extract numeric columns with where(is.numeric)

Exercise 5.4: Move the last column to the front with last_col

Section 6. Filter and select together (4 problems)

Exercise 6.1: Houston-only housing trend feed

Exercise 6.2: Premium-cut diamond pricing slice

Exercise 6.3: Summer pollution audit

Exercise 6.4: Fuel-efficient small cars shortlist

What to do next

filter & select (20 problems) Mastery

Navigate

Tidyverse packages

Deep dives

Wrangling & EDA

Statistics

Machine Learning

Time Series

By Industry

Reporting & Apps

Levels

dplyr filter() and select() Exercises: 20 Practice Problems

Section 1. Single-condition row filtering (3 problems)

Exercise 1.1: Keep the fuel-efficient cars from mtcars

Exercise 1.2: Pull a budget tier from the diamonds catalog

Exercise 1.3: Keep only compact cars from the mpg dataset

Section 2. Combining multiple conditions (4 problems)

Exercise 2.1: Pick setosa flowers with wide petals

Exercise 2.2: Climatologist flags hot and broken-sensor days

Exercise 2.3: Use %in% for four or six-cylinder cars

Exercise 2.4: Filter mid-carat diamonds with between()

Section 3. Missing values and slicing helpers (2 problems)

Exercise 3.1: Drop rows where Ozone is missing

Exercise 3.2: Top five highest horsepower cars with slice_max

Section 4. Selecting and renaming columns (3 problems)

Exercise 4.1: Keep three core columns from mtcars

Exercise 4.2: Drop one column with negative selection

Exercise 4.3: Rename columns inline while selecting

Section 5. Selection helpers (4 problems)

Exercise 5.1: Pull Sepal columns with starts_with

Exercise 5.2: Pull dimension columns with any_of

Exercise 5.3: Extract numeric columns with where(is.numeric)

Exercise 5.4: Move the last column to the front with last_col

Section 6. Filter and select together (4 problems)

Exercise 6.1: Houston-only housing trend feed

Exercise 6.2: Premium-cut diamond pricing slice

Exercise 6.3: Summer pollution audit

Exercise 6.4: Fuel-efficient small cars shortlist

What to do next

Related Tutorials

filter & select (20 problems) Mastery