r-statistics.co by Selva Prabhakaran

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R Subsetting Exercises: 10 [] vs [[]] vs $ Practice Problems

Practice R's three subsetting operators — single bracket [], double bracket [[]], and dollar sign $ — across vectors, lists, and data frames. Each exercise has an interactive solution you can run in your browser.

R's subsetting operators look similar but behave very differently. These 10 exercises force you to think carefully about which operator returns what — and why. Work through them in order: they build from simple vector indexing to tricky list and data frame scenarios.

Quick Reference

OperatorWorks onReturnsAnalogy
x[i]Vectors, lists, data framesSame type as input (subset)Pulling pages from a book — you get a smaller book
x[[i]]Lists, data framesThe element itself (extracted)Opening a box and taking the item out
x$nameLists, data framesThe element itself (like [[)Shorthand for x[["name"]]

Easy (1-4): Vector and List Basics

Exercise 1: Vector Indexing Six Ways

Given a named vector of city temperatures, extract subsets using positive indices, negative indices, logical vectors, and names.

# Exercise 1: Six ways to subset a vector temps <- c(London = 15, Paris = 18, Berlin = 14, Rome = 22, Madrid = 25, Oslo = 8) # 1. Extract the 2nd and 4th elements by position # 2. Exclude the 1st and 6th elements # 3. Extract cities warmer than 17 degrees # 4. Extract London and Madrid by name # 5. Get the first 3 elements # 6. Reverse the vector order 

  






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temps <- c(London = 15, Paris = 18, Berlin = 14, Rome = 22, Madrid = 25, Oslo = 8)

# 1. Positive integers select by position
cat("2nd and 4th:", temps[c(2, 4)], "\n")

# 2. Negative integers exclude
cat("Exclude 1st and 6th:", temps[c(-1, -6)], "\n")

# 3. Logical vector keeps TRUE positions
cat("Warmer than 17:", temps[temps > 17], "\n")

# 4. Character vector selects by name
cat("By name:", temps[c("London", "Madrid")], "\n")

# 5. Sequence for ranges
cat("First 3:", temps[1:3], "\n")

# 6. rev() or descending index
cat("Reversed:", temps[length(temps):1], "\n")

    

      
      
    

    

  





Key concept: [ with a vector always returns a vector of the same type. Positive integers select, negative integers exclude, logicals filter, and characters match names.






Exercise 2: Single Bracket vs Double Bracket on a List



A named list contains three different data types. Use [ and [[ to extract elements and observe the difference.




  

    
# Exercise 2: [ vs [[ on a list
student <- list(
  name = "Alice",
  scores = c(88, 92, 79, 95),
  passed = TRUE
)

# 1. Use [ to extract the "scores" element. What type is the result?
# 2. Use [[ to extract the "scores" element. What type is the result?
# 3. Get the 3rd score (79) in a single expression using [[
# 4. What does student[2:3] return? What about student[[2:3]]?


    

      
      
    

    

  






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student <- list(
  name = "Alice",
  scores = c(88, 92, 79, 95),
  passed = TRUE
)

# 1. Single bracket returns a LIST containing the element
result1 <- student["scores"]
cat("student['scores'] class:", class(result1), "\n")
str(result1)

# 2. Double bracket extracts the ELEMENT itself
result2 <- student[["scores"]]
cat("\nstudent[['scores']] class:", class(result2), "\n")
print(result2)

# 3. Chain [[ with [ to reach inside
cat("\n3rd score:", student[["scores"]][3], "\n")

# 4. [ can take a range; [[ cannot
cat("\nstudent[2:3]:\n")
str(student[2:3])

# student[[2:3]] would throw an error in most cases
# [[ is for extracting a SINGLE element only

    

      
      
    

    

  





Key concept: [ always returns a list (a sub-list). [[ extracts the actual element inside. Think of [ as selecting a train car (you get a smaller train) and [[ as opening the car door and pulling out the contents.






Exercise 3: The $ Operator



Use $ to access list elements and understand when it works and when it doesn't.




  

    
# Exercise 3: Using $ on lists
config <- list(
  host = "localhost",
  port = 8080,
  debug = FALSE,
  tags = c("api", "v2", "production")
)

# 1. Extract the host using $
# 2. Extract tags using $
# 3. Can you use $ with a variable? Try: key <- "port"; config$key
# 4. What does config$por return? (note: partial match)
# 5. Add a new element "timeout" with value 30 using $


    

      
      
    

    

  






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config <- list(
  host = "localhost",
  port = 8080,
  debug = FALSE,
  tags = c("api", "v2", "production")
)

# 1. $ accesses by name directly
cat("Host:", config$host, "\n")

# 2. Works with any element type
cat("Tags:", config$tags, "\n")

# 3. $ does NOT evaluate variables — it uses the literal name
key <- "port"
cat("\nconfig$key:", config$key, "\n")        # NULL — no element named "key"
cat("config[[key]]:", config[[key]], "\n")    # 8080 — [[ evaluates the variable

# 4. $ does partial matching (dangerous!)
cat("\nconfig$por:", config$por, "\n")        # Returns 8080 — matched "port"
# Use [[ for exact matching in production code

# 5. Assign with $ to add/modify elements
config$timeout <- 30
cat("\nUpdated config:\n")
str(config)

    

      
      
    

    

  





Key concept: $ is convenient but has two traps: it doesn't work with variables (use [[ instead), and it does partial name matching, which can silently return the wrong element.






Exercise 4: Data Frame Subsetting Basics



Subset a data frame using [, [[, and $ to extract rows, columns, and individual values.




  

    
# Exercise 4: Data frame subsetting
df <- data.frame(
  name = c("Alice", "Bob", "Carol", "David", "Eve"),
  age = c(25, 30, 28, 35, 22),
  score = c(88, 76, 92, 81, 95),
  stringsAsFactors = FALSE
)

# 1. Extract the "score" column as a vector (3 different ways)
# 2. Extract rows 2 through 4 (all columns)
# 3. Extract the score for Carol only
# 4. Extract rows where score > 85


    

      
      
    

    

  






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df <- data.frame(
  name = c("Alice", "Bob", "Carol", "David", "Eve"),
  age = c(25, 30, 28, 35, 22),
  score = c(88, 76, 92, 81, 95),
  stringsAsFactors = FALSE
)

# 1. Three ways to extract a column as a vector
cat("Using $:", df$score, "\n")
cat("Using [[:", df[["score"]], "\n")
cat("Using [,]:", df[, "score"], "\n")

# Note: df["score"] returns a data frame, not a vector!
cat("\ndf['score'] class:", class(df["score"]), "\n")

# 2. Row subsetting — [rows, columns]
cat("\nRows 2-4:\n")
print(df[2:4, ])

# 3. Combine row and column conditions
cat("\nCarol's score:", df[df$name == "Carol", "score"], "\n")

# 4. Logical condition on rows
cat("\nScores above 85:\n")
print(df[df$score > 85, ])

    

      
      
    

    

  





Key concept: Data frames are lists of columns. $ and [[ extract a column vector. [ with two arguments ([rows, cols]) subsets both dimensions. Leave one blank to keep all rows or all columns.






Medium (5-7): Combining Operators



Exercise 5: Nested List Extraction



Navigate a deeply nested list to extract specific values.




  

    
# Exercise 5: Nested list subsetting
company <- list(
  name = "TechCorp",
  departments = list(
    engineering = list(
      head = "Alice",
      team_size = 15,
      projects = c("API", "Frontend", "ML Pipeline")
    ),
    marketing = list(
      head = "Bob",
      team_size = 8,
      projects = c("Campaign Q1", "Brand Refresh")
    )
  ),
  founded = 2019
)

# 1. Extract the engineering department head's name
# 2. Get the 3rd project in engineering ("ML Pipeline")
# 3. Extract both department heads as a character vector
# 4. How many total projects across all departments?


    

      
      
    

    

  






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company <- list(
  name = "TechCorp",
  departments = list(
    engineering = list(
      head = "Alice",
      team_size = 15,
      projects = c("API", "Frontend", "ML Pipeline")
    ),
    marketing = list(
      head = "Bob",
      team_size = 8,
      projects = c("Campaign Q1", "Brand Refresh")
    )
  ),
  founded = 2019
)

# 1. Chain $ or [[ to navigate levels
cat("Eng head:", company$departments$engineering$head, "\n")
# Equivalent with [[:
cat("Eng head:", company[["departments"]][["engineering"]][["head"]], "\n")

# 2. Add vector indexing at the end
cat("3rd project:", company$departments$engineering$projects[3], "\n")

# 3. Use sapply to extract across list elements
heads <- sapply(company$departments, function(d) d$head)
cat("Heads:", heads, "\n")

# 4. Count total projects
total <- sum(sapply(company$departments, function(d) length(d$projects)))
cat("Total projects:", total, "\n")

    

      
      
    

    

  





Key concept: Chain $ or [[ to drill into nested lists — each level peels off one layer. Use sapply() to extract the same field from every element in a list.






Exercise 6: Logical Subsetting with Multiple Conditions



Filter data frame rows using compound logical conditions and handle edge cases.




  

    
# Exercise 6: Logical subsetting
sales <- data.frame(
  product = c("Laptop", "Mouse", "Keyboard", "Monitor", "Webcam",
              "Headset", "Tablet", "Speaker"),
  price = c(999, 25, 75, 450, 60, 120, 599, 85),
  category = c("Computer", "Accessory", "Accessory", "Computer",
               "Accessory", "Audio", "Computer", "Audio"),
  in_stock = c(TRUE, TRUE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE),
  stringsAsFactors = FALSE
)

# 1. Products that cost between 50 and 200 (inclusive)
# 2. Accessories that are in stock
# 3. Products that are NOT in the "Computer" category
# 4. In-stock items under $100 OR any computer (regardless of price/stock)
# 5. Which row numbers match category == "Audio"?


    

      
      
    

    

  






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sales <- data.frame(
  product = c("Laptop", "Mouse", "Keyboard", "Monitor", "Webcam",
              "Headset", "Tablet", "Speaker"),
  price = c(999, 25, 75, 450, 60, 120, 599, 85),
  category = c("Computer", "Accessory", "Accessory", "Computer",
               "Accessory", "Audio", "Computer", "Audio"),
  in_stock = c(TRUE, TRUE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE),
  stringsAsFactors = FALSE
)

# 1. Combine conditions with &
cat("$50-$200:\n")
print(sales[sales$price >= 50 & sales$price <= 200, ])

# 2. Two conditions: category AND stock
cat("\nIn-stock accessories:\n")
print(sales[sales$category == "Accessory" & sales$in_stock, ])

# 3. Negate with != or !
cat("\nNon-computers:\n")
print(sales[sales$category != "Computer", ])

# 4. OR condition with |
cat("\nCheap in-stock OR any computer:\n")
print(sales[(sales$in_stock & sales$price < 100) | sales$category == "Computer", ])

# 5. which() returns row numbers instead of a logical vector
cat("\nAudio row numbers:", which(sales$category == "Audio"), "\n")

    

      
      
    

    

  





Key concept: Use & (and) and | (or) to combine conditions. Wrap compound conditions in parentheses for clarity. which() converts a logical vector to integer positions.






Exercise 7: Replacing Values via Subsetting



Use subsetting on the left side of assignment to modify specific elements.




  

    
# Exercise 7: Subsetting for assignment
grades <- c(A = 95, B = 82, C = 67, D = 58, F = 41)

# 1. Boost every grade below 60 by 5 points
# 2. Cap all grades at 100 (in case any exceed it)
# 3. Replace the "F" grade with NA

inventory <- data.frame(
  item = c("Apples", "Bread", "Milk", "Eggs", "Butter"),
  qty = c(50, 0, 12, 0, 8),
  price = c(1.20, 2.50, 1.80, 3.00, 2.20),
  stringsAsFactors = FALSE
)

# 4. Set qty to 20 for all items currently at 0
# 5. Increase price by 10% for items with qty < 15


    

      
      
    

    

  






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grades <- c(A = 95, B = 82, C = 67, D = 58, F = 41)
cat("Original grades:", grades, "\n")

# 1. Subset on LEFT side of <- modifies those elements
grades[grades < 60] <- grades[grades < 60] + 5
cat("After boost:", grades, "\n")

# 2. pmin() caps at a maximum, or use conditional assignment
grades[grades > 100] <- 100
cat("After cap:", grades, "\n")

# 3. Assign NA by name
grades["F"] <- NA
cat("After NA:", grades, "\n")

# Data frame version
inventory <- data.frame(
  item = c("Apples", "Bread", "Milk", "Eggs", "Butter"),
  qty = c(50, 0, 12, 0, 8),
  price = c(1.20, 2.50, 1.80, 3.00, 2.20),
  stringsAsFactors = FALSE
)

# 4. Conditional row subsetting on left side
inventory$qty[inventory$qty == 0] <- 20
cat("\nAfter restock:\n")
print(inventory)

# 5. Modify a column conditionally
low_stock <- inventory$qty < 15
inventory$price[low_stock] <- round(inventory$price[low_stock] * 1.10, 2)
cat("\nAfter price increase:\n")
print(inventory)

    

      
      
    

    

  





Key concept: Subsetting works on both sides of <-. The pattern x[condition] <- new_value is R's idiomatic way to update specific elements without a loop.






Hard (8-10): Advanced Scenarios



Exercise 8: Matrix Subsetting



Subset matrices using row/column indices, names, and logical conditions.




  

    
# Exercise 8: Matrix subsetting
set.seed(99)
mat <- matrix(sample(1:50, 20), nrow = 4,
  dimnames = list(
    c("Q1", "Q2", "Q3", "Q4"),
    c("North", "South", "East", "West", "Central")))
print(mat)

# 1. Extract the value in row "Q2", column "East"
# 2. Extract the entire "South" column as a vector
# 3. Extract rows Q1 and Q3 for columns North and West
# 4. Find all values greater than 30 (return the values, not a logical matrix)
# 5. Which quarter had the highest Central region value?


    

      
      
    

    

  






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set.seed(99)
mat <- matrix(sample(1:50, 20), nrow = 4,
  dimnames = list(
    c("Q1", "Q2", "Q3", "Q4"),
    c("North", "South", "East", "West", "Central")))
cat("Matrix:\n"); print(mat)

# 1. [row, col] by name
cat("\nQ2 East:", mat["Q2", "East"], "\n")

# 2. Leave row blank → all rows; drop=TRUE (default) returns vector
cat("South column:", mat[, "South"], "\n")

# 3. Use vectors in both dimensions
cat("\nQ1 & Q3, North & West:\n")
print(mat[c("Q1", "Q3"), c("North", "West")])

# 4. Logical matrix subsetting returns a vector of matching values
cat("\nValues > 30:", mat[mat > 30], "\n")

# 5. which.max on a column, then get the row name
cat("Best quarter (Central):", rownames(mat)[which.max(mat[, "Central"])], "\n")

    

      
      
    

    

  





Key concept: Matrices use [row, col] subsetting. A logical matrix applied to [ returns a flat vector of matching values. Use which.max() or which.min() to find positions of extremes.






Exercise 9: Subsetting with NA Gotchas



Handle NA values that break logical subsetting — a common source of bugs.




  

    
# Exercise 9: NA in subsetting
readings <- c(23.1, NA, 19.8, NA, 25.6, 21.0, NA, 22.4)

# 1. What happens when you run: readings[readings > 20]? Why?
# 2. Extract only non-NA values above 20
# 3. Replace all NA values with the mean of non-NA values

df <- data.frame(
  city = c("A", "B", "C", "D", "E"),
  temp = c(30, NA, 25, NA, 28),
  rain = c(NA, 12, 8, NA, 5),
  stringsAsFactors = FALSE
)

# 4. Extract rows where temp is NOT NA
# 5. Extract rows where BOTH temp and rain are non-NA


    

      
      
    

    

  






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readings <- c(23.1, NA, 19.8, NA, 25.6, 21.0, NA, 22.4)

# 1. NA comparisons produce NA, which [ treats as a missing row
cat("readings[readings > 20]:", readings[readings > 20], "\n")
# Returns: 23.1 NA NA 25.6 21.0 NA 22.4
# The NAs propagate because NA > 20 is NA (not TRUE or FALSE)

# 2. Use which() to drop NAs, or combine conditions with !is.na()
cat("\nSafe filter (which):", readings[which(readings > 20)], "\n")
cat("Safe filter (& !is.na):", readings[!is.na(readings) & readings > 20], "\n")

# 3. Replace NAs
readings[is.na(readings)] <- mean(readings, na.rm = TRUE)
cat("After fill:", round(readings, 1), "\n")

# Data frame version
df <- data.frame(
  city = c("A", "B", "C", "D", "E"),
  temp = c(30, NA, 25, NA, 28),
  rain = c(NA, 12, 8, NA, 5),
  stringsAsFactors = FALSE
)

# 4. Filter non-NA rows for one column
cat("\nRows with temp:\n")
print(df[!is.na(df$temp), ])

# 5. complete.cases() checks ALL columns, or combine manually
cat("\nRows with both temp and rain:\n")
print(df[!is.na(df$temp) & !is.na(df$rain), ])

    

      
      
    

    

  





Key concept: NA propagates through comparisons — NA > 20 is NA, not FALSE. Use which() or add !is.na() to your conditions when subsetting data that might contain missing values.






Exercise 10: Putting It All Together



A real-world scenario that requires combining multiple subsetting techniques.




  

    
# Exercise 10: Real-world challenge
# A survey stores results as a nested list. Extract, clean, and summarize.

survey <- list(
  metadata = list(title = "R User Survey", year = 2026, n = 8),
  responses = list(
    list(id = 1, name = "Alice", years_r = 5, rating = 9, tools = c("ggplot2", "dplyr", "shiny")),
    list(id = 2, name = "Bob", years_r = 2, rating = 7, tools = c("ggplot2", "tidyr")),
    list(id = 3, name = "Carol", years_r = NA, rating = 8, tools = c("data.table", "ggplot2")),
    list(id = 4, name = "David", years_r = 10, rating = NA, tools = c("base", "lattice")),
    list(id = 5, name = "Eve", years_r = 3, rating = 9, tools = c("dplyr", "purrr", "ggplot2", "shiny")),
    list(id = 6, name = "Frank", years_r = 1, rating = 6, tools = c("ggplot2")),
    list(id = 7, name = "Grace", years_r = 7, rating = 10, tools = c("ggplot2", "data.table", "shiny")),
    list(id = 8, name = "Hank", years_r = 4, rating = 8, tools = c("dplyr", "stringr"))
  )
)

# 1. Extract all respondent names as a character vector
# 2. Build a data frame with columns: name, years_r, rating
# 3. Find the average rating (ignoring NAs)
# 4. Which respondent uses the most tools?
# 5. How many respondents use "ggplot2"?
# 6. Extract names of respondents with 5+ years experience AND rating >= 9


    

      
      
    

    

  






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survey <- list(
  metadata = list(title = "R User Survey", year = 2026, n = 8),
  responses = list(
    list(id = 1, name = "Alice", years_r = 5, rating = 9, tools = c("ggplot2", "dplyr", "shiny")),
    list(id = 2, name = "Bob", years_r = 2, rating = 7, tools = c("ggplot2", "tidyr")),
    list(id = 3, name = "Carol", years_r = NA, rating = 8, tools = c("data.table", "ggplot2")),
    list(id = 4, name = "David", years_r = 10, rating = NA, tools = c("base", "lattice")),
    list(id = 5, name = "Eve", years_r = 3, rating = 9, tools = c("dplyr", "purrr", "ggplot2", "shiny")),
    list(id = 6, name = "Frank", years_r = 1, rating = 6, tools = c("ggplot2")),
    list(id = 7, name = "Grace", years_r = 7, rating = 10, tools = c("ggplot2", "data.table", "shiny")),
    list(id = 8, name = "Hank", years_r = 4, rating = 8, tools = c("dplyr", "stringr"))
  )
)

resp <- survey$responses

# 1. sapply extracts one field from each list element
names_vec <- sapply(resp, function(r) r$name)
cat("Names:", names_vec, "\n")

# 2. Build data frame from list
df <- data.frame(
  name = sapply(resp, `[[`, "name"),
  years_r = sapply(resp, `[[`, "years_r"),
  rating = sapply(resp, `[[`, "rating"),
  stringsAsFactors = FALSE
)
cat("\nSurvey data frame:\n"); print(df)

# 3. Mean rating, handling NAs
cat("\nAvg rating:", round(mean(df$rating, na.rm = TRUE), 1), "\n")

# 4. Most tools — get lengths, then find the max
tool_counts <- sapply(resp, function(r) length(r$tools))
most_tools_idx <- which.max(tool_counts)
cat("Most tools:", resp[[most_tools_idx]]$name,
    "(", tool_counts[most_tools_idx], "tools )\n")

# 5. Count ggplot2 users
uses_ggplot <- sapply(resp, function(r) "ggplot2" %in% r$tools)
cat("ggplot2 users:", sum(uses_ggplot), "\n")

# 6. Combine conditions on the data frame (watch for NAs!)
experienced <- which(!is.na(df$years_r) & df$years_r >= 5 &
                     !is.na(df$rating) & df$rating >= 9)
cat("5+ yrs & rating >= 9:", df$name[experienced], "\n")

    

      
      
    

    

  





Key concept: Real data often lives in nested lists. The pattern sapply(list, function(x) x$field) extracts a field from every element. Always guard against NA when combining conditions.






Summary






ConceptWhat to Remember




x[i]Returns same type as input. Use for subsetting vectors, lists, data frames


x[[i]]Extracts a single element. Works on lists and data frames only


x$nameShorthand for x[["name"]]. Partial matches — be careful


Logical subsettingx[condition] filters elements where condition is TRUE


NA propagationNA > 5 is NA. Use which() or !is.na() for safe filtering


Assignment subsettingx[condition] <- value modifies elements in place


Nested extractionChain [[ or $ to drill into nested lists


Matrix subsettingmat[rows, cols] — leave blank for "all"






FAQ



What is the difference between [ and [[ in R?



[ returns a subset of the same type — subsetting a list with [ gives you a smaller list. [[ extracts a single element from its container — using [[ on a list gives you the actual object stored inside, not a list wrapping it.



When should I use $ vs [[?



Use $ for interactive work when you know the exact name. Use [[ in functions and scripts because it works with variables (key <- "col"; df[[key]]) and does exact matching. The $ operator does partial matching, which can cause subtle bugs.



Why does logical subsetting return NA values?



When a logical index contains NA, R doesn't know whether to include or exclude that position, so it returns NA. This happens because comparisons with NA produce NA, not FALSE. Wrap your condition in which() or add !is.na() to avoid this.



Can I use negative indices with character subsetting?



No. Negative indexing only works with numeric indices (x[-1] removes the first element). To exclude by name, use x[!names(x) %in% c("a", "b")] or the setdiff() function.



How do I subset a data frame by both rows and columns?



Use df[rows, cols] where rows and cols can be integers, names, or logical vectors. For example, df[df$age > 30, c("name", "score")] gets the name and score of people over 30.



What's Next?



    

  • R Vectors — the parent tutorial covering everything about vectors and subsetting
    • 

  • R Lists and Data Frames — deep dive into list and data frame structures
    • 

  • R Control Flow — if/else, for loops, and when to use vectorized subsetting instead
    • 




        


        
      


      

        

        
© 2016- Selva Prabhakaran.
          This work is licensed under the Creative Commons License.