# Quick Start

#### Gilles Colling

#### 2026-03-03

## 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
    )
    #> Auto-selected scaling method: standardize

    # View matched pairs
    head(result$pairs)
    #> # A tibble: 6 × 5
    #>   left_id right_id distance .age_diff .income_diff
    #>   <chr>   <chr>       <dbl>     <dbl>        <dbl>
    #> 1 1       46          0.779    -4.23        9144.
    #> 2 2       11          0.257    -0.398       3441.
    #> 3 3       52          0.594     1.36        7840.
    #> 4 4       66          0.809    -7.87       -5112.
    #> 5 5       65          0.530     1.30        6977.
    #> 6 6       36          0.130    -1.43          62.2

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)
    #> Matching Result Summary
    #> =======================
    #>
    #> Method: lap
    #> Pairs matched: 50
    #> Unmatched: 0 left, 30 right
    #>
    #> Distance Statistics:
    #>   Total: 19.6423
    #>   Mean: 0.3928
    #>   Min: 0.0615
    #>   Q1: 0.1845
    #>   Median: 0.2967
    #>   Q3: 0.5039
    #>   Max: 1.2184
    #>   SD: 0.2863
    #>
    #> Distance Percentiles:
    #>   5%: 0.0993
    #>   10%: 0.1234
    #>   25%: 0.1845
    #>   50%: 0.2967
    #>   75%: 0.5039
    #>   90%: 0.7819
    #>   95%: 1.0396

    # Or access specific info
    result$info$n_matched
    #> [1] 50

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
    )
    #> Auto-selected scaling method: standardize

    # Compare mean distances
    cat("Unscaled mean distance:", round(mean(result_unscaled$pairs$distance), 1), "\n")
    #> Unscaled mean distance: 2769.1
    cat("Scaled mean distance:", round(mean(result_scaled$pairs$distance), 3), "\n")
    #> Scaled mean distance: 0.393

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")
    #> BEFORE matching:
    cat("  Age difference:", round(mean(treatment$age) - mean(control$age), 1), "years\n")
    #>   Age difference: -3.4 years
    cat("  Income difference: $", round(mean(treatment$income) - mean(control$income), 0), "\n\n")
    #>   Income difference: $ 9266

    cat("AFTER matching:\n")
    #> AFTER matching:
    cat("  Age difference:", round(mean(matched_treatment$age) - mean(matched_control$age), 1), "years\n")
    #>   Age difference: -1.3 years
    cat("  Income difference: $", round(mean(matched_treatment$income) - mean(matched_control$income), 0), "\n")
    #>   Income difference: $ 2667

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
    )
    #> Auto-selected scaling method: standardize

    cat("Matched", result_greedy$info$n_matched, "pairs\n")
    #> Matched 2000 pairs
    cat("Mean distance:", round(mean(result_greedy$pairs$distance), 3), "\n")
    #> Mean distance: 0.201

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
    )
    #> Auto-selected scaling method: standardize

    # 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
    )
    #> Auto-selected scaling method: standardize

    cat("Without caliper:", result_loose$info$n_matched, "pairs\n")
    #> Without caliper: 50 pairs
    cat("With caliper:", result_strict$info$n_matched, "pairs\n")
    #> With caliper: 40 pairs

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
    )
    #> Auto-selected scaling method: standardize

    # Verify: matches stay within their block
    result_blocked$pairs |> count(block_id)
    #> # A tibble: 3 × 2
    #>   block_id       n
    #>   <chr>      <int>
    #> 1 Hospital A    20
    #> 2 Hospital B    20
    #> 3 Hospital C    20

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
    )
    #> Auto-selected scaling method: standardize

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

    # 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")
    #>
    #> Balance check (difference in means):
    cat("  Age:", round(mean(treated_matched$age) - mean(control_matched$age), 2), "\n")
    #>   Age: NA
    cat("  BMI:", round(mean(treated_matched$bmi) - mean(control_matched$bmi), 2), "\n")
    #>   BMI: NA
    cat("  Smoker %:", round(100*(mean(treated_matched$smoker) - mean(control_matched$smoker)), 1), "\n")
    #>   Smoker %: NA

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

    result <- lap_solve(cost)
    print(result)
    #> Assignment Result
    #> =================
    #>
    #> # A tibble: 3 × 3
    #>   source target  cost
    #>    <int>  <int> <dbl>
    #> 1      1      2     2
    #> 2      2      1     3
    #> 3      3      3     4
    #>
    #> Total cost: 9
    #> Method: bruteforce

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)
    #> Assignment Result
    #> =================
    #>
    #> # A tibble: 3 × 3
    #>   source target  cost
    #>    <int>  <int> <dbl>
    #> 1      1      1     4
    #> 2      2      2     3
    #> 3      3      3     4
    #>
    #> Total cost: 11
    #> Method: bruteforce

### 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)
    #> Assignment Result
    #> =================
    #>
    #> # A tibble: 3 × 3
    #>   source target  cost
    #>    <int>  <int> <dbl>
    #> 1      1      1     8
    #> 2      2      2     7
    #> 3      3      3     9
    #>
    #> Total cost: 24
    #> Method: bruteforce

### 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)
    #> # A tibble: 9 × 4
    #>   day   source target  cost
    #>   <chr>  <int>  <int> <dbl>
    #> 1 Mon        1      2     2
    #> 2 Mon        2      1     3
    #> 3 Mon        3      3     4
    #> 4 Tue        1      2     3
    #> 5 Tue        2      1     2
    #> 6 Tue        3      3     3
    #> 7 Wed        1      1     3
    #> 8 Wed        2      2     2
    #> 9 Wed        3      3     4

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)
    #> K-Best Assignment Results
    #> =========================
    #>
    #> Number of solutions: 3
    #>
    #> Solution costs:
    #>   Rank 1: 5.0000
    #>   Rank 2: 7.0000
    #>   Rank 3: 7.0000
    #>
    #> Assignments:
    #> # A tibble: 9 × 6
    #>    rank solution_id source target  cost total_cost
    #>   <int>       <int>  <int>  <int> <dbl>      <dbl>
    #> 1     1           1      1      1     1          5
    #> 2     1           1      2      2     3          5
    #> 3     1           1      3      3     1          5
    #> 4     2           2      1      2     2          7
    #> 5     2           2      2      1     4          7
    #> 6     2           2      3      3     1          7
    #> 7     3           3      1      1     1          7
    #> 8     3           3      2      3     2          7
    #> 9     3           3      3      2     4          7

## See Also

- `?match_couples` - Optimal matching function reference

- `?greedy_couples` - Fast approximate matching

- `?balance_diagnostics` - Formal balance assessment

- `?lap_solve` - Direct assignment problem solving