Error Triage Drills: 20 Broken R Scripts to Fix
This featured debugging collection hands you 20 broken R scripts to repair. Each drill shows the exact error R throws, from object not found and subscript out of bounds to unused argument and factor traps. Read the message, locate the fault, and fix the code so it returns the stated result. Full solutions stay hidden until you choose to reveal them.
Run the setup block once, then work through the drills in any order. Every script fails on purpose: your job is to read the error, diagnose the cause, and repair the code.
Section 1. Missing names: objects and functions (3 problems)
Exercise 1.1: Fix a misspelled object name in a sum
Task: A quick sales tally throws Error: object 'monthy' not found because the variable name inside the sum() call is misspelled. Repair the typo so the total of the three monthly figures is computed, and save the repaired result to ex_1_1.
Expected result:
#> [1] 3680Difficulty: Beginner
The value already exists in the workspace under a correctly spelled name; the error is a mismatch between what you typed and what you defined.
Change monthy to monthly inside sum() so the call sees the real vector.
Click to reveal solution
Explanation: R looks up names exactly as written, so a single dropped or transposed letter yields object not found rather than a helpful guess. The fix is never to redefine the data but to correct the reference. When this bites, print ls() to see the names actually in scope.
Exercise 1.2: Correct a mistyped column-means function call
Task: This audit line meant to average two mtcars columns fails with could not find function "colmeans". R is case-sensitive about function names, so correct the call to compute the mean of mpg and hp, and store the output in ex_1_2.
Expected result:
#> mpg hp
#> 20.09062 146.68750 Difficulty: Intermediate
The function you want is real, but R found nothing under the exact casing you used; think about how R capitalises its built-in matrix helpers.
Use colMeans() with a capital M on the two selected columns.
Click to reveal solution
Explanation: Function lookup is case-sensitive: colmeans and colMeans are different names, and only the camelCase one exists in base R. The same trap hits rowMeans, colSums, and is.na. When you see could not find function, check spelling and casing before assuming a package is missing.
Exercise 1.3: Reference a data frame column that is out of scope
Task: An analyst tries mean(disp) to get the average displacement, but R replies Error: object 'disp' not found because disp lives inside mtcars, not the global workspace. Point the call at the right column and save the mean to ex_1_3.
Expected result:
#> [1] 230.7219Difficulty: Intermediate
The name is only visible from inside the data frame, not on its own; you must tell R which table the column belongs to.
Reach the column with mtcars$disp inside mean().
Click to reveal solution
Explanation: Column names are not free-floating variables; they live inside the data frame and must be reached with $, [[, or with(). Writing bare disp searches the global environment and fails. Avoid attach() as a shortcut: it invites silent masking bugs when two frames share a column name.
Section 2. Indexing past the edge (3 problems)
Exercise 2.1: Repair a list index that runs past the end
Task: A pricing sheet holds three bids in a list, but reading bids[[4]] returns Error in bids[[4]] : subscript out of bounds because only three elements exist. Grab the last real element instead and save its value to ex_2_1.
Expected result:
#> [1] 1750Difficulty: Intermediate
You asked for a position one past the end; count how many elements the container actually holds before indexing.
Index the third element with bids[[3]].
Click to reveal solution
Explanation: Double brackets extract a single element, and asking for position 4 of a three-element list runs off the end. Always size the container with length() before indexing a computed position. Note that single brackets bids[4] would instead return a one-element list holding NULL rather than erroring.
Exercise 2.2: Fix a matrix column subscript out of bounds
Task: A 2-by-3 matrix is queried for a fourth column with m[, 4], raising Error in m[, 4] : subscript out of bounds. The matrix has only three columns, so read the correct final column and store the resulting vector in ex_2_2.
Expected result:
#> [1] 5 6Difficulty: Advanced
The requested column index exceeds the number of columns present; check the shape before you index.
Select the third column with m[, 3].
Click to reveal solution
Explanation: A matrix subscript must fall within its dimensions, and m[, 4] asks for a fourth column that does not exist. Check the shape with dim() or ncol() first. Unlike a data frame, this matrix has no column names, so positional indexing is all you have.
Exercise 2.3: Replace the dollar operator on a named vector
Task: A named numeric vector fees is accessed with fees$setup, which triggers Error in fees$setup : $ operator is invalid for atomic vectors. The dollar sign only works on lists and data frames, so use the right accessor and save the value to ex_2_3.
Expected result:
#> [1] 1250Difficulty: Intermediate
The dollar sign only reaches inside recursive structures; an atomic vector needs a different accessor for a named element.
Pull the element with fees[["setup"]] (or fees["setup"] to keep the name).
Click to reveal solution
Explanation: The $ operator is defined only for recursive objects like lists and data frames, not atomic vectors. For a named vector, use [[ for one value or [ to keep the name attached. A frequent cause is treating a vector as if it were a one-row data frame.
Section 3. Types that will not coerce (3 problems)
Exercise 3.1: Add a string and a number without erroring
Task: A form field arrives as the text "1200", and adding a 50 surcharge with "1200" + 50 fails with Error in "1200" + 50 : non-numeric argument to binary operator. Convert the string before adding so the sum is 1250, and save the corrected result to ex_3_1.
Expected result:
#> [1] 1250Difficulty: Beginner
One operand is text, and arithmetic needs both sides to be numbers, so a conversion has to happen first.
Wrap the string in as.numeric() before the +.
Click to reveal solution
Explanation: Arithmetic operators need numeric or logical operands, so a character string on either side stops the operation cold. as.numeric() performs the conversion, returning NA with a warning if the text is not a clean number. This is the classic bug when values arrive from form inputs or read.csv() columns.
Exercise 3.2: Sum numbers that were stored as a factor
Task: Three heights were stored as a factor, so sum(heights) throws Error in Summary.factor(...) : 'sum' not meaningful for factors. Convert the factor back to real numbers before adding (a bare as.numeric returns level codes, not the labels) and save the total to ex_3_2.
Expected result:
#> [1] 515Difficulty: Intermediate
A factor stores labels plus hidden integer codes; converting straight to numeric gives the codes, so you must recover the labels as text first.
Use as.numeric(as.character(heights)) before sum().
Click to reveal solution
Explanation: Summary generics like sum, mean, and max refuse factors outright. Converting with a bare as.numeric() is worse than the error: it silently returns the integer level codes (1, 2, 3), not the labels. Always route through as.character() first so the printed labels become the numbers you meant.
Exercise 3.3: Total a character vector of prices safely
Task: A vector of prices read as text triggers Error in sum(prices) : invalid 'type' (character) of argument when totalled directly. Coerce the character values to numeric first so the three prices add up correctly, and store the sum in ex_3_3.
Expected result:
#> [1] 37.49Difficulty: Intermediate
The values look like numbers but are stored as text, and totals need real numerics under the hood.
Apply as.numeric(prices) inside sum().
Click to reveal solution
Explanation: A character vector cannot be summed because addition is undefined for text, and as.numeric() restores the real values first. This commonly appears after importing data where one stray non-numeric entry forced the whole column to character. Inspect with str() to catch a text column that should have been numeric.
Section 4. Arguments the function never accepted (2 problems)
Exercise 4.1: Fix a mistyped matrix dimension argument
Task: A reshape step calls matrix(1:6, nrow = 2, ncols = 3) and fails with Error in matrix(1:6, nrow = 2, ncols = 3) : unused argument (ncols = 3). The keyword for column count is spelled differently, so fix the argument name to build the 2-by-3 matrix and save it to ex_4_1.
Expected result:
#> [,1] [,2] [,3]
#> [1,] 1 3 5
#> [2,] 2 4 6Difficulty: Intermediate
The function rejected an argument it does not define; compare the keyword you passed against the ones the function expects.
Rename ncols to ncol in the matrix() call.
Click to reveal solution
Explanation: R matches arguments by the exact names a function declares, and matrix() expects ncol, not ncols. Any unrecognised name triggers unused argument. When unsure, run args(matrix) or read ?matrix to see the real parameter names rather than guessing at a plural.
Exercise 4.2: Correct a misspelled argument name in a helper
Task: A helper defined as standardize(x, center = TRUE) is called with centre = TRUE, giving Error in standardize(...) : unused argument (centre = TRUE). Match the argument name the function actually declares so the z-scores are returned, and store them in ex_4_2.
Expected result:
#> [1] -1.1618950 -0.3872983 0.3872983 1.1618950Difficulty: Advanced
The function only accepts the argument names it declares, so a near-miss spelling is treated as unknown rather than corrected.
Pass center = TRUE to match the parameter standardize() defines.
Click to reveal solution
Explanation: Argument names must match the function signature character for character, so the British centre never binds to the American center the function declared. Partial matching would accept an unambiguous prefix like cent, but a differently spelled full word is rejected. Standardising spelling in your own APIs prevents this entirely.
Section 5. Code that will not parse (2 problems)
Exercise 5.1: Separate two statements crammed onto one line
Task: Two assignments were crammed onto one line as price <- 100 quantity <- 3, so R aborts with unexpected symbol. Put each statement on its own line so the product of price and quantity evaluates, then save the value to ex_5_1.
Expected result:
#> [1] 300Difficulty: Beginner
R could not tell where the first statement ended; two expressions cannot share a line without a separator between them.
Put price <- 100 and quantity <- 3 on separate lines (or divide them with a semicolon).
Click to reveal solution
Explanation: R parses one complete expression at a time and cannot tell where the first statement ends without a newline or semicolon, so two assignments on one line collide at the second name. Prefer newlines over semicolons for readability, and reserve ; for terse one-off console lines.
Exercise 5.2: Add the missing comma in a data frame call
Task: A data frame call is written data.frame(x = 1:3 y = 4:6) and R stops with unexpected symbol at y. A separator between the two columns is missing, so add it and build the two-column frame, saving the result to ex_5_2.
Expected result:
#> x y
#> 1 1 4
#> 2 2 5
#> 3 3 6Difficulty: Intermediate
The parser hit a second column name with nothing telling it the first argument had already ended.
Insert a comma between x = 1:3 and y = 4:6 in data.frame().
Click to reveal solution
Explanation: Inside a function call every argument must be separated by a comma, and without it the parser reaches the next name unexpectedly and stops. The caret in the error points at the first token it could not place. Scan just left of that position, where a comma or operator is usually missing.
Section 6. Conditions that meet NA and emptiness (3 problems)
Exercise 6.1: Guard an if condition against a missing mean
Task: A threshold check if (mean(threshold) > 5) errors with missing value where TRUE/FALSE needed because threshold contains an NA that poisons the mean. Make the mean ignore the missing value so the branch resolves, and save the chosen label to ex_6_1.
Expected result:
#> [1] "low"Difficulty: Intermediate
The condition evaluated to NA, and an if needs a definite TRUE or FALSE; the NA came from the summary of the data.
Compute the mean with na.rm = TRUE so the NA is dropped before the comparison.
Click to reveal solution
Explanation: An if needs a single, non-missing logical, but mean() over data containing NA returns NA, which cannot be branched on. Passing na.rm = TRUE drops the gap before averaging. The deeper lesson is to handle NA deliberately rather than letting it surface at a control-flow boundary.
Exercise 6.2: Skip NA safely inside a filtering loop
Task: A loop keeps readings above four, but if (v > 4) errors with missing value where TRUE/FALSE needed the moment it reaches the NA reading. Guard the test so the loop skips missing values, collecting the survivors into ex_6_2.
Expected result:
#> [1] 5 8Difficulty: Advanced
The comparison returns NA on the missing element, and the if cannot branch on NA, so you must exclude missing values before testing.
Test !is.na(v) && v > 4 inside the loop.
Click to reveal solution
Explanation: The comparison v > 4 yields NA when v is NA, and if cannot act on NA. Guarding with !is.na(v) first, combined with && so the second test is skipped when the first is FALSE, keeps the loop moving. A vectorised alternative is readings[!is.na(readings) & readings > 4].
Exercise 6.3: Handle an empty vector before an if test
Task: After filtering scores above 95 the result is empty, so if (top > 90) throws Error in if (top > 90) ... : argument is of length zero. Check whether any value survived the filter before testing it, and save the resulting label to ex_6_3.
Expected result:
#> [1] "good"Difficulty: Intermediate
The comparison ran on an empty vector, so the if received nothing to decide on; confirm something survived the filter first.
Gate the test with length(top) > 0 before comparing.
Click to reveal solution
Explanation: Filtering can legitimately return an empty vector, and comparing it produces a length-zero logical that if rejects. Checking length() first is the guard. This is why if on a subset is fragile; for element-wise decisions on a vector, ifelse() or any() and all() are safer.
Section 7. Data-frame pipelines that break (4 problems)
Exercise 7.1: Build the revenue column a select could not find
Task: A reporting analyst runs orders[, c("region", "revenue")] and hits Error in ... : undefined columns selected because revenue was never computed. Derive it from quantity and price, then keep region and the new column, saving the two-column result to ex_7_1.
Expected result:
#> region revenue
#> 1 West 59.97
#> 2 East 5.00
#> 3 West 99.95
#> 4 North 85.00
#> 5 East 20.00Difficulty: Intermediate
You cannot select a column that was never built; the fix is to create it, not to reference it.
Add the column with mutate(revenue = quantity * price), then keep region and revenue with select().
Click to reveal solution
Explanation: You cannot select a column that was never created, so the error is a symptom of a missing derivation rather than a typo. Building revenue with mutate() and then narrowing with select() produces exactly the two columns requested. In base R the same fix is transform() followed by column indexing.
Exercise 7.2: Fix a replacement vector of the wrong length
Task: Adding a flat discount with orders$discount <- c(0.1, 0.2) fails with Error ... replacement has 2 rows, data has 5. The replacement vector is shorter than the five-row frame, so supply a value that applies to every row and save the updated copy to ex_7_2.
Expected result:
#> region product quantity price discount
#> 1 West A 3 19.99 0.1
#> 2 East B 1 5.00 0.1
#> 3 West A 5 19.99 0.1
#> 4 North C 2 42.50 0.1
#> 5 East B 4 5.00 0.1Difficulty: Intermediate
The number of values you assign must match the number of rows, or recycle cleanly into them.
Assign a single value like 0.1 so it recycles across all five rows.
Click to reveal solution
Explanation: Assigning into a data frame column requires a vector whose length matches the rows or recycles evenly, so two values cannot spread across five rows. A single scalar recycles cleanly to every row. When you truly need per-row values, supply a full-length vector, often computed from the existing columns.
Exercise 7.3: Correct a mistyped column inside a mutate
Task: A pipeline orders |> mutate(line_total = quantop * price) breaks with object 'quantop' not found raised inside mutate(). The column name is mistyped, so correct it to multiply quantity by price per row, and store the widened frame in ex_7_3.
Expected result:
#> region product quantity price line_total
#> 1 West A 3 19.99 59.97
#> 2 East B 1 5.00 5.00
#> 3 West A 5 19.99 99.95
#> 4 North C 2 42.50 85.00
#> 5 East B 4 5.00 20.00Difficulty: Advanced
The expression references a name that does not exist in the frame; check the column spelling against the actual data.
Change quantop to quantity inside mutate().
Click to reveal solution
Explanation: Inside mutate(), unquoted names are resolved against the frame's columns, so a misspelled quantop is reported as object not found. dplyr wraps the base error to show which argument failed. Fixing the spelling to quantity lets the multiplication run row by row.
Exercise 7.4: Fix an equals sign used inside filter
Task: A filter written orders |> filter(region = "West") errors because R reads region = "West" as a named argument, not a comparison. Swap in the equality test dplyr expects to keep only the West rows, and save them to ex_7_4.
Expected result:
#> region product quantity price
#> 1 West A 3 19.99
#> 2 West A 5 19.99Difficulty: Advanced
A single equals sign inside the call is read as naming an argument, not comparing values.
Use the double-equals test region == "West" inside filter().
Click to reveal solution
Explanation: A single = inside filter() looks like you are naming an argument, so dplyr warns that you likely meant ==. Comparison always uses double equals, while single equals is assignment or argument binding. The helpful hint in the message even suggests the corrected expression for you to copy.
What to do next
You have triaged 20 broken scripts across ten of R's most common error families. Keep the momentum going with these related practice sets:
- R Debugging Exercises: step through
traceback(),browser(), and defensive coding drills. - R Beginner Exercises: shore up the fundamentals that prevent these errors in the first place.
- R Control Flow Exercises: practise the
if,for, andwhilepatterns behind the condition errors above. - R Vectors Exercises: drill the indexing and coercion rules that cause subscript and type errors.
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