Quick Start

In couplr: Optimal Pairing and Matching via Linear Assignment

knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>",
  fig.width = 7,
  fig.height = 5
)
library(couplr)
library(dplyr)

What couplr Does

couplr creates matched pairs from two groups of observations. Given a "left" group (e.g., treatment) and a "right" group (e.g., control), it finds the best one-to-one pairing based on similarity across variables you specify.

Common use cases:

• Matching treated patients to similar controls in observational studies
• Pairing survey respondents for comparison
• Creating balanced samples for causal inference

Documentation Roadmap

| Vignette | Focus | Audience | |----------|-------|----------| | Quick Start (this) | Basic matching with match_couples() | Everyone | | Matching Workflows | Full pipeline: preprocessing, blocking, diagnostics | Researchers | | Algorithms | Mathematical foundations, solver selection | Technical users | | Comparison | vs MatchIt, optmatch, designmatch | Package evaluators |

Start here, then proceed to whichever vignette matches your use case.

---

Your First Match

The simplest workflow uses match_couples():

library(couplr)
library(dplyr)

# Create example data: treatment and control groups
set.seed(123)
treatment <- tibble(
  id = 1:50,
  age = rnorm(50, mean = 45, sd = 10),
  income = rnorm(50, mean = 55000, sd = 12000)
)

control <- tibble(
  id = 1:80,
  age = rnorm(80, mean = 50, sd = 12),
  income = rnorm(80, mean = 48000, sd = 15000)
)

# Match on age and income
result <- match_couples(
  left = treatment,
  right = control,
  vars = c("age", "income"),
  auto_scale = TRUE
)

# View matched pairs
head(result$pairs)

What happened:

1. couplr calculated how similar each treatment unit is to each control unit
2. It found the optimal one-to-one pairing that minimizes total distance
3. Each treatment unit gets matched to exactly one control unit

Understanding the Output

# Quick overview with summary()
summary(result)

# Or access specific info
result$info$n_matched

The result$pairs table contains:

• left_id: Row number from the treatment group
• right_id: Row number from the control group
• distance: How different the matched units are (lower = more similar)

---

Why Scaling Matters

Without scaling, variables with larger values dominate the matching. Income (measured in thousands) would overwhelm age (measured in decades):

# BAD: Without scaling, income dominates
result_unscaled <- match_couples(
  treatment, control,
  vars = c("age", "income"),
  auto_scale = FALSE
)

# GOOD: With scaling, both variables contribute equally
result_scaled <- match_couples(
  treatment, control,
  vars = c("age", "income"),
  auto_scale = TRUE
)

# Compare mean distances
cat("Unscaled mean distance:", round(mean(result_unscaled$pairs$distance), 1), "\n")
cat("Scaled mean distance:", round(mean(result_scaled$pairs$distance), 3), "\n")

Rule of thumb: Always use auto_scale = TRUE unless you have a specific reason not to.

---

Checking Match Quality

After matching, verify that treatment and control groups are now balanced:

# Get the matched observations
matched_treatment <- treatment[result$pairs$left_id, ]
matched_control <- control[result$pairs$right_id, ]

# Compare means before and after matching
cat("BEFORE matching:\n")
cat("  Age difference:", round(mean(treatment$age) - mean(control$age), 1), "years\n")
cat("  Income difference: $", round(mean(treatment$income) - mean(control$income), 0), "\n\n")

cat("AFTER matching:\n")
cat("  Age difference:", round(mean(matched_treatment$age) - mean(matched_control$age), 1), "years\n")
cat("  Income difference: $", round(mean(matched_treatment$income) - mean(matched_control$income), 0), "\n")

For formal balance assessment, use balance_diagnostics() (covered in Matching Workflows).

Visualizing Match Quality

Use plot() to see the distribution of match distances:

plot(result)

The histogram shows how similar matched pairs are. A distribution concentrated near zero indicates good matches.

---

Large Datasets: Use Greedy Matching

For datasets larger than a few thousand observations, optimal matching becomes slow. Use greedy_couples() instead; it's 10-100x faster with nearly identical results:

# Create larger datasets
set.seed(456)
large_treatment <- tibble(
  id = 1:2000,
  age = rnorm(2000, 45, 10),
  income = rnorm(2000, 55000, 12000)
)

large_control <- tibble(
  id = 1:3000,
  age = rnorm(3000, 50, 12),
  income = rnorm(3000, 48000, 15000)
)

# Fast greedy matching
result_greedy <- greedy_couples(
  large_treatment, large_control,
  vars = c("age", "income"),
  auto_scale = TRUE,
  strategy = "row_best"  # fastest strategy
)

cat("Matched", result_greedy$info$n_matched, "pairs\n")
cat("Mean distance:", round(mean(result_greedy$pairs$distance), 3), "\n")

When to use which:

| Dataset size | Recommended function | |--------------|---------------------| | < 1,000 per group | match_couples() | | 1,000 - 5,000 | Either works; greedy is faster | | > 5,000 | greedy_couples() |

---

Setting a Maximum Distance (Caliper)

Sometimes you want to reject poor matches rather than force bad pairings. Use max_distance to set a caliper:

# Allow any match
result_loose <- match_couples(
  treatment, control,
  vars = c("age", "income"),
  auto_scale = TRUE
)

# Only allow close matches
result_strict <- match_couples(
  treatment, control,
  vars = c("age", "income"),
  auto_scale = TRUE,
  max_distance = 0.5  # reject pairs more different than this
)

cat("Without caliper:", result_loose$info$n_matched, "pairs\n")
cat("With caliper:", result_strict$info$n_matched, "pairs\n")

Stricter calipers mean fewer but better matches.

---

Matching Within Groups (Blocking)

When you have natural groups in your data (e.g., hospitals, regions, study sites), you can match within each group separately. This ensures exact balance on the grouping variable.

First, create blocks with matchmaker(), then pass the result to match_couples():

# Data from multiple hospital sites
set.seed(321)
treated <- tibble(
  id = 1:60,
  site = rep(c("Hospital A", "Hospital B", "Hospital C"), each = 20),
  age = rnorm(60, 55, 10),
  severity = rnorm(60, 5, 2)
)

controls <- tibble(
  id = 1:90,
  site = rep(c("Hospital A", "Hospital B", "Hospital C"), each = 30),
  age = rnorm(90, 52, 12),
  severity = rnorm(90, 4.5, 2.5)
)

# Step 1: Create blocks by hospital site
blocks <- matchmaker(
  left = treated,
  right = controls,
  block_type = "group",
  block_by = "site"
)

# Step 2: Match within each block
result_blocked <- match_couples(
  left = blocks$left,
  right = blocks$right,
  vars = c("age", "severity"),
  block_id = "block_id",
  auto_scale = TRUE
)

# Verify: matches stay within their block
result_blocked$pairs |> count(block_id)

Blocking guarantees that Hospital A patients are only matched to Hospital A controls, etc.

---

Complete Example

Here's a realistic workflow from start to finish:

# 1. Prepare your data
set.seed(789)
patients_treated <- tibble(
  patient_id = paste0("T", 1:100),
  age = rnorm(100, 62, 8),
  bmi = rnorm(100, 28, 4),
  smoker = sample(0:1, 100, replace = TRUE, prob = c(0.6, 0.4))
)

patients_control <- tibble(
  patient_id = paste0("C", 1:200),
  age = rnorm(200, 58, 10),
  bmi = rnorm(200, 26, 5),
  smoker = sample(0:1, 200, replace = TRUE, prob = c(0.7, 0.3))
)

# 2. Match on clinical variables
matched <- match_couples(
  left = patients_treated,
  right = patients_control,
  vars = c("age", "bmi", "smoker"),
  auto_scale = TRUE
)

# 3. Check how many matched
cat("Treated patients:", nrow(patients_treated), "\n")
cat("Successfully matched:", matched$info$n_matched, "\n")
cat("Match rate:", round(100 * matched$info$n_matched / nrow(patients_treated), 1), "%\n")

# 4. Extract matched samples for analysis
treated_matched <- patients_treated[matched$pairs$left_id, ]
control_matched <- patients_control[matched$pairs$right_id, ]

# 5. Verify balance
cat("\nBalance check (difference in means):\n")
cat("  Age:", round(mean(treated_matched$age) - mean(control_matched$age), 2), "\n")
cat("  BMI:", round(mean(treated_matched$bmi) - mean(control_matched$bmi), 2), "\n")
cat("  Smoker %:", round(100*(mean(treated_matched$smoker) - mean(control_matched$smoker)), 1), "\n")

---

Next Steps

You now know the basics of matching with couplr. Here's where to go next:

For production research workflows:

• Matching Workflows covers preprocessing, blocking, formal balance diagnostics, and publication-ready output

For understanding algorithm choices:

• Algorithms explains when different solvers are faster or more appropriate

For comparing with other packages:

• Comparison shows how couplr differs from MatchIt, optmatch, and designmatch

---

Additional: Direct Assignment Problem Solving

If you need to solve assignment problems directly (not matching workflows), couplr also provides lower-level functions.

lap_solve(): Matrix-Based Assignment

Given a cost matrix where entry (i,j) is the cost of assigning row i to column j:

# Cost matrix: 3 workers × 3 tasks
cost <- matrix(c(
  4, 2, 5,
  3, 3, 6,
  7, 5, 4
), nrow = 3, byrow = TRUE)

result <- lap_solve(cost)
print(result)

Row 1 is assigned to column 2 (cost 2), row 2 to column 1 (cost 3), row 3 to column 3 (cost 4). Total cost: 9.

Forbidden Assignments

Use NA or Inf for impossible assignments:

cost_forbidden <- matrix(c(
  4, 2, NA,   # Row 1 cannot go to column 3
  Inf, 3, 6,  # Row 2 cannot go to column 1
  7, 5, 4
), nrow = 3, byrow = TRUE)

lap_solve(cost_forbidden)

Maximization

For preference or profit maximization:

preferences <- matrix(c(
  8, 5, 3,
  4, 7, 6,
  2, 4, 9
), nrow = 3, byrow = TRUE)

lap_solve(preferences, maximize = TRUE)

Grouped Data

Solve multiple assignment problems at once using grouped data frames:

# Weekly nurse-shift scheduling: solve each day separately
schedule <- tibble(
  day = rep(c("Mon", "Tue", "Wed"), each = 9),
  nurse = rep(rep(1:3, each = 3), 3),
  shift = rep(1:3, 9),
  cost = c(4,2,5, 3,3,6, 7,5,4,   # Monday costs
           5,3,4, 2,4,5, 6,4,3,   # Tuesday costs
           3,4,5, 4,2,6, 5,5,4)   # Wednesday costs
)

# Solve all three days at once
schedule |>
  group_by(day) |>
  lap_solve(nurse, shift, cost)

This solves each day's assignment problem independently and returns all results in one tidy table.

K-Best Solutions

Find multiple near-optimal solutions:

cost <- matrix(c(1, 2, 3, 4, 3, 2, 5, 4, 1), nrow = 3, byrow = TRUE)

kbest <- lap_solve_kbest(cost, k = 3)
print(kbest)

---

See Also

• ?match_couples - Optimal matching function reference
• ?greedy_couples - Fast approximate matching
• ?balance_diagnostics - Formal balance assessment
• ?lap_solve - Direct assignment problem solving

couplr documentation built on Jan. 20, 2026, 5:07 p.m.