dplyr Joins Exercises in R: 30 Practice Problems

Thirty practice problems on dplyr joins: left, right, inner, full, semi, anti, multi-key, inequality, and rolling. Hidden solutions.

By Selva Prabhakaran · Published May 11, 2026 · Last updated May 11, 2026

RRun this once before any exercise

library(dplyr) library(tibble) library(purrr)

RSetup tables for the exercises

customers <- tibble(id = 1:5, name = c("Alice","Bob","Carol","Dan","Eve")) orders <- tibble(order_id = 101:106, customer_id = c(1, 2, 7, 3, 9, 1), amount = c(50, 80, 30, 65, 120, 25)) products <- tibble(product_id = 1:3, name = c("X","Y","Z"))

Section 1. Basic joins (8 problems)

Exercise 1.1: left_join

Difficulty: Beginner. Augment customers with orders.

Show solution

RInteractive R

customers |> left_join(orders, by = c("id" = "customer_id"))

Exercise 1.2: right_join

Difficulty: Beginner.

Show solution

RInteractive R

customers |> right_join(orders, by = c("id" = "customer_id"))

Exercise 1.3: inner_join

Difficulty: Beginner. Only matching rows.

Show solution

RInteractive R

customers |> inner_join(orders, by = c("id" = "customer_id"))

Exercise 1.4: full_join

Difficulty: Beginner.

Show solution

RInteractive R

customers |> full_join(orders, by = c("id" = "customer_id"))

Exercise 1.5: semi_join

Difficulty: Intermediate. Customers WITH at least one order; no order columns added.

Show solution

RInteractive R

customers |> semi_join(orders, by = c("id" = "customer_id"))

Exercise 1.6: anti_join

Difficulty: Intermediate. Customers without orders.

Show solution

RInteractive R

customers |> anti_join(orders, by = c("id" = "customer_id"))

Exercise 1.7: cross_join

Difficulty: Intermediate. All combinations.

Show solution

RInteractive R

tibble(a = 1:3) |> cross_join(tibble(b = c("x","y")))

Exercise 1.8: nest_join

Difficulty: Advanced. Each customer with a list-column of their orders.

Show solution

RInteractive R

customers |> nest_join(orders, by = c("id" = "customer_id"))

Section 2. Multi-key and key tricks (6 problems)

Exercise 2.1: Multi-key vector

Difficulty: Intermediate.

Show solution

RInteractive R

sales <- tibble(region = c("US","EU"), product = c("X","X"), qty = c(100,50)) prices <- tibble(region = c("US","EU"), product = c("X","X"), price = c(10,12)) sales |> left_join(prices, by = c("region","product"))

Exercise 2.2: Different key names with c()

Difficulty: Intermediate.

Show solution

RInteractive R

customers |> left_join(orders, by = c("id" = "customer_id"))

Exercise 2.3: join_by syntax

Difficulty: Intermediate.

Show solution

RInteractive R

customers |> left_join(orders, by = join_by(id == customer_id))

Exercise 2.4: Suffix on collisions

Difficulty: Intermediate.

Show solution

RInteractive R

a <- tibble(id = 1:2, name = c("A","B")) b <- tibble(id = 1:2, name = c("X","Y")) a |> inner_join(b, by = "id", suffix = c("_a","_b"))

Exercise 2.5: relationship = "one-to-one"

Difficulty: Advanced.

Show solution

RInteractive R

a <- tibble(id = 1:3, val = 10:12) b <- tibble(id = 1:3, info = c("x","y","z")) a |> left_join(b, by = "id", relationship = "one-to-one")

Exercise 2.6: relationship = "many-to-many"

Difficulty: Advanced. Acknowledge intentional explosion.

Show solution

RInteractive R

a <- tibble(id = c(1, 1, 2), x = c("a","b","c")) b <- tibble(id = c(1, 1, 2), y = c("p","q","r")) a |> left_join(b, by = "id", relationship = "many-to-many")

Section 3. Inequality and rolling (6 problems)

Exercise 3.1: Inequality join_by

Difficulty: Advanced.

Show solution

RInteractive R

txns <- tibble(id = 1:3, amount = c(50, 200, 1500)) tiers <- tibble(min_amount = c(0, 100, 1000), tier = c("micro","small","medium")) txns |> left_join(tiers, by = join_by(closest(amount >= min_amount)))

Exercise 3.2: Range overlap (between)

Difficulty: Advanced.

Show solution

RInteractive R

events <- tibble(id = 1:2, t = c(10, 50)) windows <- tibble(start = c(0, 40), end = c(20, 60), label = c("A","B")) events |> left_join(windows, by = join_by(between(t, start, end)))

Exercise 3.3: Rolling join via closest

Difficulty: Advanced.

Show solution

RInteractive R

prices <- tibble(time = c(0, 10, 20), price = c(100, 105, 110)) queries <- tibble(time = c(5, 15, 25)) queries |> left_join(prices, by = join_by(closest(time >= time)))

Exercise 3.4: Strict inequality

Difficulty: Advanced.

Show solution

RInteractive R

a <- tibble(x = c(5, 15, 25)) b <- tibble(threshold = c(10, 20), level = c("low","mid")) a |> left_join(b, by = join_by(x > threshold))

Exercise 3.5: Time-window match

Difficulty: Advanced.

Show solution

RInteractive R

a <- tibble(ts = as.POSIXct(c("2024-01-15 10:00","2024-01-15 11:00"))) b <- tibble(start = as.POSIXct(c("2024-01-15 09:00","2024-01-15 10:30")), end = as.POSIXct(c("2024-01-15 10:30","2024-01-15 12:00")), tag = c("morn","late_morn")) a |> left_join(b, by = join_by(between(ts, start, end)))

Exercise 3.6: Multi-condition join_by

Difficulty: Advanced.

Show solution

RInteractive R

a <- tibble(region = c("US","EU"), x = c(50, 200)) b <- tibble(region = c("US","US","EU"), threshold = c(10, 100, 100), tier = c("a","b","c")) a |> left_join(b, by = join_by(region == region, closest(x >= threshold)))

Section 4. Reconciliation patterns (5 problems)

Exercise 4.1: Detect added rows

Difficulty: Intermediate.

Show solution

RInteractive R

prev <- tibble(id = 1:5) curr <- tibble(id = c(2,3,5,6)) curr |> anti_join(prev, by = "id")

Exercise 4.2: Detect removed rows

Difficulty: Intermediate.

Show solution

RInteractive R

prev <- tibble(id = 1:5) curr <- tibble(id = c(2,3,5,6)) prev |> anti_join(curr, by = "id")

Exercise 4.3: Detect changes (full diff)

Difficulty: Advanced.

Show solution

RInteractive R

prev <- tibble(id = 1:3, val = c(10, 20, 30)) curr <- tibble(id = 1:3, val = c(10, 25, 30)) prev |> inner_join(curr, by = "id", suffix = c("_p","_c")) |> filter(val_p != val_c)

Exercise 4.4: Full upsert with coalesce

Difficulty: Advanced.

Show solution

RInteractive R

prev <- tibble(id = 1:3, val = c(10, 20, 30)) curr <- tibble(id = 2:4, val = c(25, NA, 40)) prev |> full_join(curr, by = "id", suffix = c("_p","_c")) |> mutate(val = coalesce(val_c, val_p)) |> select(id, val)

Exercise 4.5: Find unmatched on either side

Difficulty: Advanced.

Show solution

RInteractive R

a <- tibble(id = 1:3) b <- tibble(id = 2:4) list(only_a = anti_join(a, b, by = "id"), only_b = anti_join(b, a, by = "id"))

Section 5. Real workflows (5 problems)

Exercise 5.1: Lookup + augment

Difficulty: Intermediate.

Show solution

RInteractive R

mt <- mtcars |> tibble::rownames_to_column("car") labels <- tibble(cyl = c(4, 6, 8), label = c("small","mid","large")) mt |> left_join(labels, by = "cyl") |> head()

Exercise 5.2: Latest record per group + join

Difficulty: Advanced.

Show solution

RInteractive R

sales <- tibble(prod = rep(c("X","Y"), each = 3), date = as.Date(rep(c("2024-01-01","2024-02-01","2024-03-01"), 2)), qty = c(10, 20, 30, 40, 50, 60)) info <- tibble(prod = c("X","Y"), category = c("hw","tool")) sales |> slice_max(date, n = 1, by = prod) |> left_join(info, by = "prod")

Exercise 5.3: Order summary per customer

Difficulty: Advanced.

Show solution

RInteractive R

customer_summary <- orders |> group_by(customer_id) |> summarise(n = n(), total = sum(amount), .groups = "drop") customers |> left_join(customer_summary, by = c("id" = "customer_id"))

Exercise 5.4: Aggregate child counts on parent

Difficulty: Advanced.

Show solution

RInteractive R

orders_count <- orders |> count(customer_id, name = "n_orders") customers |> left_join(orders_count, by = c("id" = "customer_id")) |> mutate(n_orders = coalesce(n_orders, 0L))

Exercise 5.5: Multi-table reduce join

Difficulty: Advanced.

Show solution

RInteractive R

dfs <- list(tibble(id = 1:3, a = 10:12), tibble(id = 1:3, b = 20:22), tibble(id = 1:3, c = 30:32)) purrr::reduce(dfs, inner_join, by = "id")

What to do next

dplyr-Exercises (shipped), broader dplyr drills.
dplyr-Window-Functions-Exercises (coming), lead/lag/rank focus.