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This is still an early draft. Let me know if there are any errors or typos.
Keep in mind that no programmer can avoid errors. I strongly agree with this quote from “CodeAcademy” that “Errors in your code mean you’re trying to do something cool.”
https://news.codecademy.com/errors-in-code-think-differently/
Objective - Dividing the target market or customers on the basis of some significant features which could help a company sell more products in less marketing expenses.
A potentially interesting question might be are some products (or customers) more alike than the others.
Market segmentation is a strategy that divides a broad target market of customers into smaller, more similar groups, and then designs a marketing strategy specifically for each group. Clustering is a common technique for market segmentation since it automatically finds similar groups given a data set.
Imagine that you are the Director of Customer Relationships at Apple, and you might be interested in understanding consumers’ attitude towards iPhone 12 and Google’s Pixel 5. Once the product is created, the ball shifts to the marketing team’s court. As mentioned above, to understand which groups of customers will be interested in which kind of features, marketers will make use of market segmentation strategy. The cluster analysis algorithm is designed to address this problem. Doing this ensures the product is positioned to the right segment of customers with a high propensity to buy.
1.Identify the type of customers who would respond to a particular offer
2.Identify high spenders among customers who will use the e-commerce channel for festive shopping
3.Identify customers who will default on their credit obligation for a loan or credit card
The file customer_segmetation.csv contains data collected by my students in spring 2020.
Search for Rstudio Cloud, register (or set up a free user account), and log into the cloud environment with your Gmail credentials.
You will upload your dataset (.csv) from your own computer to R Studio Cloud first. Make sure the first column is id instead of a variable.
Once the dataset is uploaded, you will see the dataset available on the right pane of your cloud environment.
Now we will be using the package (readr) and the function read_csv to read the dataset.
library(readr)
mydata <-read_csv('customer_segmentation.csv')
##
## ── Column specification ────────────────────────────────────────────────────────
## cols(
## ID = col_double(),
## `CS is helpful` = col_double(),
## Recommend = col_double(),
## `Come again` = col_double(),
## `All Product I need` = col_double(),
## Profesionalism = col_double(),
## Limitation = col_double(),
## `Online grocery` = col_double(),
## delivery = col_double(),
## `Pick up sevice` = col_double(),
## `Find items` = col_double(),
## `other shops` = col_double(),
## Gender = col_double(),
## Age = col_double(),
## Education = col_double()
## )
In the following step, you will standardize your data(i.e., data with a mean of 0 and a standard deviation of 1). You can use the scale function from the R environment which is a generic function whose default method centers and/or scales the columns of a numeric matrix.
Hierarchical clustering (using the function hclust) is an informative way to visualize the data.
We will see if we could discover subgroups among the variables or among the observations.
use = scale(mydata[,-c(1)], center = TRUE, scale = TRUE)
dist = dist(use)
d <- dist(as.matrix(dist)) # find distance matrix
seg.hclust <- hclust(d) # apply hirarchical clustering
library(ggplot2) # needs no introduction
plot(seg.hclust)
Imagine if your goal is to find some profitable customers to target. Now you will be able to see the number of customers using this algorithm.
groups.3 = cutree(seg.hclust,3)
table(groups.3) #A good first step is to use the table function to see how # many observations are in each cluster
## groups.3
## 1 2 3
## 17 2 3
#In the following step, we will find the members in each cluster or group.
mydata$ID[groups.3 == 1]
## [1] 1 2 3 6 7 8 9 10 11 12 13 14 15 16 17 18 21
mydata$ID[groups.3 == 2]
## [1] 4 22
mydata$ID[groups.3 == 3]
## [1] 5 19 20
#?aggregate
aggregate(mydata,list(groups.3),median)
## Group.1 ID CS is helpful Recommend Come again All Product I need
## 1 1 11 1 1.0 1.0 2
## 2 2 13 3 2.5 1.5 3
## 3 3 19 2 1.0 3.0 3
## Profesionalism Limitation Online grocery delivery Pick up sevice Find items
## 1 1.0 1 2 2 3.0 1
## 2 1.5 2 3 3 2.5 2
## 3 2.0 1 2 3 1.0 2
## other shops Gender Age Education
## 1 2.0 1 2.0 2
## 2 1.5 1 2.5 5
## 3 3.0 2 2.0 2
aggregate(mydata,list(groups.3),mean)
## Group.1 ID CS is helpful Recommend Come again All Product I need
## 1 1 10.76471 1.294118 1.117647 1.235294 1.823529
## 2 2 13.00000 3.000000 2.500000 1.500000 3.000000
## 3 3 14.66667 2.333333 1.666667 2.666667 3.000000
## Profesionalism Limitation Online grocery delivery Pick up sevice Find items
## 1 1.235294 1.352941 2.235294 2.235294 2.705882 1.294118
## 2 1.500000 2.000000 3.000000 3.000000 2.500000 2.000000
## 3 2.333333 2.000000 2.000000 3.000000 1.000000 2.000000
## other shops Gender Age Education
## 1 2.647059 1.176471 2.411765 3.117647
## 2 1.500000 1.000000 2.500000 5.000000
## 3 3.000000 2.000000 2.666667 2.333333
aggregate(mydata[,-1],list(groups.3),median)
## Group.1 CS is helpful Recommend Come again All Product I need Profesionalism
## 1 1 1 1.0 1.0 2 1.0
## 2 2 3 2.5 1.5 3 1.5
## 3 3 2 1.0 3.0 3 2.0
## Limitation Online grocery delivery Pick up sevice Find items other shops
## 1 1 2 2 3.0 1 2.0
## 2 2 3 3 2.5 2 1.5
## 3 1 2 3 1.0 2 3.0
## Gender Age Education
## 1 1 2.0 2
## 2 1 2.5 5
## 3 2 2.0 2
aggregate(mydata[,-1],list(groups.3),mean)
## Group.1 CS is helpful Recommend Come again All Product I need Profesionalism
## 1 1 1.294118 1.117647 1.235294 1.823529 1.235294
## 2 2 3.000000 2.500000 1.500000 3.000000 1.500000
## 3 3 2.333333 1.666667 2.666667 3.000000 2.333333
## Limitation Online grocery delivery Pick up sevice Find items other shops
## 1 1.352941 2.235294 2.235294 2.705882 1.294118 2.647059
## 2 2.000000 3.000000 3.000000 2.500000 2.000000 1.500000
## 3 2.000000 2.000000 3.000000 1.000000 2.000000 3.000000
## Gender Age Education
## 1 1.176471 2.411765 3.117647
## 2 1.000000 2.500000 5.000000
## 3 2.000000 2.666667 2.333333
cluster_means <- aggregate(mydata[,-1],list(groups.3),mean)
write.csv(groups.3, "clusterID.csv")
write.csv(cluster_means, "cluster_means.csv")
First, select the files (“clusterID.csv” & “cluster_means.csv”) and put a checkmark before each file.
Second, click the gear icon on the right side of your pane and export the data.
Imagine if your goal is to find some profitable customers to target. Now using the mean function or the median function, you will be able to see the characteristics of each sub-group. Now it is time to use your domain expertise.
How many observations do we have in each cluster? Answer: Your answer here:
We can look at the medians (or means) for the variables in each cluster. Why is this important?
Answer: Your answer here:
Answer: Your answer here:
Answer: Your answer here:
Answer: Your answer here:
O. The aggregate function is well suited for this task. Should we use mydata or mydata[,-1] along with the aggregate function? Why? Hint: see the results on my tutorial.
Cluster analysis - reading (p.385-p.399) https://faculty.marshall.usc.edu/gareth-james/ISL/ISLR%20Seventh%20Printing.pdf
Comparison of similarity coefficients used for cluster analysis with dominant markers in maize (Zea mays L) https://www.scielo.br/scielo.php?script=sci_arttext&pid=S1415-47572004000100014&lng=en&nrm=iso
Principal Component Methods in R: Practical Guide http://www.sthda.com/english/articles/31-principal-component-methods-in-r-practical-guide/118-principal-component-analysis-in-r-prcomp-vs-princomp/