Missing Data in R Exercises: 18 Real-World NA Practice Problems

Eighteen hands-on exercises drill NA detection, removal, and imputation in R, from base is.na() audits to grouped median imputation and end-to-end audit pipelines. Solutions stay hidden until you click reveal so you actually try each one first.

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

How to use this hub

Read each task, type your attempt into the "Your turn" box, then click reveal to compare. Every exercise saves its answer into a variable named ex_<section>_<number> so you can inspect intermediate results without overwriting the source datasets. All code runs in one shared session, so the setup block must run once before any exercise.

The exercises lean on the built-in airquality dataset (153 rows of New York weather and air-quality measurements with real NAs in Ozone and Solar.R), plus a small inline clinic tibble for grouped and sentinel-value problems. If is.na(), complete.cases(), and na.omit() are unfamiliar, skim the parent Missing Values in R tutorial first.

RRun this once before any exercise

library(dplyr) library(tidyr) # airquality has 44 NAs: 37 in Ozone, 7 in Solar.R data(airquality) # clinic: small inline tibble used for grouped/sentinel exercises clinic <- tibble( patient_id = 1:10, arm = c("drug", "drug", "drug", "drug", "drug", "placebo", "placebo", "placebo", "placebo", "placebo"), bp_systolic = c(132, NA, 128, 140, 135, 145, NA, 150, NA, 138), weight_kg = c(72, 80, NA, 68, 75, 90, 85, NA, 78, 82), visit_score = c(8, 7, 9, NA, 8, 6, 7, 0, 6, 7), diagnosis = c("HT", NA, "HT", "HT", NA, "HT", "HT", "HT", "HT", NA) ) clinic #> # A tibble: 10 x 6 #> patient_id arm bp_systolic weight_kg visit_score diagnosis #> <int> <chr> <dbl> <dbl> <dbl> <chr> #> 1 1 drug 132 72 8 HT #> 2 2 drug NA 80 7 NA #> 3 3 drug 128 NA 9 HT #> 4 4 drug 140 68 NA HT #> 5 5 drug 135 75 8 NA #> 6 6 placebo 145 90 6 HT #> 7 7 placebo NA 85 7 HT #> 8 8 placebo 150 NA 0 HT #> 9 9 placebo NA 78 6 HT #> 10 10 placebo 138 82 7 NA

The visit_score of 0 for patient 8 is a deliberate sentinel: in this fictional protocol, 0 is the code clinicians enter when the visit was missed but the row still had to be created. Exercise 5.3 asks you to recover that as NA before any summary touches it.

Section 1. Detect and quantify missingness (3 problems)

Exercise 1.1: Compute the total NA count across airquality

Task: An onboarding analyst auditing the airquality dataset wants a single headline number for the report: how many missing values does the entire frame contain across every column? Use is.na() plus sum() on the whole frame and save the integer to ex_1_1.

Expected result:

#> ex_1_1
#> [1] 44

Difficulty: Beginner

RYour turn

ex_1_1 <- # your code here ex_1_1

Click to reveal solution

RSolution

ex_1_1 <- sum(is.na(airquality)) ex_1_1 #> [1] 44

Explanation: is.na() returns a logical matrix the same shape as airquality, and sum() coerces TRUE to 1 and FALSE to 0. The total is 44 missing cells out of 153 rows by 6 columns. A common slip is writing sum(airquality == NA), which silently returns NA because any comparison with NA propagates. Always use is.na() for NA tests, never ==.

Exercise 1.2: Count NAs in every column with colSums

Task: Before deciding which columns to impute and which to drop, the analyst needs a per-column NA count for airquality. Use is.na() combined with colSums() to produce a named integer vector and save it to ex_1_2.

Expected result:

#>   Ozone Solar.R    Wind    Temp   Month     Day
#>      37       7       0       0       0       0

Difficulty: Beginner

RYour turn

ex_1_2 <- # your code here ex_1_2

Click to reveal solution

RSolution

ex_1_2 <- colSums(is.na(airquality)) ex_1_2 #> Ozone Solar.R Wind Temp Month Day #> 37 7 0 0 0 0

Explanation: is.na(airquality) returns a 153 by 6 logical matrix, and colSums() sums each column treating TRUE as 1. The result keeps the column names automatically, which is why this idiom is preferred over sapply(airquality, function(x) sum(is.na(x))). With only Ozone and Solar.R carrying any NAs, you can move straight to imputation on those two and leave the rest alone.

Exercise 1.3: Report the percentage missing per column

Task: A reporting analyst wants the per-column NA count expressed as a percentage of total rows, rounded to two decimals, so the cleaning summary slide reads cleanly. Build a named numeric vector of percentages for every column of airquality and save it to ex_1_3.

Expected result:

#>   Ozone Solar.R    Wind    Temp   Month     Day
#>   24.18    4.58    0.00    0.00    0.00    0.00

Difficulty: Intermediate

RYour turn

ex_1_3 <- # your code here ex_1_3

Click to reveal solution

RSolution

ex_1_3 <- round(colMeans(is.na(airquality)) * 100, 2) ex_1_3 #> Ozone Solar.R Wind Temp Month Day #> 24.18 4.58 0.00 0.00 0.00 0.00

Explanation: colMeans() on a logical matrix returns the proportion of TRUE per column, which is exactly the NA rate. Multiplying by 100 converts to a percentage, and round(..., 2) formats it for reporting. The alternative colSums(is.na(airquality)) / nrow(airquality) * 100 is equivalent but more keystrokes. At 24% missing, Ozone is on the borderline where many practitioners switch from imputation to a "missing as feature" flag.

Section 2. Locate and pattern-find NAs (3 problems)

Exercise 2.1: Find row indices that contain any NA

Task: A data engineer building a quarantine pipeline needs the row positions in airquality that have at least one NA anywhere in the row, so downstream consumers can choose to skip or repair them. Return an integer vector of those row indices and save it to ex_2_1.

Expected result:

#>  [1]   5   6  10  11  25  26  27  32  33  34  35  36  37  39  42  43  45  46  52
#> [20]  53  54  55  56  57  58  59  60  61  65  72  75  83  84 102 103 111
#> ... (some indices omitted)
#> length(ex_2_1)
#> [1] 42

Difficulty: Intermediate

RYour turn

ex_2_1 <- # your code here ex_2_1 length(ex_2_1)

Click to reveal solution

RSolution

ex_2_1 <- which(!complete.cases(airquality)) length(ex_2_1) #> [1] 42 head(ex_2_1, 10) #> [1] 5 6 10 11 25 26 27 32 33 34

Explanation: complete.cases() returns TRUE for rows with no NAs, so !complete.cases() flips the test and which() converts the logical vector to positional indices. The 42 incomplete rows include 2 rows where BOTH Ozone and Solar.R are missing (which is why the total NA count was 44 but only 42 distinct rows are affected). For just dropping incomplete rows use na.omit(); for collecting them for inspection, which(!complete.cases(...)) is the right tool.

Exercise 2.2: Find rows where Ozone is missing but Solar.R is not

Task: A climatologist wants to study days where the ozone sensor failed but the solar radiation sensor still recorded, since those rows are the best candidates for model-based imputation of Ozone from Solar.R. Subset airquality to those rows and save the resulting data frame to ex_2_2.

Expected result:

#> # First few rows of ex_2_2
#>     Ozone Solar.R Wind Temp Month Day
#> 5      NA      NA 14.3   56     5   5
#> 10     NA     194  8.6   69     5  10
#> 25     NA      66 16.6   57     5  25
#> ...
#> nrow(ex_2_2)
#> [1] 35

Difficulty: Intermediate

RYour turn

ex_2_2 <- # your code here nrow(ex_2_2)

Click to reveal solution

RSolution

ex_2_2 <- airquality[is.na(airquality$Ozone) & !is.na(airquality$Solar.R), ] nrow(ex_2_2) #> [1] 35 head(ex_2_2, 3) #> Ozone Solar.R Wind Temp Month Day #> 10 NA 194 8.6 69 5 10 #> 25 NA 66 16.6 57 5 25 #> 26 NA 266 14.9 58 5 26

Explanation: The two NA conditions are combined with &, which keeps only rows where the first is TRUE and the second is TRUE. Beginners often try airquality$Ozone == NA, which always returns NA and therefore filters nothing. The dplyr equivalent reads more naturally: filter(airquality, is.na(Ozone) & !is.na(Solar.R)). Of the 37 Ozone NAs, 35 still have a usable Solar.R reading, which is a strong basis for regression-based imputation.

Exercise 2.3: Build a missingness flag matrix

Task: Before fitting a missing-not-at-random model, a statistician wants a logical matrix the same shape as airquality where every cell is TRUE if the original cell was NA. This shadow matrix becomes a feature set in some downstream models. Produce it and save to ex_2_3.

Expected result:

#>      Ozone Solar.R  Wind  Temp Month   Day
#> [1,] FALSE   FALSE FALSE FALSE FALSE FALSE
#> [2,] FALSE   FALSE FALSE FALSE FALSE FALSE
#> [3,] FALSE   FALSE FALSE FALSE FALSE FALSE
#> [4,] FALSE   FALSE FALSE FALSE FALSE FALSE
#> [5,]  TRUE    TRUE FALSE FALSE FALSE FALSE
#> ... 148 more rows
#> dim(ex_2_3)
#> [1] 153   6
#> sum(ex_2_3)
#> [1] 44

Difficulty: Advanced

RYour turn

ex_2_3 <- # your code here dim(ex_2_3) sum(ex_2_3)

Click to reveal solution

RSolution

ex_2_3 <- is.na(airquality) dim(ex_2_3) #> [1] 153 6 sum(ex_2_3) #> [1] 44 head(ex_2_3, 5) #> Ozone Solar.R Wind Temp Month Day #> [1,] FALSE FALSE FALSE FALSE FALSE FALSE #> [2,] FALSE FALSE FALSE FALSE FALSE FALSE #> [3,] FALSE FALSE FALSE FALSE FALSE FALSE #> [4,] FALSE FALSE FALSE FALSE FALSE FALSE #> [5,] TRUE TRUE FALSE FALSE FALSE FALSE

Explanation: Applied to a data frame, is.na() returns a logical matrix preserving dimensions and column names. This shadow matrix has many uses: bind it as suffixed columns (Ozone_was_na) to enable a "missingness indicator" feature, or pass it to imputation packages that need the original mask. Note that sum(ex_2_3) recovers the headline NA count from Exercise 1.1: the two views are consistent.

Section 3. Remove missing data safely (3 problems)

Exercise 3.1: Drop every row with any NA using na.omit

Task: A junior analyst preparing data for a linear model that cannot tolerate NAs needs a row-complete version of airquality. Use na.omit() to drop any row containing at least one NA in any column, preserving column order, and save the cleaned frame to ex_3_1.

Expected result:

#> head(ex_3_1)
#>    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
#> 4     18     313 11.5   62     5   4
#> 7     23     299  8.6   65     5   7
#> 8     19      99 13.8   59     5   8
#> nrow(ex_3_1)
#> [1] 111

Difficulty: Beginner

RYour turn

ex_3_1 <- # your code here nrow(ex_3_1)

Click to reveal solution

RSolution

ex_3_1 <- na.omit(airquality) nrow(ex_3_1) #> [1] 111

Explanation: na.omit() deletes any row with at least one NA, attaches a na.action attribute listing the dropped row numbers, and otherwise leaves the frame untouched. Going from 153 to 111 rows is a 27% loss, which is steep, so this is only safe when downstream models genuinely cannot handle NAs and the missingness is plausibly random. For column-targeted dropping, prefer tidyr::drop_na(col1, col2) (Exercise 3.2). To recover the dropped row indices, inspect attr(ex_3_1, "na.action").

Exercise 3.2: Drop rows only when specific columns are missing

Task: A retailer fitting a model that needs Ozone and Wind but is fine with NAs in Solar.R wants to drop rows only where Ozone is missing, keeping every row that has an Ozone value even if Solar.R is NA. Use tidyr::drop_na() on the Ozone column and save to ex_3_2.

Expected result:

#> # A tibble: 116 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
#> ... 113 more rows
#> nrow(ex_3_2)
#> [1] 116
#> sum(is.na(ex_3_2$Solar.R))
#> [1] 5

Difficulty: Intermediate

RYour turn

ex_3_2 <- # your code here nrow(ex_3_2) sum(is.na(ex_3_2$Solar.R))

Click to reveal solution

RSolution

ex_3_2 <- airquality |> drop_na(Ozone) nrow(ex_3_2) #> [1] 116 sum(is.na(ex_3_2$Solar.R)) #> [1] 5

Explanation: drop_na(Ozone) removes only the 37 rows where Ozone is NA, leaving rows with other NAs intact. The 5 remaining Solar.R NAs prove that NAs in other columns survived. Bare drop_na() with no arguments is equivalent to na.omit() and would drop 42 rows. The named-column form is almost always what you want, since it lets you keep partial information for variables you do not need in this model.

Exercise 3.3: Keep rows only where Wind is non-missing

Task: Imagine Wind is your model's response variable, so rows with missing Wind are useless, but partial predictors are fine. Filter airquality to keep only rows where Wind is NOT NA (in this dataset all 153 rows qualify, so the result equals the input) and use dplyr::filter() with !is.na(). Save to ex_3_3 and report the row count.

Expected result:

#> # A tibble: 153 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
#> ... 151 more rows
#> nrow(ex_3_3)
#> [1] 153

Difficulty: Advanced

RYour turn

ex_3_3 <- # your code here nrow(ex_3_3)

Click to reveal solution

RSolution

ex_3_3 <- airquality |> filter(!is.na(Wind)) nrow(ex_3_3) #> [1] 153

Explanation: Even when no rows are dropped, this filter belongs in the pipeline as a defensive guard. If a future data feed introduces NAs in Wind, the filter quietly removes them rather than letting them propagate into the model fit (where they would either error or, worse, silently bias the result). The idiom filter(!is.na(col)) is preferred over filter(col != NA) because the latter is always NA and removes every row. Defensive code earns its keep on data drift.

Section 4. Simple imputation (4 problems)

Exercise 4.1: Mean-impute Ozone

Task: A first-pass workflow needs Ozone to be NA-free so a downstream linear regression runs, accepting the bias that single-value imputation introduces. Replace every NA in airquality$Ozone with the column mean (rounded to 2 decimals) using base R indexing and save the patched vector (not the whole frame) to ex_4_1.

Expected result:

#> head(ex_4_1, 10)
#>  [1] 41.00 36.00 12.00 18.00 42.13 28.00 23.00 19.00  8.00 42.13
#> any(is.na(ex_4_1))
#> [1] FALSE
#> length(ex_4_1)
#> [1] 153

Difficulty: Beginner

RYour turn

ex_4_1 <- # your code here head(ex_4_1, 10) any(is.na(ex_4_1))

Click to reveal solution

RSolution

ex_4_1 <- airquality$Ozone ex_4_1[is.na(ex_4_1)] <- round(mean(ex_4_1, na.rm = TRUE), 2) head(ex_4_1, 10) #> [1] 41.00 36.00 12.00 18.00 42.13 28.00 23.00 19.00 8.00 42.13 any(is.na(ex_4_1)) #> [1] FALSE

Explanation: Logical indexing on the left of <- overwrites only the positions that satisfy the condition. na.rm = TRUE is mandatory: without it mean() returns NA and you replace NAs with NAs. Mean imputation shrinks the variable's variance and pulls correlations toward zero, so it is acceptable as a quick fix but rarely as a final strategy. Note we returned the vector, not the frame, so the source airquality is untouched.

Exercise 4.2: Median-impute Solar.R for robustness

Task: Because solar radiation is right-skewed and has a few very high readings, the team prefers the median over the mean to avoid pulling imputed values upward. Replace the 7 NAs in airquality$Solar.R with the column median, keep the original 153-row length, and save the imputed numeric vector to ex_4_2.

Expected result:

#> sum(is.na(ex_4_2))
#> [1] 0
#> median(ex_4_2)
#> [1] 205
#> length(ex_4_2)
#> [1] 153

Difficulty: Intermediate

RYour turn

ex_4_2 <- # your code here sum(is.na(ex_4_2)) median(ex_4_2)

Click to reveal solution

RSolution

ex_4_2 <- airquality$Solar.R ex_4_2[is.na(ex_4_2)] <- median(ex_4_2, na.rm = TRUE) sum(is.na(ex_4_2)) #> [1] 0 median(ex_4_2) #> [1] 205

Explanation: The median is unaffected by extreme values, which is the right call when the distribution is skewed or has heavy tails. Note that imputing with the median 7 times will slightly bias future medians toward the original median (since you now have 7 extra copies of it), so downstream variance estimators should account for the imputation. For production, multiple imputation packages like mice produce several plausible imputations and pool the analyses to recover honest standard errors.

Exercise 4.3: Mode-impute the diagnosis column

Task: For the clinic$diagnosis character column, 3 patients have NA. Best practice is mode imputation: replace NAs with the most frequent non-NA category. Compute the mode of diagnosis and use it to fill the NAs, returning the fully populated character vector saved to ex_4_3.

Expected result:

#> ex_4_3
#>  [1] "HT" "HT" "HT" "HT" "HT" "HT" "HT" "HT" "HT" "HT"
#> sum(is.na(ex_4_3))
#> [1] 0

Difficulty: Intermediate

RYour turn

ex_4_3 <- # your code here ex_4_3 sum(is.na(ex_4_3))

Click to reveal solution

RSolution

mode_val <- names(sort(table(clinic$diagnosis), decreasing = TRUE))[1] ex_4_3 <- clinic$diagnosis ex_4_3[is.na(ex_4_3)] <- mode_val ex_4_3 #> [1] "HT" "HT" "HT" "HT" "HT" "HT" "HT" "HT" "HT" "HT" sum(is.na(ex_4_3)) #> [1] 0

Explanation: Base R has no built-in mode() for the statistical mode (the function named mode() returns storage type), so the idiom is names(sort(table(x), decreasing = TRUE))[1]. table() excludes NAs by default, which is exactly what you want here. Mode imputation underestimates the rare categories' frequencies, so for low-cardinality vars consider a "Missing" explicit level instead, especially when missingness itself carries signal.

Exercise 4.4: Forward-fill Ozone within each month

Task: A time-series convention for sensor data is to carry the last observed value forward until a new reading arrives. Use tidyr::fill() to forward-fill NAs in airquality$Ozone within each Month, so a missing day inherits the previous day's Ozone reading from the same month. Save the result to ex_4_4 as a tibble.

Expected result:

#> ex_4_4 |> filter(Month == 5) |> head(10)
#> # A tibble: 10 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    18      NA  14.3    56     5     5
#>  6    18      NA  14.9    66     5     6
#>  7    23     299   8.6    65     5     7
#>  8    19      99  13.8    59     5     8
#>  9     8      19  20.1    61     5     9
#> 10     8     194   8.6    69     5    10

Difficulty: Advanced

RYour turn

ex_4_4 <- # your code here ex_4_4 |> filter(Month == 5) |> head(10)

Click to reveal solution

RSolution

ex_4_4 <- airquality |> group_by(Month) |> fill(Ozone, .direction = "down") |> ungroup() ex_4_4 |> filter(Month == 5) |> head(10) #> # A tibble: 10 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 18 NA 14.3 56 5 5 #> 6 18 NA 14.9 66 5 6 #> 7 23 299 8.6 65 5 7

Explanation: fill(.direction = "down") carries the last non-NA value forward; "up" walks backward, and "downup" does both. Grouping by Month prevents May's last value from leaking into June. This last-observation-carried-forward (LOCF) is standard in clinical trials and sensor pipelines but biases analyses when values trend (since a long missing run inherits a stale value). For trended series, linear interpolation (zoo::na.approx) is usually fairer.

Section 5. Grouped and conditional imputation (3 problems)

Exercise 5.1: Impute Ozone with the per-month median

Task: A more honest single-value strategy is to impute Ozone NAs with the median for that observation's month, since ozone has a strong seasonal cycle and the May median (around 11) is very different from August's (around 39). Add an imputed Ozone column on top of the existing structure of airquality, computed per Month, and save to ex_5_1.

Expected result:

#> ex_5_1 |> group_by(Month) |> summarise(median_ozone = median(Ozone))
#> # A tibble: 5 x 2
#>   Month median_ozone
#>   <int>        <dbl>
#> 1     5         11
#> 2     6         20.5
#> 3     7         60
#> 4     8         39.5
#> 5     9         23
#> sum(is.na(ex_5_1$Ozone))
#> [1] 0

Difficulty: Intermediate

RYour turn

ex_5_1 <- # your code here ex_5_1 |> group_by(Month) |> summarise(median_ozone = median(Ozone)) sum(is.na(ex_5_1$Ozone))

Click to reveal solution

RSolution

ex_5_1 <- airquality |> group_by(Month) |> mutate(Ozone = ifelse(is.na(Ozone), median(Ozone, na.rm = TRUE), Ozone)) |> ungroup() ex_5_1 |> group_by(Month) |> summarise(median_ozone = median(Ozone)) #> # A tibble: 5 x 2 #> Month median_ozone #> <int> <dbl> #> 1 5 11 #> 2 6 20.5 #> 3 7 60 #> 4 8 39.5 #> 5 9 23 sum(is.na(ex_5_1$Ozone)) #> [1] 0

Explanation: group_by(Month) |> mutate(...) computes the median once per month and broadcasts it back to every row in that month. ifelse() then swaps NAs for the group median and leaves observed values untouched. Per-group imputation preserves between-group differences (here, seasonality) where global imputation would smear them. Always remember to ungroup() after group_by() inside a pipeline so downstream operations are not silently grouped.

Exercise 5.2: Conditional imputation with case_when

Task: A trial statistician wants three rules for clinic$weight_kg: if arm == "drug" and weight_kg is NA, fill with the drug-arm mean; if arm == "placebo" and NA, fill with the placebo-arm mean; otherwise keep the observed value. Implement with dplyr::case_when() and save the patched clinic tibble to ex_5_2.

Expected result:

#> ex_5_2 |> select(patient_id, arm, weight_kg)
#> # A tibble: 10 x 3
#>    patient_id arm     weight_kg
#>         <int> <chr>       <dbl>
#>  1          1 drug         72
#>  2          2 drug         80
#>  3          3 drug         73.75
#>  4          4 drug         68
#>  5          5 drug         75
#>  6          6 placebo      90
#>  7          7 placebo      85
#>  8          8 placebo      83.75
#>  9          9 placebo      78
#> 10         10 placebo      82
#> sum(is.na(ex_5_2$weight_kg))
#> [1] 0

Difficulty: Advanced

RYour turn

ex_5_2 <- # your code here ex_5_2 |> select(patient_id, arm, weight_kg) sum(is.na(ex_5_2$weight_kg))

Click to reveal solution

RSolution

ex_5_2 <- clinic |> group_by(arm) |> mutate(weight_kg = case_when( !is.na(weight_kg) ~ weight_kg, arm == "drug" ~ mean(weight_kg, na.rm = TRUE), arm == "placebo" ~ mean(weight_kg, na.rm = TRUE) )) |> ungroup() ex_5_2 |> select(patient_id, arm, weight_kg) #> # A tibble: 10 x 3 #> patient_id arm weight_kg #> <int> <chr> <dbl> #> 1 1 drug 72 #> 2 2 drug 80 #> 3 3 drug 73.75 #> 4 4 drug 68 #> 5 5 drug 75 #> 6 6 placebo 90 #> 7 7 placebo 85 #> 8 8 placebo 83.75 #> 9 9 placebo 78 #> 10 10 placebo 82 sum(is.na(ex_5_2$weight_kg)) #> [1] 0

Explanation: case_when() is evaluated top-to-bottom, so listing !is.na(weight_kg) ~ weight_kg first lets observed values short-circuit before any imputation rule fires. Because the data is grouped by arm, mean(weight_kg, na.rm = TRUE) inside mutate() returns the within-arm mean. The drug-arm mean of (72+80+68+75)/4 = 73.75 fills patient 3; the placebo-arm mean of (90+85+78+82)/4 = 83.75 fills patient 8. Without grouping, you would impute every NA with the global mean and erase the treatment effect.

Exercise 5.3: Recode sentinel zeros as NA before imputing

Task: Patient 8 in clinic has visit_score = 0, which the protocol uses as a "missed visit" sentinel rather than a real score of zero. Detect and convert that zero (and any other zeros that might creep in) to NA, then impute with the column median. Save the corrected visit_score vector to ex_5_3.

Expected result:

#> ex_5_3
#>  [1] 8.0 7.0 9.0 7.0 8.0 6.0 7.0 7.0 6.0 7.0
#> any(is.na(ex_5_3))
#> [1] FALSE

Difficulty: Advanced

RYour turn

ex_5_3 <- # your code here ex_5_3 any(is.na(ex_5_3))

Click to reveal solution

RSolution

ex_5_3 <- clinic$visit_score ex_5_3[ex_5_3 == 0] <- NA ex_5_3[is.na(ex_5_3)] <- median(ex_5_3, na.rm = TRUE) ex_5_3 #> [1] 8 7 9 7 8 6 7 7 6 7 any(is.na(ex_5_3)) #> [1] FALSE

Explanation: Sentinel detection (treating special codes like 0, -99, 999, or "N/A" as missingness) is one of the most common silent data-quality bugs. If you skip the recoding step, mean() and median() quietly use the sentinel as a real number and bias every summary. After recoding, the median of the seven observed scores (8,7,9,8,6,7,6) is 7, which fills both the protocol NA at row 4 and the sentinel at row 8. Always document sentinel codes in a data dictionary and apply this two-step recode early in the pipeline.

Section 6. End-to-end pipelines (3 problems)

Exercise 6.1: Audit, impute, and verify in one pipeline

Task: An ops engineer wants a single, idempotent pipeline that takes airquality, reports per-column NA counts, mean-imputes Ozone, median-imputes Solar.R, and returns the cleaned tibble plus a "before" summary attribute. Build the pipeline using dplyr verbs and save the final cleaned tibble to ex_6_1. The attr(ex_6_1, "na_before") should hold the pre-imputation NA counts.

Expected result:

#> attr(ex_6_1, "na_before")
#>   Ozone Solar.R    Wind    Temp   Month     Day
#>      37       7       0       0       0       0
#> colSums(is.na(ex_6_1))
#>   Ozone Solar.R    Wind    Temp   Month     Day
#>       0       0       0       0       0       0
#> nrow(ex_6_1)
#> [1] 153

Difficulty: Advanced

RYour turn

ex_6_1 <- # your code here attr(ex_6_1, "na_before") colSums(is.na(ex_6_1)) nrow(ex_6_1)

Click to reveal solution

RSolution

na_before <- colSums(is.na(airquality)) ex_6_1 <- airquality |> mutate( Ozone = ifelse(is.na(Ozone), mean(Ozone, na.rm = TRUE), Ozone), Solar.R = ifelse(is.na(Solar.R), median(Solar.R, na.rm = TRUE), Solar.R) ) attr(ex_6_1, "na_before") <- na_before attr(ex_6_1, "na_before") #> Ozone Solar.R Wind Temp Month Day #> 37 7 0 0 0 0 colSums(is.na(ex_6_1)) #> Ozone Solar.R Wind Temp Month Day #> 0 0 0 0 0 0 nrow(ex_6_1) #> [1] 153

Explanation: Attaching the pre-imputation NA counts as an attribute keeps the audit trail next to the data, so anyone inspecting ex_6_1 later can answer "how much of this came from imputation?" without re-reading the source. The pipeline is idempotent: running it twice yields the same result, since the second pass finds no NAs to impute. In production, log the na_before to a metrics store so you can alert when an upstream feed starts losing data.

Exercise 6.2: Compare drop vs mean-impute vs median-impute row counts

Task: Before committing to a strategy, the team wants a side-by-side comparison: how many rows survive (a) na.omit(), (b) mean-imputing Ozone and Solar.R, and (c) median-imputing the same? Build a 3-row tibble with columns strategy and rows_retained and save to ex_6_2.

Expected result:

#> # A tibble: 3 x 2
#>   strategy        rows_retained
#>   <chr>                   <int>
#> 1 drop_na                   111
#> 2 mean_impute               153
#> 3 median_impute             153

Difficulty: Intermediate

RYour turn

ex_6_2 <- # your code here ex_6_2

Click to reveal solution

RSolution

drop_n <- nrow(na.omit(airquality)) mean_imp <- airquality |> mutate(across(c(Ozone, Solar.R), ~ ifelse(is.na(.x), mean(.x, na.rm = TRUE), .x))) median_imp <- airquality |> mutate(across(c(Ozone, Solar.R), ~ ifelse(is.na(.x), median(.x, na.rm = TRUE), .x))) ex_6_2 <- tibble( strategy = c("drop_na", "mean_impute", "median_impute"), rows_retained = c(drop_n, nrow(mean_imp), nrow(median_imp)) ) ex_6_2 #> # A tibble: 3 x 2 #> strategy rows_retained #> <chr> <int> #> 1 drop_na 111 #> 2 mean_impute 153 #> 3 median_impute 153

Explanation: across(c(col1, col2), ~ ifelse(...)) is the modern dplyr idiom for applying the same imputation rule to several columns without duplication. The headline number (111 vs 153) frames the tradeoff: dropping costs 27% of your sample but introduces no synthetic values, while imputing retains all 153 rows but injects 44 fabricated cells. The right call depends on whether the analysis is sensitive to bias (favor drop) or to sample size (favor impute).

Exercise 6.3: Build a one-line missingness report

Task: A code reviewer asks for a tidy missingness report on airquality with columns column, n_missing, and pct_missing (to 2 decimals), sorted descending by n_missing so the worst offenders sit at the top. Produce it with dplyr and save the result to ex_6_3.

Expected result:

#> # A tibble: 6 x 3
#>   column  n_missing pct_missing
#>   <chr>       <int>       <dbl>
#> 1 Ozone          37       24.18
#> 2 Solar.R         7        4.58
#> 3 Wind            0        0
#> 4 Temp            0        0
#> 5 Month           0        0
#> 6 Day             0        0

Difficulty: Intermediate

RYour turn

ex_6_3 <- # your code here ex_6_3

Click to reveal solution

RSolution

ex_6_3 <- airquality |> summarise(across(everything(), ~ sum(is.na(.x)))) |> pivot_longer(everything(), names_to = "column", values_to = "n_missing") |> mutate(pct_missing = round(n_missing / nrow(airquality) * 100, 2)) |> arrange(desc(n_missing)) ex_6_3 #> # A tibble: 6 x 3 #> column n_missing pct_missing #> <chr> <int> <dbl> #> 1 Ozone 37 24.18 #> 2 Solar.R 7 4.58 #> 3 Wind 0 0 #> 4 Temp 0 0 #> 5 Month 0 0 #> 6 Day 0 0

Explanation: across(everything(), ~ sum(is.na(.x))) produces a wide one-row tibble of NA counts; pivot_longer() reshapes it into a tidy two-column frame. The ~ .x formula-style anonymous function keeps the code compact. This idiom generalises: swap sum(is.na(.x)) for any per-column summary (mean, n_distinct, min, etc.) to get a tidy diagnostic table you can write to disk or render in a report.

What to do next

You worked through 18 missing-data problems. Now consolidate with these companion hubs:

dplyr Exercises: general dplyr verbs including grouped mutate, the workhorse for imputation pipelines.
tidyr Exercises: more practice with drop_na(), fill(), pivot_longer(), and missingness reshaping.
Data Cleaning Exercises: broader cleaning patterns including outlier detection and type coercion alongside NA handling.
EDA Exercises: the diagnostic phase where missingness audits live in a real workflow.

Navigate

Tidyverse packages

Deep dives

Wrangling & EDA

Statistics

Machine Learning

Time Series

By Industry

Reporting & Apps

Levels

Missing Data in R Exercises: 18 Real-World NA Practice Problems

How to use this hub

Section 1. Detect and quantify missingness (3 problems)

Exercise 1.1: Compute the total NA count across airquality

Exercise 1.2: Count NAs in every column with colSums

Exercise 1.3: Report the percentage missing per column

Section 2. Locate and pattern-find NAs (3 problems)

Exercise 2.1: Find row indices that contain any NA

Exercise 2.2: Find rows where Ozone is missing but Solar.R is not

Exercise 2.3: Build a missingness flag matrix

Section 3. Remove missing data safely (3 problems)

Exercise 3.1: Drop every row with any NA using na.omit

Exercise 3.2: Drop rows only when specific columns are missing

Exercise 3.3: Keep rows only where Wind is non-missing

Section 4. Simple imputation (4 problems)

Exercise 4.1: Mean-impute Ozone

Exercise 4.2: Median-impute Solar.R for robustness

Exercise 4.3: Mode-impute the diagnosis column

Exercise 4.4: Forward-fill Ozone within each month

Section 5. Grouped and conditional imputation (3 problems)

Exercise 5.1: Impute Ozone with the per-month median

Exercise 5.2: Conditional imputation with case_when

Exercise 5.3: Recode sentinel zeros as NA before imputing

Section 6. End-to-end pipelines (3 problems)

Exercise 6.1: Audit, impute, and verify in one pipeline

Exercise 6.2: Compare drop vs mean-impute vs median-impute row counts

Exercise 6.3: Build a one-line missingness report

What to do next

Missing Data (18 problems) Mastery

Navigate

Tidyverse packages

Deep dives

Wrangling & EDA

Statistics

Machine Learning

Time Series

By Industry

Reporting & Apps

Levels

Missing Data in R Exercises: 18 Real-World NA Practice Problems

How to use this hub

Section 1. Detect and quantify missingness (3 problems)

Exercise 1.1: Compute the total NA count across airquality

Exercise 1.2: Count NAs in every column with colSums

Exercise 1.3: Report the percentage missing per column

Section 2. Locate and pattern-find NAs (3 problems)

Exercise 2.1: Find row indices that contain any NA

Exercise 2.2: Find rows where Ozone is missing but Solar.R is not

Exercise 2.3: Build a missingness flag matrix

Section 3. Remove missing data safely (3 problems)

Exercise 3.1: Drop every row with any NA using na.omit

Exercise 3.2: Drop rows only when specific columns are missing

Exercise 3.3: Keep rows only where Wind is non-missing

Section 4. Simple imputation (4 problems)

Exercise 4.1: Mean-impute Ozone

Exercise 4.2: Median-impute Solar.R for robustness

Exercise 4.3: Mode-impute the diagnosis column

Exercise 4.4: Forward-fill Ozone within each month

Section 5. Grouped and conditional imputation (3 problems)

Exercise 5.1: Impute Ozone with the per-month median

Exercise 5.2: Conditional imputation with case_when

Exercise 5.3: Recode sentinel zeros as NA before imputing

Section 6. End-to-end pipelines (3 problems)

Exercise 6.1: Audit, impute, and verify in one pipeline

Exercise 6.2: Compare drop vs mean-impute vs median-impute row counts

Exercise 6.3: Build a one-line missingness report

What to do next

Related Tutorials

Missing Data (18 problems) Mastery