Activity 4.2 - Kmeans, PAM, and DBSCAN clustering

SUBMISSION INSTRUCTIONS

1. Render to html
2. Publish your html to RPubs
3. Submit a link to your published solutions

Loading required packages:

library(cluster)
library(dbscan)
library(factoextra)
library(tidyverse)
library(patchwork)
library(ggrepel)

Question 1

Reconsider the three data sets below. We will now compare kmeans, PAM, and DBSCAN to cluster these data sets.

three_spheres <- read.csv('Data/cluster_data1.csv')
ring_moon_sphere <- read.csv('Data/cluster_data2.csv')
two_spirals_sphere <- read.csv('Data/cluster_data3.csv')

A)

With kmeans and PAM, we can specify that we want 3 clusters. But recall with DBSCAN we select minPts and eps, and the number of clusters is determined accordingly. Use k-nearest-neighbor distance plots to determine candidate epsilon values for each data set if minPts = 4. Add horizontal line(s) to each plot indicating your selected value(s) of \(\epsilon.\)

kNNdistplot(three_spheres, minPts = 4)
abline(h = .21)

kNNdistplot(ring_moon_sphere, minPts = 4)
abline(h = .22)

kNNdistplot(two_spirals_sphere, minPts = 4)
abline(h = 1.05)

B)

Write a function called plot_dbscan_results(df, eps, minPts). This function takes a data frame, epsilon value, and minPts as arguments and does the following:

• Runs DBSCAN on the inputted data frame df, given the eps and minPts values;
• Creates a scatterplot of the data frame with points color-coded by assigned cluster membership. Make sure the title of the plot includes the value of eps and minPts used to create the clusters!!

Using this function, and your candidate eps values from A) as a starting point, implement DBSCAN to correctly identify the 3 cluster shapes in each of the three data sets. You will likely need to revise the eps values until you settle on a “correct” solution.

plot_dbscan_results <- function(df, eps, minPts) {
  
  db <- dbscan(df, eps = eps, minPts = minPts)
  
  df_plot <- df %>%
    mutate(cluster = factor(db$cluster))
  
  ggplot(df_plot, aes(x = x, y = y, color = cluster)) +
    geom_point(size = 2, alpha = 0.9) +
    labs(
      title = paste0("DBSCAN Results (eps = ", eps, ", minPts = ", minPts, ")"),
      color = "Cluster"
    ) +
    theme_minimal()
}

plot_dbscan_results(three_spheres, eps = 0.21, minPts = 4)

plot_dbscan_results(ring_moon_sphere, eps = 0.3, minPts = 4)

plot_dbscan_results(two_spirals_sphere, eps = 1.2, minPts = 4)

C)

Compare your DBSCAN solutions to the 3-cluster solutions from k-means and PAM. Use the patchwork package and your function from B) to produce a 3x3 grid of plots: one plot per method/data set combo. Comment on your findings.

plot_dbscan_results <- function(df, eps, minPts, name) {
  db <- dbscan(df, eps = eps, minPts = minPts)
  df$cluster <- as.factor(db$cluster)

  ggplot(df, aes(x, y, color = cluster)) +
    geom_point(size = 2) +
    ggtitle(paste0(name, " – DBSCAN \n(eps = ", eps, ", minPts = ", minPts, ")")) +
    theme_minimal() +
    theme(
      plot.title = element_text(size = 6)
    )
}

plot_kmeans <- function(df, k, name) {
  km <- kmeans(df, centers = k)
  df$cluster <- as.factor(km$cluster)

  ggplot(df, aes(x, y, color = cluster)) +
    geom_point(size = 2) +
    ggtitle(paste0(name, " – k-means (k = ", k, ")")) +
    theme_minimal() +
    theme(
      plot.title = element_text(size = 6)
    )
}

plot_pam <- function(df, k, name) {
  pam_res <- pam(df, k)
  df$cluster <- as.factor(pam_res$clustering)

  ggplot(df, aes(x, y, color = cluster)) +
    geom_point(size = 2) +
    ggtitle(paste0(name, " – PAM (k = ", k, ")")) +
    theme_minimal() +
    theme(
      plot.title = element_text(size = 6)
    )
}

p1 <- plot_dbscan_results(df = three_spheres, eps = .21, minPts = 4, 'Three Spheres')
p2 <- plot_kmeans(three_spheres, 3, 'Three Spheres')
p3 <- plot_pam(three_spheres, 3, 'Three Spheres')

p4 <- plot_dbscan_results(ring_moon_sphere, eps = .3, minPts = 4, 'Ring Moon Sphere')
p5 <- plot_kmeans(ring_moon_sphere, 3, 'Ring Moon Sphere')
p6 <- plot_pam(ring_moon_sphere, 3, 'Ring Moon Sphere')

p7 <- plot_dbscan_results(two_spirals_sphere, eps = 1.2, minPts = 4, 'Two Sprial Spheres')
p8 <- plot_kmeans(two_spirals_sphere, 3, 'Two Sprial Spheres')
p9 <- plot_pam(two_spirals_sphere, 3, 'Two Sprial Spheres')

(p1 | p2 | p3) /
(p4 | p5 | p6) /
(p7 | p8 | p9)

DBSCAN is the best algorithm for finding the outliers and identifying the shapes in the data set. It correctly identified the clusters in the three spheres data set and identified the clusters. In the ring moon sphere data set it correctly identified the ring, moon, and sphere as separated clusters. Then in the spiral data set is correctly identified the spirals and found outliers.

Kmeans and pam work well when the clusters are well separated and convex. When it comes to other chapes it doesn’t do very well. For example neither algorithms did a good job at identifying the moon shape in the ring moon sphere data set.

Question 2

In this question we will apply cluster analysis to analyze economic development indicators (WDIs) from the World Bank. The data are all 2020 indicators and include:

• life_expectancy: average life expectancy at birth
• gdp: GDP per capita, in 2015 USD
• co2: CO2 emissions, in metric tons per capita
• fert_rate: annual births per 1000 women
• health: percentage of GDP spent on health care
• imports and exports: imports and exports as a percentage of GDP
• internet and electricity: percentage of population with access to internet and electricity, respectively
• infant_mort: infant mortality rate, infant deaths per 1000 live births
• inflation: consumer price inflation, as annual percentage
• income: annual per-capita income, in 2020 USD

wdi <- read.csv('Data/wdi_extract_clean.csv') 
head(wdi)

      country life_expectancy        gdp      co2 fert_rate    health internet
1 Afghanistan        61.45400   527.8346 0.180555     5.145 15.533614  17.0485
2     Albania        77.82400  4437.6535 1.607133     1.371  7.503894  72.2377
3     Algeria        73.25700  4363.6853 3.902928     2.940  5.638317  63.4727
4      Angola        63.11600  2433.3764 0.619139     5.371  3.274885  36.6347
5   Argentina        75.87800 11393.0506 3.764393     1.601 10.450306  85.5144
6     Armenia        73.37561  4032.0904 2.334560     1.700 12.240562  76.5077
  infant_mort electricity  imports inflation  exports    income
1        55.3        97.7 36.28908  5.601888 10.42082  475.7181
2         8.1       100.0 36.97995  1.620887 22.54076 4322.5497
3        20.4        99.7 24.85456  2.415131 15.53520 2689.8725
4        42.3        47.0 27.62749 22.271539 38.31454 1100.2175
5         8.7       100.0 13.59828 42.015095 16.60541 7241.0303
6        10.2       100.0 39.72382  1.211436 29.76499 3617.0320

Focus on using kmeans for this problem.

A)

My claim: 3-5 clusters appear optimal for this data set. Support or refute my claim using appropriate visualizations.

wdi_numeric <- wdi %>% select_if(is.numeric)
wdi_scaled <- scale(wdi_numeric)

fviz_nbclust(
  wdi_scaled, 
  kmeans, 
  method = "wss"
) + 
  ggtitle("Elbow Method for Optimal k")

fviz_nbclust(
  wdi_scaled, 
  kmeans, 
  method = "silhouette"
) + 
  ggtitle("Silhouette Scores for k")

Looking at the elbow in the wss plot it can be seen that 3, 4, or 5 clusters would be optimal. Next looking at the silhouette plot it appears that 4 clusters is the most optimal. It also appears that 3 clusters is almost the same average silhouette width as 4 clusters.

B)

Use k-means to identify 4 clusters. Characterize the 4 clusters using a dimension reduction technique. Provide examples of countries that are representative of each cluster. Be thorough.

countries <- wdi %>% select(country)
kmeans4 <- kmeans(wdi_scaled, centers = 4, nstart = 10)
cluster <- factor(kmeans4$cluster)
wdi$cluster <- factor(kmeans4$cluster)

pca_res <- prcomp(wdi_scaled, center = TRUE, scale. = FALSE)
pca_scores <- as.data.frame(pca_res$x)
pca_scores$cluster <- wdi$cluster
pca_scores$country <- wdi$country

centroids <- pca_scores %>%
  group_by(cluster) %>%
  summarise(
    c1 = mean(PC1),
    c2 = mean(PC2)
  )

distances <- pca_scores %>%
  left_join(centroids, by = "cluster") %>%
  mutate(distance = sqrt((PC1 - c1)^2 + (PC2 - c2)^2))

representatives <- distances %>%
  group_by(cluster) %>%
  slice_min(order_by = distance, n = 3) %>%
  select(cluster, country, PC1, PC2, distance)

representatives

# A tibble: 12 × 5
# Groups:   cluster [4]
   cluster country           PC1    PC2 distance
   <fct>   <chr>           <dbl>  <dbl>    <dbl>
 1 1       Angola        -3.61   -1.16    0.350 
 2 1       Togo          -3.31   -0.515   0.366 
 3 1       Cote d'Ivoire -3.51   -0.385   0.439 
 4 2       Cabo Verde     0.0747  0.493   0.0279
 5 2       Azerbaijan     0.127   0.596   0.102 
 6 2       Mexico         0.115   0.389   0.139 
 7 3       Austria        3.17    0.101   0.229 
 8 3       Sweden         3.27    0.392   0.407 
 9 3       Finland        3.05    0.626   0.489 
10 4       Singapore      5.22   -5.59    0.393 
11 4       Ireland        4.68   -3.94    1.92  
12 4       Luxembourg     6.94   -7.31    2.15  

ggplot(pca_scores, aes(PC1, PC2, color = cluster)) +
  geom_point(alpha = 0.8) +
  geom_text_repel(
    data = representatives,
    aes(label = country),
    color = "black",
    size = 3,
    fontface = "bold",
    nudge_y = 0.2
  ) +
  theme_minimal() +
  ggtitle("PCA Biplot of WDI Data with Cluster Representatives (k = 4)")

The representatives in the final plot are the countries that best represent the cluster. They represent the most typical country in that cluster. This is done by getting the countries that are closest to the centroids of the cluster. The 3 closest countries to the centroid are selected for the final plot.

In cluster 1 the countries that best represent that cluster are Ireland, Singapore, and Luxembourg.

In cluster 2 the countries that best represent that cluster are Finland, Sweden, and Austria.

In cluster 3 the countries that best represent that cluster are Mexico, Azerbaijan, and Cabo Verde.

In cluster 4 the countries that best represent that cluster are Cot d’Ivoire, Togo, Angola.

C)

Remove Ireland, Singapore, and Luxembourg from the data set. Use k-means to find 4 clusters again, with these three countries removed. How do the cluster definitions change?

wdi_removed <- wdi %>%
  filter(!country %in% c("Ireland", "Singapore", "Luxembourg"))

wdi_removed_numeric <- wdi_removed %>% select_if(is.numeric)

wdi_removed_scaled <- scale(wdi_removed_numeric)

km_removed <- kmeans(wdi_removed_scaled, centers = 4)

wdi_removed$cluster <- factor(km_removed$cluster)

pca_removed <- prcomp(wdi_removed_scaled, center = TRUE, scale. = FALSE)

scores_removed <- as.data.frame(pca_removed$x[, 1:2])
colnames(scores_removed) <- c("PC1", "PC2")

scores_removed$country <- wdi_removed$country
scores_removed$cluster <- wdi_removed$cluster

ggplot(scores_removed, aes(PC1, PC2, color = cluster, label = country)) +
  geom_point(size = 3) +
  theme_minimal() +
  labs(
    title = "K-Means Clusters (4 clusters) after removing Ireland, Singapore, Luxembourg",
    x = "PC1",
    y = "PC2"
  )

The clusters are now different. Instead of cluster 1 from the previous problem being three countries that were outliers, it is now a cluster with more countries that aren’t all outliers of the entire data set. Other than cluster 1 being changed heavily from the previous problem the clusters are relatively in the same location.