
Introduction
Stratton AE-Banking is a newly founded online bank in the US market. The E-banking service is a joint venture of a young fintech start-up and the long-time standing New York Stratton & Fils private banking house. The joint venture was founded in 2020 and has since then enjoyed great interest by providing digital private banking services. It profits from an AI driven recommender engine that uses past investment information together with a market and finance machine learning engine to derive investment tips and portfolio suggestions for its customers. So far, the fintech startup was well able to successfully approach young investors and customers. After the joint venture with Stratton & Fils, the fintech hopes to also attract existing customers from the established bank.
However, the conservative bank management of Stratton & Fils is extremely worried about simply approaching all of its customers, as it fears that the data driven and digital customer experience of Stratton AE may disturb some of its long-standing customers and may harm the longtime established and very intimate customer relations, which are believed to be an essential success factor in the bank’s success history.
The management thus approaches you as the head of the data science team and asks you to conduct a segmentation analysis of the bank’s existing customer base and to identify suitable customer segments, which might be open to try out Stratton & Fils joint venture. As a base for your segmentation analysis, the CRM manager provides you with the following data.
import pandas as pd
file_path = 'Markdown/Data/tabledesc.csv'
# Read the CSV file into a DataFrame
df = pd.read_csv(file_path)
# Display the DataFrame
print(df)
Variable Description \
0 Age Customer Age
1 Income Household Net Income
2 HouseholdSize Number of People Living in Household
3 CityAreaSize City or Main Area Population
4 MeanCityIncome Average Income on ZIP-Code and Street Level
5 MeanCityHousePrize Average House Prizes on ZIP-Code and Street Le...
6 MeanCityHouseholdSize Average Household Size on ZIP-Code and Street ...
7 MeanCitySqFtPrice Average Prizes per Square Foot on ZIP-Code and...
8 NumbCars Number of registered cars of customer
9 InternetTrafficVolume Volume of Internet Traffic per customer household
10 MortageVolume Mortage to be paid by Customer
11 AccountSpending Monthly average spending from bank account
12 CreditCardSpending Monthly average spending from Credit Card
13 HelpHotlineTime Number of Minutes with Banking Hotline
14 CustomerSince Time since opening bank account
15 GrocerySpending Average grocery related spendings from bank ac...
16 StockVolume Stock Investment
17 CreditVolume Credits with the bank
18 NASDAQInvest Amount of money invested in NASDAQ listed comp...
19 USAXSFundInvest Amount of money invested in Stratton owned sha...
20 BranchVisits Number of recorded branch visits within the la...
21 AppLogins Number of customer logins in mobile banking ap...
22 ATMVisitis Number of times customer used an ATM service p...
23 TimeOnlineBanking Time logged into the Online Banking System
24 ServiceFees Extra Fees paid for banking services
25 SocialMediaInter Number of Finance Specific Social Media Profil...
26 Bitcoins Number of Bitcoins hold by customer
27 NFT Number of NFTs bought by customer
Measurement
0 Age in Years
1 Net Income in USD
2 Integer number
3 Integer number
4 Average Income in USD
5 Average Prizes in USD
6 Average Number Inhabitants
7 Yes/No
8 Number of Cars
9 GB
10 USD
11 USD
12 USD
13 Minutes
14 Months
15 USD
16 USD
17 USD
18 USD
19 USD
20 Integer number
21 Integer number
22 Integer number
23 Minutes
24 USD
25 Integer number
26 Number
27 Integer number
file_path = 'Markdown/Data/StrattonAEBankingCRM.csv'
# Read the CSV file into a DataFrame
BankinCRMData = pd.read_csv(file_path)
# Display summary statistics of the DataFrame
print(BankinCRMData.describe())
Age Income HouseholdSize CityAreaSize \
count 10750.000000 10750.000000 10750.000000 10750.000000
mean 35.156651 84699.601953 2.809674 372196.129023
std 13.542091 49337.110399 1.325119 228010.188666
min 18.000000 35202.000000 1.000000 61613.000000
25% 23.000000 42803.250000 2.000000 121703.750000
50% 30.000000 71268.000000 3.000000 450100.000000
75% 45.000000 125870.000000 4.000000 459418.000000
max 74.000000 181863.000000 8.000000 708729.000000
MeanCityIncome MeanCityHousePrize MeanCityHouseHoldSize \
count 10750.000000 1.075000e+04 10750.000000
mean 163706.020279 9.425050e+05 3.022977
std 64581.642188 7.136861e+05 1.338834
min 35372.000000 1.250110e+05 1.000000
25% 116252.750000 4.448165e+05 2.000000
50% 140457.500000 6.146005e+05 3.000000
75% 235000.000000 1.849915e+06 4.000000
max 286996.000000 1.850000e+06 8.000000
MeanCitySqFtPrice NumberCars InternetTrafficVolume ... \
count 10750.000000 10750.000000 10750.000000 ...
mean 5318.269023 1.383814 67.565023 ...
std 2394.522297 0.925567 33.096996 ...
min 1871.000000 0.000000 6.000000 ...
25% 2627.000000 1.000000 45.000000 ...
50% 5778.500000 1.000000 60.000000 ...
75% 6741.000000 2.000000 86.000000 ...
max 9886.000000 4.000000 118.000000 ...
NASDAQInvest USAXSFundInvest BranchVisits AppLogins \
count 10750.000000 10750.000000 10750.000000 10750.000000
mean 1828.480162 761.645087 3.912558 55.704558
std 1447.337174 836.464178 3.219362 34.641660
min 228.352966 69.947903 0.000000 1.000000
25% 401.400337 149.795783 2.000000 18.000000
50% 1498.098994 313.824135 3.000000 64.000000
75% 3056.049762 1060.232555 5.000000 82.000000
max 4532.364575 3396.609135 20.000000 130.000000
ATMVisits TimeOnlineBanking ServiceFees SocialMediaInter \
count 10750.000000 10750.000000 10750.000000 10750.000000
mean 4.928093 113.878607 40.938245 19.028837
std 2.324207 59.183105 31.086657 17.050131
min 0.000000 22.769131 0.134308 0.000000
25% 3.000000 69.281909 17.844189 5.000000
50% 5.000000 88.256085 27.238595 16.000000
75% 7.000000 152.915443 50.165175 31.000000
max 11.000000 232.214599 124.261270 60.000000
Bitcoins NFTs
count 10750.000000 10750.000000
mean 0.193681 3.317395
std 0.222972 2.691006
min 0.000000 0.000000
25% 0.000500 1.000000
50% 0.099800 3.000000
75% 0.400500 4.000000
max 0.601400 12.000000
[8 rows x 28 columns]
To identify segments of similar customers, let us first focus on the question how to measure similarity. Table 2 shows us some observations for customers from another banking database. The columns show the values of some customer related attributes. We can use the individual attribute characteristics to now calculate a so called distance measure, which shows how similar or dissimilar customers are. The higher the distance, the more dissimilar they are. For continuous variables, we can use the basic Euclidean Distance measure to derive similarities. The Euclidean Distance between two customers A and B can be expressed by the following equation.
$$ED_{A,B}= \sqrt{(f_{1,A}-f_{1,B})^{2}+(f_{2,A}-f_{2,B})^{2}+...+(f_{n,A}-f_{n,B})^{2}} $$Table2_1 = pd.read_csv("Markdown/Data/mashcustomer.csv")
Table2_1.head()
| Customer | Age | Income | Debt | Household Size | |
|---|---|---|---|---|---|
| 0 | Hawkeye | 32 | 45,000 | 25,000 | 1 |
| 1 | Potter | 64 | 75,000 | 10,000 | 3 |
| 2 | Burns | 49 | 42,000 | 20,000 | 5 |
| 3 | Hotlips | 33 | 22,000 | 2,000 | 1 |
| 4 | Klinger | 29 | 16,000 | 6,000 | 4 |
We can now use the formula of the Euclidean Distance to calculate for example the distance between Hawkeye and Potter.
import math
ED_Hawkeye_Potter = math.sqrt((32-64)**2 + (45-75)**2 + (25-10)**2 + (1-3)**2)
print(ED_Hawkeye_Potter)
46.400431032480725
Question: Repeat the calculations for Hawkeye and Burns as well as Hawkeye and Hotlips.
ED_Hawkeye_Burns = math.sqrt((32-49)**2 + (45-22)**2+ (25-20)**2 + (1-5)**2)
ED_Hawkeye_Burns
29.30870177950569
ED_Hawkeye_Hotlips = math.sqrt((32-33)**2 + (45-42)**2+ (25-2)**2 + (1-1)**2)
ED_Hawkeye_Hotlips
23.2163735324878
While this is a great exercise, it will be impossible to calculate the distances amongst all members of a large customer data base with e.g. 200,000 entries. However in this case we can also use R's function for Euclidean Distances. We simply need to give the function a data frame with all observations we would like to compare, and R will return a table with the corresponding distances.
from scipy.spatial.distance import euclidean
from scipy.spatial.distance import pdist, squareform
# Convert 'Income' and 'Debt' columns to numeric
Table2_1['Income'] = pd.to_numeric(Table2_1['Income'].str.replace(',', ''))
Table2_1['Debt'] = pd.to_numeric(Table2_1['Debt'].str.replace(',', ''))
# Extracting columns 2 to 5 (Age, Income, Debt, Household Size) for distance calculation
subset_data = Table2_1.iloc[:, 1:5]
# Calculate pairwise Euclidean distances
distances = pdist(subset_data, metric='euclidean')
# Convert to squareform for better readability
dist_matrix = squareform(distances)
# Create a DataFrame with the distances
dist_df = pd.DataFrame(dist_matrix, index=Table2_1['Customer'], columns=Table2_1['Customer'])
# Display the result
print(f"Metric: 'euclidean'; comparing: {len(subset_data)} vectors.")
print(dist_df)
Metric: 'euclidean'; comparing: 5 vectors. Customer Hawkeye Potter Burns Hotlips Klinger Customer Hawkeye 0.000000 33541.034987 5830.978048 32526.911950 34669.871906 Potter 33541.034987 0.000000 34481.882620 53600.382135 59135.448134 Burns 5830.978048 34481.882620 0.000000 26907.253149 29529.652910 Hotlips 32526.911950 53600.382135 26907.253149 0.000000 7211.104284 Klinger 34669.871906 59135.448134 29529.652910 7211.104284 0.000000
While the distances help us with understanding similarities and dissimilarities they do not yet help us with forming subgroups, ad only from the distances, you do not know which threshold determines similarity/dissimilarity. Hawkeye may be closest to Burns, but is 18 still a great distance? Or actually already pretty similar? Who should be paired with whom?
This implies that grouping consumers in homogenous subgroups requires a lot of attention and balance and some more information than just similarity measures. In addition, we realized with our 5 customers, that grouping takes us some time and effort and will certainly prevent us from forming larger groups or segmenting larger data sets with hundreds of thousands of customers. Therefore, it is time to discover a method that uses intersubject distances to automatically form groups. Such methods are commonly referred to as cluster analysis. Cluster analysis are a well-known and established statistical method, that is used for the last 30-40 years in marketing research. With the advent of machine learning and artificial intelligence applications, cluster analysis became again popular in data science, where it is often referred to as an unsupervised learning algorithm.
k-mean cluster analysis uses distances to form clusters within data. Once the user determined the k number of clusters the algorithm should define, the cluster randomly assigns k starting points within the data (Step1). It continues then to calculate the distance of each observation in the data to each starting point. As pointed out in in below's Figure the algorithm then assigns each observation according to the distance to the closest starting point (Step2). This leads to an initial cluster solution. For each of these clusters the algorithm then calculates the new center point of the cluster, called centroid (Step3). The centroid can be interpreted as the mean of all observations within this cluster. Step4 now repeats the procedure of step2. The new centroids are used to again calculate all distances between all observations and all centroids. Then again, the observations are assigned to the closest centroid. This may lead to changes in cluster membership and lead to new forms of clusters. In the subsequent step, the algorithm continues to calculate the resulting new centroids (Step5), to then re-calculate the distances and re-assigning observations to clusters. The algorithm stops once no observation can be re-assigned to another cluster or after a to be specified number of iterations.

One thing we may mind before running a cluster analysis, is scale heterogeneity. Especially k-mean clustering is sensitive to data that comes at different scale levels. Having variables at very different levels, thus creates problems, which may ultimately lead to biased results. A quick fix is to standardize the variables so that they all share a similar range. This procedure is commonly referred to as standardization.
R can standardize all variables for us with the help of the scale() function. When we now inspect the resulting new data frame scaled.crm with the head() function.
from sklearn.preprocessing import StandardScaler
# Extracting numeric columns for scaling
numeric_columns = BankinCRMData.select_dtypes(include=['int64', 'float64']).columns
numeric_data = BankinCRMData[numeric_columns]
# Scale the numeric data
scaler = StandardScaler()
scaled_crm = scaler.fit_transform(numeric_data)
# Create a DataFrame with the scaled data
scaled_crm_df = pd.DataFrame(scaled_crm, columns=numeric_columns)
# Display the head of the scaled DataFrame
print(scaled_crm_df.head(6))
###CHECK : The values are not an exact match with that in the R-Code. They are rounded off.
Age Income HouseholdSize CityAreaSize MeanCityIncome \ 0 0.357668 -0.102901 -0.611049 0.361798 -1.130994 1 0.136126 -0.265200 1.653005 0.352057 -0.107838 2 2.646934 -0.125177 -1.365733 0.370167 -1.722273 3 1.317683 -0.299213 0.143636 0.370167 -0.701223 4 0.357668 -0.211932 -1.365733 0.350035 -1.974012 5 -0.233110 -0.181649 3.162374 0.368189 -1.121269 MeanCityHousePrize MeanCityHouseHoldSize MeanCitySqFtPrice NumberCars \ 0 1.271589 -0.017163 -0.628659 0.665770 1 1.271058 1.476744 -0.022665 0.665770 2 1.271554 -0.017163 -0.799056 -0.414699 3 1.270641 0.729791 -1.326952 -1.495169 4 1.270564 2.223698 -1.319017 -0.414699 5 1.271276 -0.017163 -0.264060 -0.414699 InternetTrafficVolume ... NASDAQInvest USAXSFundInvest BranchVisits \ 0 -0.289013 ... -0.235347 -0.341807 -0.283472 1 -0.953757 ... -0.224107 -0.324460 0.027162 2 -0.319229 ... -0.226961 -0.354929 -0.283472 3 0.103790 ... -0.229419 -0.327110 0.027162 4 -0.863110 ... -0.226898 -0.372217 -0.283472 5 -0.470307 ... -0.225932 -0.231325 0.027162 AppLogins ATMVisits TimeOnlineBanking ServiceFees SocialMediaInter \ 0 -1.319414 1.752037 -0.715268 0.028782 0.467535 1 -1.059599 1.321763 -0.789812 0.361236 0.350228 2 -1.203941 1.321763 -0.928995 0.440483 0.526188 3 -1.203941 1.321763 -0.894188 0.027947 0.878108 4 -1.146204 1.752037 -0.789528 0.520945 0.878108 5 -1.232809 1.321763 -0.976702 0.518330 0.291575 Bitcoins NFTs 0 -0.854321 -0.489578 1 -0.852078 -0.861203 2 -0.807676 -0.861203 3 -0.861497 -0.861203 4 -0.835035 -1.232829 5 -0.855218 -0.117952 [6 rows x 28 columns]
As you see, all variables now range in similar areas. We can thus proceed with our analysis
We can now start with the cluster analysis. Let us first try out different solutions with different numbers of clusters. To ensure that we start with the same centroids, we use the set.seed function. This ensures that every time we run this code, we end up with the same results. If you do not use set.seed ahead of the cluster analysis, you will receive different solutions, which will be close to each other but not identical. We can run a k-mean cluster analysis with NumPy's kmean function (the library is imported at the begining) We tell the kmean function simply which data.frame contains our customer data and specify the k number of clusters we want to be included. Here we set k to 4.
To see how many customers are assigned to each cluster, we furthermore plot the cluster sizes with the help of ggplot and a simple barplot.
import numpy as np
from sklearn.cluster import KMeans
import matplotlib.pyplot as plt
import seaborn as sns
np.random.seed(13)
# Perform k-means clustering
kmeans = KMeans(n_clusters=4, random_state=13, init='k-means++', n_init=10)
StrattonCluster_4k = kmeans.fit_predict(scaled_crm_df)
# Get cluster sizes
sizes4k = pd.DataFrame({'Size': np.bincount(StrattonCluster_4k), 'Cluster': [f'Cluster{i+1}' for i in range(4)]})
# Plot the cluster sizes
plt.figure(figsize=(10, 6))
sns.barplot(x='Cluster', y='Size', data=sizes4k, color='turquoise')
plt.xlabel('Cluster')
plt.ylabel('Size')
plt.title('Cluster sizes k-means 4-cluster solution')
for i, size in enumerate(sizes4k['Size']):
plt.text(i, size + 0.1, str(size), ha='center', va='bottom')
plt.show()
print(sizes4k)
Size Cluster 0 4500 Cluster1 1 2250 Cluster2 2 1500 Cluster3 3 2500 Cluster4
We can now inspect the different clusters and check their mean values. We achieve this with the following code, that first matches the estimated cluster to each observation in our data frame. Subsequently, we use dyplr’s group_by command to calculate the mean of each variable per cluster. You can then inspect the resulting data frame. You will see that some of the clusters show substantially different mean values for specific variables, while in other cases the means do not vary across the clusters.
# Add the cluster labels to BankinCRMData
BankinCRMData['k4Cluster'] = StrattonCluster_4k
# Group by k4Cluster and calculate mean for numeric columns
summarystats_percluster_4k = BankinCRMData.groupby('k4Cluster').agg('mean')
# Display the head of the summary statistics
print(summarystats_percluster_4k.head())
Age Income HouseholdSize CityAreaSize \
k4Cluster
0 32.132667 85293.995333 2.605556 239607.172667
1 39.363111 71856.288444 2.703556 452187.049778
2 57.965333 180029.459333 3.488000 120068.719333
3 23.128800 37990.761600 2.865600 690140.867600
MeanCityIncome MeanCityHousePrize MeanCityHouseHoldSize \
k4Cluster
0 115856.230667 8.628481e+05 2.985333
1 179344.236444 3.085286e+05 2.963111
2 139990.476667 6.200123e+05 3.366667
3 249990.573200 1.849962e+06 2.938400
MeanCitySqFtPrice NumberCars InternetTrafficVolume ... \
k4Cluster ...
0 3528.396667 1.480889 51.697556 ...
1 4458.828444 0.803111 115.006222 ...
2 6497.381333 2.770000 15.015333 ...
3 8606.068400 0.900000 84.959200 ...
NASDAQInvest USAXSFundInvest BranchVisits AppLogins ATMVisits \
k4Cluster
0 1117.043945 767.378607 4.030222 41.641778 4.518000
1 4055.420171 1870.605450 1.404000 103.894222 4.928889
2 3003.141149 100.033983 9.985333 10.025333 4.036000
3 400.022754 150.227085 2.314800 65.054400 6.200800
TimeOnlineBanking ServiceFees SocialMediaInter Bitcoins \
k4Cluster
0 113.403867 21.637420 12.269333 0.134875
1 202.818078 94.546763 46.492889 0.544434
2 30.036049 44.874469 2.502667 0.000102
3 84.993150 25.070329 16.394000 0.099999
NFTs
k4Cluster
0 1.969333
1 7.879111
2 1.000000
3 3.028800
[4 rows x 28 columns]
Another approach to assess the quality of our segmentation, is to plot the different clusters. A key challenge here is dimensionality. Given that our clusters depend on a multitude of variables, we cannot plot them all together. To come to a solution that we can plot, we need to reduce the dimensions to two main factors, which then allows us to plot the points in two-dimensional space. A common technique to achieve this is a principal component analysis (PCA) that reduces all variables to two main factors, which we can subsequently plot. The plot will then allow us to better see if clusters overlap or if we end up with a meaningful separation between the different identified clusters. R’s factoextra package offers various functions, which achieve this with a single command that does not require us to code the PCA nor the plot.
import numpy as np
from sklearn.decomposition import PCA
from sklearn.mixture import GaussianMixture
from matplotlib.patches import Ellipse
# Define the plot_ellipse function
def plot_ellipse(ax, mean, covariances, color, alpha):
v, w = np.linalg.eigh(covariances)
v = 2.0 * np.sqrt(2.0) * np.sqrt(v)
u = w[0] / np.linalg.norm(w[0])
angle = np.arctan(u[1] / u[0])
angle = 180.0 * angle / np.pi
ell = Ellipse(mean, width=v[0], height=v[1], angle=angle, color=color, alpha=alpha)
ax.add_patch(ell)
# Use PCA for dimensionality reduction (optional)
pca = PCA(n_components=2)
reduced_data = pca.fit_transform(scaled_crm_df)
# Create a DataFrame with cluster labels and reduced data (optional)
cluster_data = pd.DataFrame({'Cluster': StrattonCluster_4k, 'PC1': reduced_data[:, 0], 'PC2': reduced_data[:, 1]})
# Plot the clusters using seaborn
plt.figure(figsize=(10, 6))
sns.scatterplot(x='PC1', y='PC2', hue='Cluster', data=cluster_data, palette='viridis', s=100, alpha=0.7)
plt.xlabel('Principal Component 1')
plt.ylabel('Principal Component 2')
plt.title('Cluster Visualization k-means 4-cluster solution')
# Plot ellipsoids around each cluster
for i in range(4):
cluster_points = reduced_data[StrattonCluster_4k == i]
# Fit a Gaussian Mixture Model to estimate the covariance matrix
gmm = GaussianMixture(n_components=1, covariance_type='full')
gmm.fit(cluster_points)
# Get the covariance matrix and mean of the cluster
covariances = gmm.covariances_
mean = gmm.means_
# Plot the ellipsoid
plot_ellipse(ax=plt.gca(), mean=mean[0], covariances=covariances[0], color='black', alpha=0.3)
# Adjusting the legend
handles, labels = plt.gca().get_legend_handles_labels()
labels = [str(int(label) + 1) for label in labels] # Incrementing cluster numbers in the legend
plt.legend(handles, labels, title='Cluster')
plt.show()
A quick inspection of the plot already reveals that our 4k cluster approach is not optimal, as we see some more separable groups of close to each other observations. Especially in case of Clusters 0 and 1 (the larger ones) it looks like we can still split these groups into two more subgroups each.
Question: Repeat the cluster analysis with k = 5 and k = 6
# k = 5
kmeans = KMeans(n_clusters=5, random_state=13, init='k-means++', n_init=10)
StrattonCluster_5k = kmeans.fit_predict(scaled_crm_df)
# Get cluster sizes
sizes5k = pd.DataFrame({'Size': np.bincount(StrattonCluster_5k), 'Cluster': [f'Cluster{i+1}' for i in range(5)]})
# Plot the cluster sizes
plt.figure(figsize=(10, 6))
sns.barplot(x='Cluster', y='Size', data=sizes5k, color='turquoise')
plt.xlabel('Cluster')
plt.ylabel('Size')
plt.title('Cluster sizes k-means 5-cluster solution')
for i, size in enumerate(sizes5k['Size']):
plt.text(i, size + 0.1, str(size), ha='center', va='bottom')
plt.show()
print(sizes5k)
Size Cluster 0 3000 Cluster1 1 2500 Cluster2 2 2250 Cluster3 3 1500 Cluster4 4 1500 Cluster5
# k = 6
kmeans = KMeans(n_clusters=6, random_state=13, init='k-means++', n_init=10)
StrattonCluster_6k = kmeans.fit_predict(scaled_crm_df)
# Get cluster sizes
sizes6k = pd.DataFrame({'Size': np.bincount(StrattonCluster_6k), 'Cluster': [f'Cluster{i+1}' for i in range(6)]})
# Plot the cluster sizes
plt.figure(figsize=(10, 6))
sns.barplot(x='Cluster', y='Size', data=sizes6k, color='turquoise')
plt.xlabel('Cluster')
plt.ylabel('Size')
plt.title('Cluster sizes k-means 6-cluster solution')
for i, size in enumerate(sizes6k['Size']):
plt.text(i, size + 0.1, str(size), ha='center', va='bottom')
plt.show()
print(sizes6k)
Size Cluster 0 3000 Cluster1 1 2500 Cluster2 2 1250 Cluster3 3 1500 Cluster4 4 1500 Cluster5 5 1000 Cluster6
# k = 7
kmeans = KMeans(n_clusters=7, random_state=13, init='k-means++', n_init=10)
StrattonCluster_7k = kmeans.fit_predict(scaled_crm_df)
# Get cluster sizes
sizes7k = pd.DataFrame({'Size': np.bincount(StrattonCluster_7k), 'Cluster': [f'Cluster{i+1}' for i in range(7)]})
# k = 8
kmeans = KMeans(n_clusters=8, random_state=13, init='k-means++', n_init=10)
StrattonCluster_8k = kmeans.fit_predict(scaled_crm_df)
# Get cluster sizes
sizes8k = pd.DataFrame({'Size': np.bincount(StrattonCluster_8k), 'Cluster': [f'Cluster{i+1}' for i in range(8)]})
How to determine the right number of clusters
Trying out different solutions may point you onto something, still you will realize that determining the right number of clusters can be tricky.
To find the "best" number of cluster, there are different approaches and measures available. Before we discuss these, let us first reflect again on what we want to achieve with a cluster analysis.
We want to obtain subgroups that are homogenous within. So, to say we try to maximize within-group homogeneity, which means we try to reduce the level of variance between members of a cluster. The overall level of within-cluster-variances across all identified clusters can thus be used to describe the total degree of homogeneity obtained with a specific cluster solution. This gives us a chance to compare different cluster analyses with different numbers of clusters, as we can try to minimize the overall variances.
Using the within-cluster-variance values, we can determine which solution works best and then focus on this cluster analysis. To do so, we first estimate n cluster solutions with cluster numbers from 1 to k. Subsequently, we can then plot the within-cluster variance sums for each cluster solution.
Again Python can do this for us with some short lines of code. Below you find two measures for within-cluster variances. We can now ask Python to estimate kmean models with k values from 2 to 15 and to then plot the within variances of each solution. Don't worry, if this takes some time.
# If you do not have the yellowbrick library installed uncomment the line below and run it
# pip install yellowbrick
from sklearn.impute import SimpleImputer
from yellowbrick.cluster import KElbowVisualizer
from sklearn.metrics import silhouette_score
imputer = SimpleImputer(strategy="mean")
scaled_crm_imputed = imputer.fit_transform(scaled_crm_df)
# Elbow Method with explicit n_init
model = KMeans(n_init=10) # You can set the value based on your preference
visualizer = KElbowVisualizer(model, k=(2, 15), metric='distortion', timings=False)
visualizer.fit(scaled_crm_imputed)
visualizer.show()
# Average Silhouette Width for k=2 to k=15 without visualization
avg_silhouette_scores = []
for k in range(2, 16):
model = KMeans(n_clusters=k, random_state=13, n_init=10)
labels = model.fit_predict(scaled_crm_imputed)
avg_silhouette_scores.append(silhouette_score(scaled_crm_imputed, labels))
# Plotting Average Silhouette Width
plt.figure(figsize=(10, 6))
plt.plot(range(2, 16), avg_silhouette_scores, marker='o')
plt.xlabel('Number of Clusters (k)')
plt.ylabel('Average Silhouette Width')
plt.title('Average Silhouette Width for Different Values of k')
plt.show()
The Elbow plot (1st plot), shows the total within sum of cluster variances for all estimated 15 solutions. The rule of thumb states, that the optimal cluster number lays within the “elbow” of the plot. This seems to be here rather tricky. As the function drops immediately and shows very low summed variances for clusters 2 to 15. Therefore, we rely on a second method, the Silhouette plot. The silhouette coefficient measures of how close an object is to its own cluster centroid, compared to the one of other clusters. The coefficient ranges from −1 to +1. High values indicate strong separation. Low values indicate poor separation. We thus want to select the cluster solution with the highest silhouette coefficient. In our case, the plot suggests 8 clusters. Looking again at the Elbow plot on the left, 8 seems rather high, especially as the “Elbow” – lays somewhere between 5 and 7. The silhouette plot suggests that the 7-cluster solution is inferior to the 6- and 8-cluster solutions. We may thus enrich our insights by plotting all three solutions with the following command.
# Create a DataFrame with cluster labels and reduced data (optional)
cluster_data = pd.DataFrame({'Cluster': StrattonCluster_5k, 'PC1': reduced_data[:, 0], 'PC2': reduced_data[:, 1]})
# Plot the clusters using seaborn
plt.figure(figsize=(10, 6))
sns.scatterplot(x='PC1', y='PC2', hue='Cluster', data=cluster_data, palette='viridis', s=100, alpha=0.7)
plt.xlabel('Principal Component 1')
plt.ylabel('Principal Component 2')
plt.title('Cluster Visualization k-means 5-cluster solution')
handles, labels = plt.gca().get_legend_handles_labels()
labels = [str(int(label) + 1) for label in labels] # Incrementing cluster numbers in the legend
plt.legend(handles, labels, title='Cluster')
plt.show()
# Create a DataFrame with cluster labels and reduced data (optional)
cluster_data = pd.DataFrame({'Cluster': StrattonCluster_6k, 'PC1': reduced_data[:, 0], 'PC2': reduced_data[:, 1]})
# Plot the clusters using seaborn
plt.figure(figsize=(10, 6))
sns.scatterplot(x='PC1', y='PC2', hue='Cluster', data=cluster_data, palette='viridis', s=100, alpha=0.7)
plt.xlabel('Principal Component 1')
plt.ylabel('Principal Component 2')
plt.title('Cluster Visualization k-means 6-cluster solution')
handles, labels = plt.gca().get_legend_handles_labels()
labels = [str(int(label) + 1) for label in labels] # Incrementing cluster numbers in the legend
plt.legend(handles, labels, title='Cluster')
plt.show()
# Create a DataFrame with cluster labels and reduced data (optional)
cluster_data = pd.DataFrame({'Cluster': StrattonCluster_7k, 'PC1': reduced_data[:, 0], 'PC2': reduced_data[:, 1]})
# Plot the clusters using seaborn
plt.figure(figsize=(10, 6))
sns.scatterplot(x='PC1', y='PC2', hue='Cluster', data=cluster_data, palette='viridis', s=100, alpha=0.7)
plt.xlabel('Principal Component 1')
plt.ylabel('Principal Component 2')
plt.title('Cluster Visualization k-means 7-cluster solution')
handles, labels = plt.gca().get_legend_handles_labels()
labels = [str(int(label) + 1) for label in labels] # Incrementing cluster numbers in the legend
plt.legend(handles, labels, title='Cluster')
plt.show()
# Create a DataFrame with cluster labels and reduced data (optional)
cluster_data = pd.DataFrame({'Cluster': StrattonCluster_8k, 'PC1': reduced_data[:, 0], 'PC2': reduced_data[:, 1]})
# Plot the clusters using seaborn
plt.figure(figsize=(10, 6))
sns.scatterplot(x='PC1', y='PC2', hue='Cluster', data=cluster_data, palette='viridis', s=100, alpha=0.7)
plt.xlabel('Principal Component 1')
plt.ylabel('Principal Component 2')
plt.title('Cluster Visualization k-means 8-cluster solution')
handles, labels = plt.gca().get_legend_handles_labels()
labels = [str(int(label) + 1) for label in labels] # Incrementing cluster numbers in the legend
plt.legend(handles, labels, title='Cluster')
plt.show()
Interpretation of Output
Let us first start by looking in more detail at our 7 -cluster k-mean model and see how big each cluster is, with the following code.
# 8 cluster k-mean cluster size plot
# Plot the cluster sizes
plt.figure(figsize=(10, 6))
sns.barplot(x='Cluster', y='Size', data=sizes7k, color='turquoise')
plt.xlabel('Cluster')
plt.ylabel('Size')
plt.title('Cluster sizes k-means 7-cluster solution')
for i, size in enumerate(sizes7k['Size']):
plt.text(i, size + 0.1, str(size), ha='center', va='bottom')
plt.show()
print(sizes7k)
Size Cluster 0 1500 Cluster1 1 2500 Cluster2 2 1250 Cluster3 3 1500 Cluster4 4 1500 Cluster5 5 1500 Cluster6 6 1000 Cluster7
To gain deeper insights related to spending behavior as well as the digital affinity of the different segments, we want to plot the means of the different variables. To achieve this we first again assemble a descriptive data set with all variable means per cluster with the help of dplyr’s group_by function.
# Build Mean per Cluster DataFrame
BankinCRMData['k7Cluster'] = StrattonCluster_7k
# Group by k4Cluster and calculate mean for numeric columns
summarystats_percluster_7k = BankinCRMData.groupby('k7Cluster').agg('mean')
We can now generate barplots of the different variables of interest and see if we find promising segments of Stratton & Fils customers, who might be open and suitable for Stratton AE Banking. Let us first focus on spending behavior, as indicated by the service fee variable. Note, that we adapted some of the commands in ggplot. By leaving geom_col() blank we do not specify a color and the plot remains in grey. In addition, we ask ggplot in geom_text to add labels with the two-digit rounded values of ServiceFees in white color and in font size 2. With the position_stack command we put the values in the middle of the barplot.
import matplotlib.pyplot as plt
# Barplot of Service Fees
plt.bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['ServiceFees'], color='grey')
plt.xlabel("Clusters")
plt.ylabel("Spending")
plt.title("Average Spending in Service Fees per Cluster")
# Adding text labels with values on top of each bar
for i, value in enumerate(summarystats_percluster_7k['ServiceFees']):
plt.text(i+1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
plt.show()
A visual inspection indicates that clusters 3 and 7 show the highest spending behavior, with clusters 5 and 6 following, while the remaining clusters show rather low service fee spendings. This makes at least the four high spending segments attractive for AE Banking. However, to be sure that the rather novel and highly digital app service appeals to these segments, we need to understand how digitally active and interested these segments are.
Let us first focus on the latest developments in fintech such as Bitcoin and NFT investments. We can again compare the segment-specific means for both variables. This time we want to combine the plots of Bitcoins and NFTs in one plot. We can arrange this with ggplot’s facet_wrap function that allows us to combine plots of different variables. The only “complication” we need to address is that we need to re-arrange the data set we want to plot. We can again use dplyr for this. We first select the variables of interest (cluster, NFTs and Bitcoins) and then transpose the data frame from a wide to a narrow format. We can then use again ggplot. This time we use the geom_bar command instead of the geom_col command. Facet_wrap will now tell ggplot to make two plots and combine them under each other (col =1). By setting scales to “free_y” we allow different y-axis levels, given that scales substantially vary across the two different variables.
# Barplot of BitCoins
plt.bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['Bitcoins'], color='grey')
plt.xlabel("Clusters")
plt.ylabel("Value")
plt.title("Average Bitcoins per Cluster")
# Adding text labels with values on top of each bar
for i, value in enumerate(summarystats_percluster_7k['Bitcoins']):
plt.text(i+1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
plt.show()
# Barplot of NFTs
plt.bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['NFTs'], color='grey')
plt.xlabel("Clusters")
plt.ylabel("Value")
plt.title("Average NFTs per Cluster")
# Adding text labels with values on top of each bar
for i, value in enumerate(summarystats_percluster_7k['NFTs']):
plt.text(i+1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
plt.show()
From the inspection, we can see that clusters 3 and 7 show both most activity in NFTs acquisitions and are also most invested in Bitcoins, which makes them even more suitable for AE Banking. Let us now look at digital activities and compare digital and offline activities.
With the following code we can inspect the means for BranchVisits, AppLogins, ATMVisits, TimeOnlineBanking, SocialMediaInter, InternetTrafficVolume. As you can see from facet_wrap we now include two columns.
import matplotlib.pyplot as plt
# Create a 2x3 grid of subplots
fig, axs = plt.subplots(3, 2, figsize=(20, 15))
# Barplot of AppLogins
axs[0, 0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['AppLogins'], color='grey')
axs[0, 0].set(xlabel="Clusters", ylabel="Value", title="Average AppLogins per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['AppLogins']):
axs[0, 0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of BranchVisits
axs[0, 1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['BranchVisits'], color='grey')
axs[0, 1].set(xlabel="Clusters", ylabel="Value", title="Average BranchVisits per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['BranchVisits']):
axs[0, 1].text(i, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of SocialMediaInter
axs[1, 0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['SocialMediaInter'], color='grey')
axs[1, 0].set(xlabel="Clusters", ylabel="Value", title="Average SocialMediaInter per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['SocialMediaInter']):
axs[1, 0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of ATMVisits
axs[1, 1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['ATMVisits'], color='grey')
axs[1, 1].set(xlabel="Clusters", ylabel="Value", title="Average ATMVisits per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['ATMVisits']):
axs[1, 1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of InternetTrafficVolume
axs[2, 0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['InternetTrafficVolume'], color='grey')
axs[2, 0].set(xlabel="Clusters", ylabel="Value", title="Average InternetTrafficVolume per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['InternetTrafficVolume']):
axs[2, 0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of TimeOnlineBanking
axs[2, 1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['TimeOnlineBanking'], color='grey')
axs[2, 1].set(xlabel="Clusters", ylabel="Value", title="Average TimeOnlineBanking per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['TimeOnlineBanking']):
axs[2, 1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Adjust layout to prevent clipping of titles
plt.tight_layout()
# Show the plots
plt.show()
The plot further confirms the strong digital affinity of clusters 3 and 7. Both show the lowest number of branch visits, while showing strong activity in online baking, internet traffic, social media interest, and banking app logins. While we can now be sure that customers from segments 3 and 7 are highly digital affine and are thus likely to be interested in AE Banking, we should in the next step control the financial situation of these customers. Let us first focus on average age, income and household sizes.
# Create a 3x1 grid of subplots
fig, axs = plt.subplots(3, 1, figsize=(5, 7.5))
# Barplot of Age
axs[0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['Age'], color='grey')
axs[0].set(xlabel="Clusters", ylabel="Value", title="Average Age per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['Age']):
axs[0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of HouseholdSize
axs[1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['HouseholdSize'], color='grey')
axs[1].set(xlabel="Clusters", ylabel="Value", title="Average HouseholdSize per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['HouseholdSize']):
axs[1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of Income
axs[2].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['Income'], color='grey')
axs[2].set(xlabel="Clusters", ylabel="Value", title="Average Income per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['Income']):
axs[2].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Adjust layout to prevent clipping of titles
plt.tight_layout()
# Show the plots
plt.show()
The plots reveal the problems with socio-economic clustering, as the results for age and household size do not vary too much across the 7 clusters. We see some variation for income, where clusters 3 and 7 remain close to the total mean of the dataset, indicating, that the digital affine users, we identified, are neither poor nor rich, making them still a suitable target group. Age-wise, we similarly see that both segments are well-established adults in their end 30s or early 40s. Given that the socio-economic information indicates that the digital affine users profit from stable incomes, we should in the next steps focus on spending and investment behavior to understand, whether these segments allow sufficient business volume and growth potential.
#Investment Cluster Means
# Create a 2x3 grid of subplots
fig, axs = plt.subplots(2, 2, figsize=(17, 12))
# Barplot of MortageVolume
axs[0, 0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['MortageVolume'], color='grey')
axs[0, 0].set(xlabel="Clusters", ylabel="Value", title="Average MortageVolume per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['MortageVolume']):
axs[0, 0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of NASDAQInvest
axs[0, 1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['NASDAQInvest'], color='grey')
axs[0, 1].set(xlabel="Clusters", ylabel="Value", title="Average NASDAQInvest per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['NASDAQInvest']):
axs[0, 1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of StockVolume
axs[1, 0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['StockVolume'], color='grey')
axs[1, 0].set(xlabel="Clusters", ylabel="Value", title="Average StockVolume per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['StockVolume']):
axs[1, 0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of USAXSFundInvest
axs[1, 1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['USAXSFundInvest'], color='grey')
axs[1, 1].set(xlabel="Clusters", ylabel="Value", title="Average USAXSFundInvest per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['USAXSFundInvest']):
axs[1, 1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Adjust layout to prevent clipping of titles
plt.tight_layout()
# Show the plots
plt.show()
#Plots for Spending and Investments
# Create a 3x1 grid of subplots
fig, axs = plt.subplots(3, 1, figsize=(5, 7.5))
# Barplot of AccountSpending
axs[0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['AccountSpending'], color='grey')
axs[0].set(xlabel="Clusters", ylabel="Value", title="Average AccountSpending per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['AccountSpending']):
axs[0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of CreditCardSpending
axs[1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['CreditCardSpending'], color='grey')
axs[1].set(xlabel="Clusters", ylabel="Value", title="Average CreditCardSpending per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['CreditCardSpending']):
axs[1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of GrocerySpending
axs[2].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['GrocerySpending'], color='grey')
axs[2].set(xlabel="Clusters", ylabel="Value", title="Average GrocerySpending per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['GrocerySpending']):
axs[2].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Adjust layout to prevent clipping of titles
plt.tight_layout()
# Show the plots
plt.show()
From the inspection of the two plots, it becomes evident that clusters 3 and 7 are more invested in stocks than their counterparts, and compared to the other clusters also share lower levels of mortgages. Looking at the types of investments, we see that cluster 3 is more invested in NASDAQ listed companies than all other clusters, while cluster 7 is strongly invested in Stratton’s fund for small and mid-size US companies. Spending behavior information tells us that both segments belong to the less spending customers, with cluster 3 showing the lowest credit card turnover of all clusters. In case of grocery expenditures, we see cluster 7 being the cluster with the second-highest average spending behavior.
Last, we can enrich our insights, by looking at the living conditions of the different segments and see where the different segments are located. To achieve this, we finally compare residential information.
# Create a 2x2 grid of subplots
fig, axs = plt.subplots(2, 2, figsize=(17, 12))
# Barplot of MortageVolume
axs[0, 0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['CityAreaSize'], color='grey')
axs[0, 0].set(xlabel="Clusters", ylabel="Value", title="Average CityAreaSize per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['CityAreaSize']):
axs[0, 0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of MeanCityHouseHoldSize
axs[0, 1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['MeanCityHouseHoldSize'], color='grey')
axs[0, 1].set(xlabel="Clusters", ylabel="Value", title="Average MeanCityHouseHoldSize per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['MeanCityHouseHoldSize']):
axs[0, 1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of MeanCityIncome
axs[1, 0].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['MeanCityIncome'], color='grey')
axs[1, 0].set(xlabel="Clusters", ylabel="Value", title="Average MeanCityIncome per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['MeanCityIncome']):
axs[1, 0].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Barplot of MeanCitySqFtPrice
axs[1, 1].bar([cluster + 1 for cluster in summarystats_percluster_7k.index], summarystats_percluster_7k['MeanCitySqFtPrice'], color='grey')
axs[1, 1].set(xlabel="Clusters", ylabel="Value", title="Average MeanCitySqFtPrice per Cluster")
# Add text labels to each bar
for i, value in enumerate(summarystats_percluster_7k['MeanCitySqFtPrice']):
axs[1, 1].text(i + 1, value, f"{round(value, 2)}", ha='center', va='bottom', color='black', size=8)
# Adjust layout to prevent clipping of titles
plt.tight_layout()
# Show the plots
plt.show()
From the plot we learn that clusters 3 and 7 both prefer city areas with mid-to-lower levels of population. In case of cluster 3 the average household sizes in the residential areas are rather small, while in case of cluster 7 we observe larger compounds with by average 4 members living in one household. Looking at income distributions and the area’s soil values, we learn that cluster 3 lives in rather richer neighborhoods with higher soil prices, whereas cluster 7 members prefer middle-class neighborhoods with affordable, low soil prices.
Question: Combining the information at hand, how do you depict members of clusters 3 and 7 and how do you believe they differ from each other? Can you similarly come up with personae for other clusters?
Taking Actions from Insights
The results of the cluster analysis allow Stratton AE Banking to take several important marketing actions. First, the profound understanding of the different available market segments, allows the joint-venture to understand the different types of customers available and to determine, which segments in the existing customer base should build the base for future marketing activities.
To develop suitable positionings for each cluster and subsequently develop communication campaigns, one can use the further insights from the cluster analysis and the comparison of the cluster-specific means of the remaining variables.
Furthermore, the results of the cluster analysis can be used to also predict the interests and preferences of newly incoming customers. Here, one may use the existing information available and calculate the Euclidean distances between the new customer and the centers (i.e. the means of each dimension) of each cluster. The customer will likely belong to the cluster, with the lowest distance.