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Apply Family Exercises in R: 20 Real-World Practice Problems
Twenty practice problems on the base R apply family: lapply, sapply, vapply, apply, mapply, and tapply. Light on warm-ups, heavy on the intermediate cases where the trap is picking the right family member. Solutions are hidden behind reveal toggles so you try first.
Section 1. lapply foundations (3 problems)
Exercise 1.1: Audit column classes of iris with lapply
Task: A junior analyst onboarding to a new project wants a quick audit of column types in the iris dataset before joining it with other tables. Use lapply() together with the class function to return one element per column of iris. Save the result to ex_1_1 and print it.
Expected result:
#> $Sepal.Length
#> [1] "numeric"
#>
#> $Sepal.Width
#> [1] "numeric"
#>
#> $Petal.Length
#> [1] "numeric"
#>
#> $Petal.Width
#> [1] "numeric"
#>
#> $Species
#> [1] "factor"Difficulty: Beginner
Click to reveal solution
Explanation: A data frame is a list of columns, so lapply() walks each column and applies class(). The return type is always a list, which is what you want when elements could differ in length or type. Using sapply() here would still work because every result is length-one character, but lapply() is the right idiom when you intend a list and want it to stay that way regardless of input.
Exercise 1.2: Count distinct values per mtcars column
Task: A data engineer is profiling mtcars to decide which columns are categorical-ish enough to convert to factors. Use lapply() with an anonymous function that calls length(unique(x)) on each column. Save the named list to ex_1_2 and print it.
Expected result:
#> $mpg
#> [1] 25
#>
#> $cyl
#> [1] 3
#>
#> $disp
#> [1] 27
#>
#> $hp
#> [1] 22
#>
#> $drat
#> [1] 22
#>
#> $wt
#> [1] 29
#>
#> $qsec
#> [1] 30
#>
#> $vs
#> [1] 2
#>
#> $am
#> [1] 2
#>
#> $gear
#> [1] 3
#>
#> $carb
#> [1] 6Difficulty: Intermediate
Click to reveal solution
Explanation: Columns with only a handful of distinct values (cyl, vs, am, gear) are factor candidates. The anonymous function pattern function(x) length(unique(x)) is the workhorse here. You could simplify the output with sapply() to get a named integer vector, but keeping it as a list keeps the door open for richer per-column metadata later (e.g. returning both count and the values themselves).
Exercise 1.3: Per-column quantile summary on airquality
Task: An environmental analyst wants a five-number quantile breakdown for the four measurement columns of airquality (Ozone, Solar.R, Wind, Temp), ignoring missing values. Use lapply() with the quantile function, passing na.rm = TRUE as an extra argument. Save the result to ex_1_3.
Expected result:
#> $Ozone
#> 0% 25% 50% 75% 100%
#> 1.00 18.00 31.50 63.25 168.00
#>
#> $Solar.R
#> 0% 25% 50% 75% 100%
#> 7.00 115.75 205.00 258.75 334.00
#>
#> $Wind
#> 0% 25% 50% 75% 100%
#> 1.7 7.4 9.7 11.5 20.7
#>
#> $Temp
#> 0% 25% 50% 75% 100%
#> 56 72 79 85 97Difficulty: Intermediate
Click to reveal solution
Explanation: Trailing arguments to lapply() are forwarded to the function being applied, so na.rm = TRUE reaches quantile() for every column. Subsetting airquality[, 1:4] drops the Month and Day index columns where quantiles are meaningless. The list-of-vectors shape is exactly right for do.call(rbind, ex_1_3) to flatten into a tidy matrix later.
Section 2. sapply for vector-shaped returns (3 problems)
Exercise 2.1: Mean of every numeric column with sapply
Task: Use sapply() to compute the mean of every column of the built-in mtcars dataset, all 11 columns of which are numeric. Save the resulting named numeric vector to ex_2_1 and print it. This is the workhorse one-line audit you will return to often.
Expected result:
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> 20.09063 6.18750 230.72188 146.68750 3.59656 3.21725 17.84875 0.43750 0.40625 3.68750 2.81250Difficulty: Beginner
Click to reveal solution
Explanation: Because every column of mtcars is numeric and mean() returns a length-one numeric, sapply() simplifies the list to a named numeric vector. The names come from the column names, so you can index ex_2_1["mpg"] directly. If a non-numeric column were present, you would get NA and a warning instead of a hard failure, which is exactly the sloppy ergonomics vapply() was added to fix.
Exercise 2.2: Range of each numeric column of airquality
Task: A climate reviewer wants the min and max of the four measurement columns in airquality (Ozone, Solar.R, Wind, Temp), ignoring missing values. Use sapply() with the range function and na.rm = TRUE. Save the resulting 2-by-4 matrix to ex_2_2 and print it.
Expected result:
#> Ozone Solar.R Wind Temp
#> [1,] 1.0 7 1.7 56
#> [2,] 168.0 334 20.7 97Difficulty: Intermediate
Click to reveal solution
Explanation: When the applied function returns a length-N vector (here 2 for min and max) and N is the same for every input, sapply() simplifies to an N-by-K matrix rather than a list. That is the key shape rule: same-length numeric returns become a matrix, mixed-length stays a list. You can transpose with t(ex_2_2) to get rows-as-columns if a tidy layout works better downstream.
Exercise 2.3: Five-number summary matrix across mtcars
Task: Build a wide 5-by-11 summary table where each column is an mtcars variable and the rows are Tukey's five-number summary (minimum, lower hinge, median, upper hinge, maximum). Use sapply() with the base fivenum function. Save the matrix to ex_2_3 and print it.
Expected result:
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> [1,] 10.4000 4.0 71.100 52.00 2.760 1.5130 14.500 0.0 0.0 3.0 1.00
#> [2,] 15.4250 4.0 120.825 96.50 3.080 2.5425 16.892 0.0 0.0 3.0 2.00
#> [3,] 19.2000 6.0 196.300 123.00 3.695 3.3250 17.710 0.0 0.0 4.0 2.00
#> [4,] 22.8000 8.0 326.000 180.00 3.920 3.6500 18.900 1.0 1.0 4.0 4.00
#> [5,] 33.9000 8.0 472.000 335.00 4.930 5.4240 22.900 1.0 1.0 5.0 8.00Difficulty: Advanced
Click to reveal solution
Explanation: fivenum() returns a length-5 numeric vector, so sapply() stacks the results column-wise into a 5-by-11 numeric matrix. The hinges from fivenum() differ slightly from quantile()'s default Q1/Q3 because the algorithms use different interpolation schemes; pick fivenum() when you specifically want Tukey's boxplot hinges. To prepend row labels, assign rownames(ex_2_3) <- c("min", "Q1", "median", "Q3", "max") after the call.
Section 3. vapply for type-safe returns (3 problems)
Exercise 3.1: Type-safe column means with vapply
Task: Repeat the column-mean audit from Exercise 2.1, but this time enforce that every return value is a single numeric. Use vapply() on mtcars with the mean function and the template numeric(1). Save the resulting named numeric vector to ex_3_1. The template makes the call fail loudly if any column ever returns a non-numeric or a different shape.
Expected result:
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> 20.09063 6.18750 230.72188 146.68750 3.59656 3.21725 17.84875 0.43750 0.40625 3.68750 2.81250Difficulty: Intermediate
Click to reveal solution
Explanation: The third argument is a template, not a value. numeric(1) says "I expect a length-one numeric for each input"; if any iteration returns something else (a character, length 2, a list), vapply() errors immediately. That is the entire point of the function: trade a tiny bit of typing for a contract that protects pipelines from silent shape drift. Prefer vapply() over sapply() for anything you intend to ship.
Exercise 3.2: Per-column NA counts with vapply
Task: A data engineer is preparing airquality for a model and needs the count of missing values in every column. Use vapply() with an anonymous function that calls sum(is.na(x)), enforcing an integer(1) return template. Save the resulting named integer vector to ex_3_2 and print it.
Expected result:
#> Ozone Solar.R Wind Temp Month Day
#> 37 7 0 0 0 0Difficulty: Intermediate
Click to reveal solution
Explanation: sum() over a logical vector returns numeric, but the template integer(1) forces coercion (and errors if anything overflows). The named vector tells you immediately that Ozone is the column with the biggest missingness problem (24% missing), which is a useful gate before regression. A common alternative is colSums(is.na(airquality)), which is more idiomatic in base R but loses the per-column type guarantee.
Exercise 3.3: Min and max in one vapply pass
Task: Return a 2-by-4 matrix where each column is one of the airquality measurement variables (Ozone, Solar.R, Wind, Temp) and the two rows are the min and max with names. Use vapply() with a named numeric(2) template so the output rows are labelled. Save the result to ex_3_3.
Expected result:
#> Ozone Solar.R Wind Temp
#> min 1.0 7 1.7 56
#> max 168.0 334 20.7 97Difficulty: Advanced
Click to reveal solution
Explanation: The named FUN.VALUE template c(min = 0, max = 0) does two things at once: it enforces a length-2 numeric return and seeds the row names of the output matrix. This is the killer feature of vapply() over sapply(): you can predeclare both the shape AND the labels for free. Drop the names from the template and you get an unnamed 2-row matrix that is much harder to read downstream.
Section 4. apply on matrices and rectangular data (4 problems)
Exercise 4.1: Row totals across a quiz scorecard
Task: A course instructor has four quiz scores for four students stored as a matrix. Use apply() with MARGIN = 1 to compute the total points each student earned across all quizzes. Save the named numeric vector of totals to ex_4_1 and print it.
Expected result:
#> Alex Brio Casey Devi
#> 337 285 366 254Difficulty: Intermediate
Click to reveal solution
Explanation: MARGIN = 1 walks the rows; MARGIN = 2 would walk the columns. Because each row is a length-4 numeric, sum() collapses it to length-one, and the result simplifies to a named numeric vector keyed by rownames(scores). The row-walk pattern is also how you implement per-record scoring rules: pass any function that takes a numeric vector and returns a scalar (mean, max, custom weighted formulas) instead of sum.
Exercise 4.2: Column standard deviations of mtcars
Task: A reviewer comparing variable spread in mtcars wants the standard deviation of every column. Convert mtcars to a matrix with as.matrix() and use apply() with MARGIN = 2. Save the named numeric vector to ex_4_2. Note that sapply(mtcars, sd) would work too; the point here is the matrix path.
Expected result:
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> 6.0269481 1.7859216 123.9386938 68.5628685 0.5346787 0.9784574 1.7869432 0.5040161 0.4989909 0.7378041 1.6152000Difficulty: Intermediate
Click to reveal solution
Explanation: apply() strictly needs an array or matrix, which is why as.matrix(mtcars) is part of the idiom. For a data frame, sapply() is the cleaner equivalent and skips the coercion. Pick apply() when the data already lives in a matrix (image data, distance matrices, model output matrices) and sapply() when it lives in a data frame. Mixing them is a common source of accidental character coercion in the wild.
Exercise 4.3: Top score and the column where it occurred
Task: A tournament organiser stores athlete scores across four events as a 5-by-4 matrix. For each athlete (row), find the maximum score AND the event name where it occurred. Use apply() with MARGIN = 1 and a custom function that returns a 2-element character vector. Save the resulting 2-by-5 character matrix to ex_4_3.
Expected result:
#> Aria Bohan Calix Doris Eshan
#> max "97" "88" "92" "85" "94"
#> event "Run" "Lift" "Swim" "Run" "Jump"Difficulty: Intermediate
Click to reveal solution
Explanation: When the inner function returns a length-2 vector, apply() stacks results as columns, producing a 2-by-N matrix. Because the value column is numeric and the event column is character, R coerces everything to character, which is why you see quote marks in the output. If you need the numeric to stay numeric, return a data.frame row instead and rbind the results. The which.max() + names() trick is the canonical "best label" pattern.
Exercise 4.4: Centre every column of mtcars by its mean
Task: Many models need predictors centred at zero. Use apply() with MARGIN = 2 and an anonymous function function(x) x - mean(x) to subtract the column mean from every entry of mtcars. The result will be a 32-by-11 numeric matrix whose every column has mean zero. Save it to ex_4_4.
Expected result:
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Mazda RX4 0.90938 -0.187 -70.722 -36.687 0.3934 -0.59725 -1.4488 -0.4375 0.5938 0.3125 1.1875
#> Mazda RX4 Wag 0.90938 -0.187 -70.722 -36.687 0.3934 -0.34225 -0.3488 -0.4375 0.5938 0.3125 1.1875
#> Datsun 710 2.70938 -2.187 -122.722 -53.687 0.2934 -0.89725 0.7713 0.5625 0.5938 0.3125 -1.8125
#> ...
#> # 29 more rows hidden; colMeans(ex_4_4) is effectively zeroDifficulty: Advanced
Click to reveal solution
Explanation: The anonymous function receives one column at a time and returns the same-length centred column. Because every return is length 32, apply() rebuilds the result back into a 32-by-11 matrix with the original row names attached. The scale() function is the production-grade equivalent and also handles standardisation by sd; the manual apply() version is useful when you want full control over the centring statistic (median, trimmed mean, group mean, etc).
Section 5. mapply for multi-argument vectorisation (3 problems)
Exercise 5.1: Range size from paired low and high vectors
Task: A reporting analyst has two aligned vectors: lows and highs representing the low and high bound of four intervals. Use mapply() to compute the width of each interval as high - low. Save the resulting numeric vector to ex_5_1 and print it. The output should have length four with one entry per interval.
Expected result:
#> [1] 5 5 8 7Difficulty: Intermediate
Click to reveal solution
Explanation: mapply() is the multivariate cousin of sapply(). It zips together as many vectors as you give it and calls the function once per aligned tuple. Here the zip produces (10, 15), (20, 25), (30, 38), (40, 47) and the function returns the difference each time. Of course highs - lows is the better one-liner for this exact case; mapply() earns its keep when the function does something genuinely non-vectorised, like the next two exercises.
Exercise 5.2: Generate samples of varying size and scale
Task: A simulation team needs three random samples of different sizes drawn from normals with different means and standard deviations. Use mapply() with the rnorm function and three aligned argument vectors n = c(3, 5, 4), mean = c(0, 10, 100), sd = c(1, 2, 5). Set SIMPLIFY = FALSE to keep the result as a list (the lengths differ). Save the list to ex_5_2.
Expected result:
#> [[1]]
#> [1] 1.3709584 -0.5646982 0.3631284
#>
#> [[2]]
#> [1] 11.2649671 10.3261099 8.7345924 9.5050700 9.2671352
#>
#> [[3]]
#> [1] 109.4933665 96.4729259 100.7798643 104.6489295Difficulty: Intermediate
Click to reveal solution
Explanation: Without SIMPLIFY = FALSE you would get a list anyway here because the three return vectors have different lengths and mapply() cannot pack them into a matrix. Setting the argument explicitly documents intent and protects you against the day all three sizes accidentally coincide and the output silently changes shape to a matrix. The trio of named arguments (n, mean, sd) maps directly to rnorm()'s signature.
Exercise 5.3: Random walks with varying length and volatility
Task: Build three random walks where each walk has a different length and a different per-step standard deviation. Use mapply() with an anonymous function that calls cumsum(rnorm(n, sd = sigma)), pass n = c(5, 8, 6) and sigma = c(0.5, 1.0, 2.0), and set SIMPLIFY = FALSE. Save the list of three numeric vectors to ex_5_3.
Expected result:
#> [[1]]
#> [1] 0.6854 -0.0668 -0.0668 -0.3998 -1.1232
#>
#> [[2]]
#> [1] -0.1058 -1.1366 -0.6131 -2.0193 -2.0193 -2.0193 -0.6193 -2.3193
#>
#> [[3]]
#> [1] -1.6500 -3.3000 0.5800 0.8200 -0.4400 -1.7000
#> # exact values depend on set.seed(42); shape is what matters: lengths 5, 8, 6Difficulty: Advanced
Click to reveal solution
Explanation: mapply() shines when the inner function is genuinely scalar in its arguments but vector in its return. Here each call generates one walk of length n with step volatility sigma. Because cumsum() returns a length-n vector and the three lengths differ, the result must be a list (forced explicit via SIMPLIFY = FALSE). For uniform-length output (e.g. always 100 steps) you would get an N-by-3 matrix instead, which is also frequently what you want for plotting trajectories.
Section 6. tapply and grouped summaries (4 problems)
Exercise 6.1: Mean mpg by cylinder count
Task: Use tapply() to compute the mean mpg of cars in mtcars grouped by cyl. The first argument is the values to summarise, the second is the grouping factor, the third is the function. Save the named numeric vector (one entry per cylinder count) to ex_6_1 and print it.
Expected result:
#> 4 6 8
#> 26.66364 19.74286 15.10000Difficulty: Beginner
Click to reveal solution
Explanation: tapply() is the base R version of "group by, then summarise". The first argument supplies values, the second the grouping vector (treated as a factor automatically), and the third the summary function. The names of the returned vector come from the factor levels. For a single grouping variable returning a single number, the output is a named vector; for multiple groupings or vector-valued returns the shape generalises to higher-dimensional arrays.
Exercise 6.2: Mean Sepal.Length by Species
Task: A biologist comparing flower morphology wants the average sepal length for each of the three Iris species. Use tapply() on iris$Sepal.Length grouped by iris$Species with the mean function. Save the named numeric vector (length 3) to ex_6_2 and print it.
Expected result:
#> setosa versicolor virginica
#> 5.006 5.936 6.588Difficulty: Intermediate
Click to reveal solution
Explanation: Species is already a factor with three levels, so tapply() slices the 150 measurements into three groups of 50 and applies mean() to each. The output's names match levels(iris$Species) in their stored order, not alphabetical order if the factor was hand-built. This is the canonical "group means" pattern; the dplyr equivalent is iris %>% group_by(Species) %>% summarise(mean(Sepal.Length)).
Exercise 6.3: Two-way tapply on ChickWeight
Task: An animal nutrition lab tracks chick weight by Diet and Time in the ChickWeight dataset. Use tapply() with a list of two grouping factors list(ChickWeight$Diet, ChickWeight$Time) to compute mean weight at each Diet-by-Time combination. Save the resulting 4-by-12 matrix to ex_6_3 and print it. NA appears where no chicks remain at that timepoint.
Expected result:
#> 0 2 4 6 8 10 12 14 16 18 20 21
#> 1 41.40000 47.2500 56.47368 66.78947 79.68421 93.05263 108.5263 123.3889 144.6471 158.9412 170.4118 177.7500
#> 2 40.70000 49.4000 59.80000 75.40000 91.70000 108.500 131.3000 141.9000 164.7000 187.7000 205.6000 214.7000
#> 3 40.80000 50.4000 62.20000 77.90000 98.40000 117.100 144.4000 164.5000 197.4000 233.1000 262.9000 270.3000
#> 4 41.00000 51.8000 64.50000 83.90000 105.6000 126.000 151.4000 161.8000 182.0000 202.9000 233.8889 238.5556Difficulty: Intermediate
Click to reveal solution
Explanation: Passing a list of grouping factors to tapply() turns the output into a multi-dimensional array (here, a 4-by-12 matrix), where rows are Diet levels and columns are Time levels. Some Diet-by-Time cells may be NA if every chick on that diet dropped out before that timepoint, which is exactly the "no observation" signal you want preserved rather than silently zero-filled. The closest tidyverse equivalent is pivot_wider() after a group_by() summarise.
Exercise 6.4: Coefficient of variation of mpg by cylinder
Task: A fleet analyst wants the coefficient of variation (sd divided by mean, expressed as a percentage) of mpg within each cyl group in mtcars. Use tapply() with a custom anonymous function function(x) sd(x) / mean(x) * 100. Save the named numeric vector to ex_6_4 and print it.
Expected result:
#> 4 6 8
#> 16.911185 7.353099 16.951474Difficulty: Advanced
Click to reveal solution
Explanation: Custom functions extend tapply() beyond the obvious mean/median/sum trio: any function that takes a numeric vector and returns a scalar is valid. The 6-cylinder group has the lowest relative spread (CV around 7%), meaning its mpg is the most predictable, while the 4-cyl and 8-cyl groups are both around 17%. Reporting CV instead of raw sd is the right move whenever the groups have very different means and you want a fair comparison of spread.
What to do next
You have worked through every member of the base R apply family. Pick the next step that matches what you want to harden:
- base apply in R: the parent tutorial that compares the whole family side by side
- Functional Programming in R: step up from apply to map/reduce idioms via purrr
- dplyr Exercises in R: the tidyverse companion to grouped summaries and column-wise operations
- R Functional Programming Exercises: broader functional-style practice across closures, higher-order functions, and recursion
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