DANA 4840 — Worksheet Ch.1: Simple Matching, Jaccard, shoppers

Study sheet for DANA 4840 — Chapter 1: similarity for binary-coded shoppers (Simple Matching, optional daisy() check), plus the lecture block on similarity vs distance and Gower for mixed data.

DANA 4840 — Classification II — Cluster Analysis — Part 2

Use cases of cluster analysis

Cluster analysis is used in many settings, for example:

• City planners segment neighborhoods (for example property-tax bands or service planning).
• A large firm may group thousands of employees to target training or promotion decisions.

Similarity vs dissimilarity measurements

• Similarity increases when two objects are more alike (e.g. Jaccard, Dice, Simple Matching on a 0–1 scale).
• Dissimilarity / distance decreases when two objects are more alike (e.g. Euclidean or Manhattan distance).

Methods to calculate similarity and dissimilarity measurements

(a) Binary data (0/1 coding) — usually similarity measures:

1. Simple Matching (SMC)
2. Jaccard
3. Dice

(b) Quantitative data — usually dissimilarities / distances:

1. Euclidean distance
2. Manhattan distance
3. Correlation-based distances

(c) Mixed data (quantitative and categorical together): Gower distance. In R, use cluster::daisy() with explicit variable types.

Turning similarity into dissimilarity (and why)

Many clustering algorithms expect a dissimilarity matrix or work directly with distances. If a similarity s is scaled between 0 and 1 with 1 = identical, a common transform is:

d = 1 − s

so d = 0 for identical objects and larger d means more different. Always check the scale of s before applying a transform.

Simple Matching, Jaccard, and Dice (definitions)

For two objects i and j and p binary attributes, count four kinds of position-wise pairs:

and p = a + b + c + d.

Simple Matching (SMC) — similarity from all agreements (both 1–1 and 0–0):

Jaccard — ignores double zeros (only joint “presence” counts as a strong agreement):

Dice — another presence-based similarity; differs from Jaccard in how double zeros are weighted:

For the shoppers worksheet we use SMC with p = 7 bits (ethnicity encoded with two dummies).

Dataset (shoppers)

Binary coding (for parts (a) and (b))

Six survey items become seven binary columns: ethnicity with three categories needs two dummy bits (Caucasian reference (0,0); Chinese (1,0); Indian (0,1)).

Binary matrix 5 × 7 (rows = shoppers 1…5):

     [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]    0    0    0    0    1    1    1
[2,]    1    1    1    0    0    1    0
[3,]    0    1    0    1    1    1    0
[4,]    0    0    1    0    0    1    1
[5,]    1    1    1    0    0    0    0

(a)

Worksheet Ch 1 — (a)

By introducing appropriate binary variables, calculate Simple Matching similarity coefficient for all pairs of shoppers. Present your final answer in a 5 × 5 matrix.

(You will need extra writing space to complete your work as this part requires lots of working.)

Worked solution (sketch): For binary rows i and j, Simple Matching is (matches) / p with p = 7 bits. Diagonal entries are 1.

Each cell shows the exact fraction k/7. Hover a cell to see the decimal in the browser tooltip (title).

Numeric matrix (4 decimals), equivalent:

        S1     S2     S3     S4     S5
S1  1.0000 0.2857 0.5714 0.7143 0.1429
S2  0.2857 1.0000 0.4286 0.5714 0.8571
S3  0.5714 0.4286 1.0000 0.2857 0.2857
S4  0.7143 0.5714 0.2857 1.0000 0.4286
S5  0.1429 0.8571 0.2857 0.4286 1.0000

(b)

Worksheet Ch 1 — (b)

Use R’s daisy() to validate that your answer in part (a) is correct.

Worked notes: With symmetric binary columns, Gower dissimilarity equals the fraction of positions where the two rows disagree. Therefore:

Declare all seven columns as symm in daisy().

Note: outer() + Vectorize() — odd-looking, but not “matrix subtraction”

The optional snippet

outer(1:5, 1:5, Vectorize(function(i, j) mean(shop_bin[i, ] == shop_bin[j, ])))

is not mysterious matrix algebra: it visits every pair of row indices (i, j) and stores the fraction of matching columns (that is the SMC). outer(X, Y, FUN) builds the grid of combinations; Vectorize() wraps f(i, j) so outer can call it scalar-by-scalar and return the 5×5 matrix. A double for (i) for (j) loop would give the same numbers.

Note: max(abs(S_daisy - S_manual)) — what it reports

S_daisy and S_manual are both 5×5. S_daisy - S_manual is element-wise subtraction (not matrix multiplication). abs(...) takes absolute differences and max(...) is the largest of the 25 cells.

You will rarely see exactly 0: R uses floating-point arithmetic, and different code paths (mean(…==…) vs daisy plus subtraction) can leave differences around 10⁻¹⁶ (e.g. 1.110223e-16). That is machine noise, not a mistake in the worksheet. For equality testing, use isTRUE(all.equal(S_daisy, S_manual)), which should return TRUE.

# =========================
# Worksheet Ch 1 — (b)
# =========================
library(cluster)

shop_bin <- matrix(
  c(
    0, 0, 0, 0, 1, 1, 1,
    1, 1, 1, 0, 0, 1, 0,
    0, 1, 0, 1, 1, 1, 0,
    0, 0, 1, 0, 0, 1, 1,
    1, 1, 1, 0, 0, 0, 0
  ),
  nrow = 5,
  byrow = TRUE
)

colnames(shop_bin) <- c(
  "tall", "heavy", "eth_chinese", "eth_indian",
  "purchase", "alone", "fashion"
)
rownames(shop_bin) <- paste0("Shopper", 1:5)

p <- ncol(shop_bin)

D <- as.matrix(daisy(shop_bin, type = list(symm = 1:p)))

S_daisy <- 1 - D

round(S_daisy, 4)

# SMC “a mano”: misma idea que un doble bucle for (i) for (j).
# outer() + Vectorize(): ver nota arriba en el post.
S_manual <- outer(
  1:5,
  1:5,
  Vectorize(function(i, j) mean(shop_bin[i, ] == shop_bin[j, ]))
)
round(S_manual, 4)

# Matriz de errores |S_daisy - S_manual| celda a celda; max = peor discrepancia.
# Suele ser ~1e-16 (no 0): coma flotante, no un error del worksheet.
print(max(abs(S_daisy - S_manual)))
# Comparación “estadística” recomendada en R (tolerancia por .Machine$double.eps^0.5):
isTRUE(all.equal(S_daisy, S_manual))

        Shopper1 Shopper2 Shopper3 Shopper4 Shopper5
Shopper1   1.0000   0.2857   0.5714   0.7143   0.1429
Shopper2   0.2857   1.0000   0.4286   0.5714   0.8571
Shopper3   0.5714   0.4286   1.0000   0.2857   0.2857
Shopper4   0.7143   0.5714   0.2857   1.0000   0.4286
Shopper5   0.1429   0.8571   0.2857   0.4286   1.0000

        Shopper1 Shopper2 Shopper3 Shopper4 Shopper5
Shopper1   1.0000   0.2857   0.5714   0.7143   0.1429
Shopper2   0.2857   1.0000   0.4286   0.5714   0.8571
Shopper3   0.5714   0.4286   1.0000   0.2857   0.2857
Shopper4   0.7143   0.5714   0.2857   1.0000   0.4286
Shopper5   0.1429   0.8571   0.2857   0.4286   1.0000

[1] 1.110223e-16
[1] TRUE

(c)

Worksheet Ch 1 — (c)

Which two shoppers are the most similar based on your answer in part (a)?

Answer: Shoppers 2 and 5 — similarity 6/7 ≈ 0.857 (they match on 6 of 7 attributes; they differ only on shop alone).

(d)

Worksheet Ch 1 — (d)

Which two shoppers are the least similar?

Answer: Shoppers 1 and 5 — similarity 1/7 ≈ 0.143 (only one matching bit out of seven).

(e)

Worksheet Ch 1 — (e)

If you were to divide these 5 shoppers into two relatively homogenous subgroups (clusters) based on the similarity numbers calculated in part (a), how will you form the two subgroups?

Answer (one reasonable split):

• Cluster A — Shoppers 2 and 5: strongest pair (6/7); both Tall, Heavy, Chinese, no purchase, not fashion conscious (they differ only on shopping alone).
• Cluster B — Shoppers 1, 3, and 4: all Short; internal similarity is weaker (e.g. 3 vs 4 is only 2/7), but the split contrasts “all Short” profiles against the “Tall + Heavy + Chinese” block.

(Other partitions are defensible if the course prescribes a different clustering rule.)

(f)

Worksheet Ch 1 — (f)

Would Simple Matching similarity coefficient or Jaccard similarity coefficient be more suitable for this dataset? Explain.

Answer: Simple Matching fits better here: every 0 and 1 is meaningful, and 0–0 agreements should count.

Jaccard for binary data often drops double zeros (it emphasizes joint “presence”). That makes sense when 0 is not comparable to 1. In this worksheet both levels are symmetric categories, so SMC matches the question better.