In this project, you will analyze a dataset containing data on various customers' annual spending amounts (reported in monetary units) of diverse product categories for internal structure. One goal of this project is to best describe the variation in the different types of customers that a wholesale distributor interacts with. Doing so would equip the distributor with insight into how to best structure their delivery service to meet the needs of each customer.
The dataset for this project can be found on the UCI Machine Learning Repository. For the purposes of this project, the features 'Channel' and 'Region' will be excluded in the analysis — with focus instead on the six product categories recorded for customers.
Run the code block below to load the wholesale customers dataset, along with a few of the necessary Python libraries required for this project. You will know the dataset loaded successfully if the size of the dataset is reported.
# Import libraries necessary for this project
import numpy as np
import pandas as pd
import renders as rs
import seaborn as sns
from IPython.display import display # Allows the use of display() for DataFrames
# Show matplotlib plots inline (nicely formatted in the notebook)
%matplotlib inline
# Load the wholesale customers dataset
try:
data = pd.read_csv("customers.csv")
data.drop(['Region', 'Channel'], axis = 1, inplace = True)
print "Wholesale customers dataset has {} samples with {} features each.".format(*data.shape)
except:
print "Dataset could not be loaded. Is the dataset missing?"Wholesale customers dataset has 440 samples with 6 features each.
In this section, you will begin exploring the data through visualizations and code to understand how each feature is related to the others. You will observe a statistical description of the dataset, consider the relevance of each feature, and select a few sample data points from the dataset which you will track through the course of this project.
Run the code block below to observe a statistical description of the dataset. Note that the dataset is composed of six important product categories: 'Fresh', 'Milk', 'Grocery', 'Frozen', 'Detergents_Paper', and 'Delicatessen'. Consider what each category represents in terms of products you could purchase.
# Display a description of the dataset
display(data.describe())| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| count | 440.000000 | 440.000000 | 440.000000 | 440.000000 | 440.000000 | 440.000000 |
| mean | 12000.297727 | 5796.265909 | 7951.277273 | 3071.931818 | 2881.493182 | 1524.870455 |
| std | 12647.328865 | 7380.377175 | 9503.162829 | 4854.673333 | 4767.854448 | 2820.105937 |
| min | 3.000000 | 55.000000 | 3.000000 | 25.000000 | 3.000000 | 3.000000 |
| 25% | 3127.750000 | 1533.000000 | 2153.000000 | 742.250000 | 256.750000 | 408.250000 |
| 50% | 8504.000000 | 3627.000000 | 4755.500000 | 1526.000000 | 816.500000 | 965.500000 |
| 75% | 16933.750000 | 7190.250000 | 10655.750000 | 3554.250000 | 3922.000000 | 1820.250000 |
| max | 112151.000000 | 73498.000000 | 92780.000000 | 60869.000000 | 40827.000000 | 47943.000000 |
To get a better understanding of the customers and how their data will transform through the analysis, it would be best to select a few sample data points and explore them in more detail. In the code block below, add three indices of your choice to the indices list which will represent the customers to track. It is suggested to try different sets of samples until you obtain customers that vary significantly from one another.
# Select three indices of your choice you wish to sample from the dataset
indices = [71,221,323]
# Create a DataFrame of the chosen samples
samples = pd.DataFrame(data.loc[indices], columns = data.keys()).reset_index(drop = True)
print "Chosen samples of wholesale customers dataset:"
display(samples)
# Generate a heatmap of the percentiles
pcts = rs.percentile_heatmap(indices, data)
# Display the percentiles data
print "Percentiles Data:"
display(pcts)Chosen samples of wholesale customers dataset:
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 0 | 18291 | 1266 | 21042 | 5373 | 4173 | 14472 |
| 1 | 5396 | 7503 | 10646 | 91 | 4167 | 239 |
| 2 | 13360 | 944 | 11593 | 915 | 1679 | 573 |
Percentiles Data:
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 71 | 78.9 | 20.0 | 92.5 | 83.9 | 77.0 | 99.5 |
| 221 | 35.9 | 76.8 | 75.0 | 2.7 | 76.8 | 13.9 |
| 323 | 68.0 | 12.6 | 78.6 | 33.6 | 61.8 | 33.6 |
Consider the total purchase cost of each product category and the statistical description of the dataset above for your sample customers.
What kind of establishment (customer) could each of the three samples you've chosen represent?
Hint: Examples of establishments include places like markets, cafes, and retailers, among many others. Avoid using names for establishments, such as saying "McDonalds" when describing a sample customer as a restaurant.
Answer:
-
Choice 1 (index = 71): Retailer or Market
- Fresh: > mean, > 75%
- Milk: < mean, ~= 25%
- Grocery: > mean, > 75%
- Frozen: > mean, > 75%
- Detergents Paper: > mean, ~= 75%
- Delicatessen: > mean, ~= 75% (x3 std deviations)
-
Choice 2 (index = 221): Cafe
- Fresh: < mean, between 25% and 50%
- Milk: > mean, ~= 75%
- Grocery: > mean, ~= 75%
- Frozen: < mean, < 75%
- Detergents Paper: > mean, ~= 75%
- Delicatessen: < mean, ~= 25%
-
Choice 3 (index = 323): Market
- Fresh: > mean, ~= 65%
- Milk: < mean, ~= 15%
- Grocery: > mean, > 75%
- Frozen: < mean, ~= 25%
- Detergents Paper: < mean, < 75%, > 50%
- Delicatessen: < mean, > 25%, < 50%
One interesting thought to consider is if one (or more) of the six product categories is actually relevant for understanding customer purchasing. That is to say, is it possible to determine whether customers purchasing some amount of one category of products will necessarily purchase some proportional amount of another category of products? We can make this determination quite easily by training a supervised regression learner on a subset of the data with one feature removed, and then score how well that model can predict the removed feature.
In the code block below, you will need to implement the following:
- Assign
new_dataa copy of the data by removing a feature of your choice using theDataFrame.dropfunction. - Use
sklearn.cross_validation.train_test_splitto split the dataset into training and testing sets.- Use the removed feature as your target label. Set a
test_sizeof0.25and set arandom_state.
- Use the removed feature as your target label. Set a
- Import a decision tree regressor, set a
random_state, and fit the learner to the training data. - Report the prediction score of the testing set using the regressor's
scorefunction.
from sklearn.cross_validation import train_test_split
from sklearn.tree import DecisionTreeRegressor
column_to_drop = 'Grocery'
# Make a copy of the DataFrame, using the 'drop' function to drop the given feature
new_data = data.drop([column_to_drop], axis = 1)
# Split the data into training and testing sets using the given feature as the target
X_train, X_test, y_train, y_test = train_test_split(new_data, data[column_to_drop],
test_size=0.25, random_state=42)
# Create a decision tree regressor and fit it to the training set
regressor = DecisionTreeRegressor(random_state=0)
fit = regressor.fit(X_train, y_train)
# Report the score of the prediction using the testing set
score = fit.score(X_test, y_test)
# Check the feature importances & score
print "The R^2 score when dropping the '{}' column is {}".format(column_to_drop, score)
# Visually inspect the correlations in the data
rs.correlations_plot(data)The R^2 score when dropping the 'Grocery' column is 0.699248196675
Which feature did you attempt to predict? What was the reported prediction score? Is this feature is necessary for identifying customers' spending habits?
Hint: The coefficient of determination, R^2, is scored between 0 and 1, with 1 being a perfect fit. A negative R^2 implies the model fails to fit the data.
Answer:
The selected feature is Grocery. This feature was selected because it has an R^2 score of 0.67 which indicates that we are closer to a perfect fit when predicting / trying to identify customers spending habits.
A higher R^2 score means we can drop that feature and still be able to predict it from the other features.
Other features, such as 'Fresh' and 'Milk', were tested against and either had a low or negative score. Another feature that had a high fit score is 'Detergents_Paper'.
To get a better understanding of the dataset, we can construct a scatter matrix of each of the six product features present in the data. If you found that the feature you attempted to predict above is relevant for identifying a specific customer, then the scatter matrix below may not show any correlation between that feature and the others. Conversely, if you believe that feature is not relevant for identifying a specific customer, the scatter matrix might show a correlation between that feature and another feature in the data. Run the code block below to produce a scatter matrix.
# Produce a scatter matrix for each pair of features in the data
# (changed original codebase to use seaborn)
mx_plot = sns.pairplot(data, diag_kind="kde", size=1.6)
mx_plot.set(xticklabels=[])<seaborn.axisgrid.PairGrid at 0x11bab9dd0>
Are there any pairs of features which exhibit some degree of correlation? Does this confirm or deny your suspicions about the relevance of the feature you attempted to predict? How is the data for those features distributed?
Hint: Is the data normally distributed? Where do most of the data points lie?
Answer:
Detergents Paper and Grocery exhibit a higher degree of correlation that other figures shown in the scatter matrix above.
This confirms the previous answer about the relevance of 'Grocery' and 'Detergents_Paper' features. The data however doesn't seem to be normally distributed.
Other interesting figures include 'Milk' and 'Detergents_Paper' or 'Grocery' and 'Milk' but the correlation degrees are not as strong.
In this section, you will preprocess the data to create a better representation of customers by performing a scaling on the data and detecting (and optionally removing) outliers. Preprocessing data is often times a critical step in assuring that results you obtain from your analysis are significant and meaningful.
If data is not normally distributed, especially if the mean and median vary significantly (indicating a large skew), it is most often appropriate to apply a non-linear scaling — particularly for financial data. One way to achieve this scaling is by using a Box-Cox test, which calculates the best power transformation of the data that reduces skewness. A simpler approach which can work in most cases would be applying the natural logarithm.
In the code block below, you will need to implement the following:
- Assign a copy of the data to
log_dataafter applying a logarithm scaling. Use thenp.logfunction for this. - Assign a copy of the sample data to
log_samplesafter applying a logrithm scaling. Again, usenp.log.
from scipy import stats
# Scale the data using the natural logarithm
log_data = np.log(data)
# Scale the sample data using the natural logarithm
log_samples = np.log(samples)
# Produce a scatter matrix for each pair of newly-transformed features
# (changed original codebase to use seaborn)
sns.pairplot(log_data, diag_kind="kde", size=1.6)<seaborn.axisgrid.PairGrid at 0x123736e50>
After applying a natural logarithm scaling to the data, the distribution of each feature should appear much more normal. For any pairs of features you may have identified earlier as being correlated, observe here whether that correlation is still present (and whether it is now stronger or weaker than before).
Run the code below to see how the sample data has changed after having the natural logarithm applied to it.
# Display the log-transformed sample data
display(log_samples)| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 0 | 9.814164 | 7.143618 | 9.954276 | 8.589142 | 8.336390 | 9.579971 |
| 1 | 8.593413 | 8.923058 | 9.272940 | 4.510860 | 8.334952 | 5.476464 |
| 2 | 9.500020 | 6.850126 | 9.358157 | 6.818924 | 7.425954 | 6.350886 |
Detecting outliers in the data is extremely important in the data preprocessing step of any analysis. The presence of outliers can often skew results which take into consideration these data points. There are many "rules of thumb" for what constitutes an outlier in a dataset. Here, we will use Tukey's Method for identfying outliers: An outlier step is calculated as 1.5 times the interquartile range (IQR). A data point with a feature that is beyond an outlier step outside of the IQR for that feature is considered abnormal.
In the code block below, you will need to implement the following:
- Assign the value of the 25th percentile for the given feature to
Q1. Usenp.percentilefor this. - Assign the value of the 75th percentile for the given feature to
Q3. Again, usenp.percentile. - Assign the calculation of an outlier step for the given feature to
step. - Optionally remove data points from the dataset by adding indices to the
outlierslist.
NOTE: If you choose to remove any outliers, ensure that the sample data does not contain any of these points!
Once you have performed this implementation, the dataset will be stored in the variable good_data.
from collections import Counter
# OPTIONAL: Select the indices for data points you wish to remove
outliers = []
# For each feature find the data points with extreme high or low values
for feature in log_data.keys():
# Calculate Q1 (25th percentile of the data) for the given feature
Q1 = np.percentile(log_data[feature], 25)
# Calculate Q3 (75th percentile of the data) for the given feature
Q3 = np.percentile(log_data[feature], 75)
# Use the interquartile range to calculate an outlier step (1.5 times the interquartile range)
step = 1.5 * (Q3 - Q1)
# Display the outliers
print "Data points considered outliers for the feature '{}':".format(feature)
feature_outliers = log_data[~((log_data[feature] >= Q1 - step) & (log_data[feature] <= Q3 + step))]
display(feature_outliers)
# Add the outliers indices to the list of outliers for removal later on
outliers.extend(list(feature_outliers.index.values))
# Find the data points that where considered outliers for more than one feature
multi_feature_outliers = (Counter(outliers) - Counter(set(outliers))).keys()
# Remove the outliers, if any were specified
good_data = log_data.drop(log_data.index[multi_feature_outliers]).reset_index(drop = True)Data points considered outliers for the feature 'Fresh':
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 65 | 4.442651 | 9.950323 | 10.732651 | 3.583519 | 10.095388 | 7.260523 |
| 66 | 2.197225 | 7.335634 | 8.911530 | 5.164786 | 8.151333 | 3.295837 |
| 81 | 5.389072 | 9.163249 | 9.575192 | 5.645447 | 8.964184 | 5.049856 |
| 95 | 1.098612 | 7.979339 | 8.740657 | 6.086775 | 5.407172 | 6.563856 |
| 96 | 3.135494 | 7.869402 | 9.001839 | 4.976734 | 8.262043 | 5.379897 |
| 128 | 4.941642 | 9.087834 | 8.248791 | 4.955827 | 6.967909 | 1.098612 |
| 171 | 5.298317 | 10.160530 | 9.894245 | 6.478510 | 9.079434 | 8.740337 |
| 193 | 5.192957 | 8.156223 | 9.917982 | 6.865891 | 8.633731 | 6.501290 |
| 218 | 2.890372 | 8.923191 | 9.629380 | 7.158514 | 8.475746 | 8.759669 |
| 304 | 5.081404 | 8.917311 | 10.117510 | 6.424869 | 9.374413 | 7.787382 |
| 305 | 5.493061 | 9.468001 | 9.088399 | 6.683361 | 8.271037 | 5.351858 |
| 338 | 1.098612 | 5.808142 | 8.856661 | 9.655090 | 2.708050 | 6.309918 |
| 353 | 4.762174 | 8.742574 | 9.961898 | 5.429346 | 9.069007 | 7.013016 |
| 355 | 5.247024 | 6.588926 | 7.606885 | 5.501258 | 5.214936 | 4.844187 |
| 357 | 3.610918 | 7.150701 | 10.011086 | 4.919981 | 8.816853 | 4.700480 |
| 412 | 4.574711 | 8.190077 | 9.425452 | 4.584967 | 7.996317 | 4.127134 |
Data points considered outliers for the feature 'Milk':
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 86 | 10.039983 | 11.205013 | 10.377047 | 6.894670 | 9.906981 | 6.805723 |
| 98 | 6.220590 | 4.718499 | 6.656727 | 6.796824 | 4.025352 | 4.882802 |
| 154 | 6.432940 | 4.007333 | 4.919981 | 4.317488 | 1.945910 | 2.079442 |
| 356 | 10.029503 | 4.897840 | 5.384495 | 8.057377 | 2.197225 | 6.306275 |
Data points considered outliers for the feature 'Grocery':
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 75 | 9.923192 | 7.036148 | 1.098612 | 8.390949 | 1.098612 | 6.882437 |
| 154 | 6.432940 | 4.007333 | 4.919981 | 4.317488 | 1.945910 | 2.079442 |
Data points considered outliers for the feature 'Frozen':
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 38 | 8.431853 | 9.663261 | 9.723703 | 3.496508 | 8.847360 | 6.070738 |
| 57 | 8.597297 | 9.203618 | 9.257892 | 3.637586 | 8.932213 | 7.156177 |
| 65 | 4.442651 | 9.950323 | 10.732651 | 3.583519 | 10.095388 | 7.260523 |
| 145 | 10.000569 | 9.034080 | 10.457143 | 3.737670 | 9.440738 | 8.396155 |
| 175 | 7.759187 | 8.967632 | 9.382106 | 3.951244 | 8.341887 | 7.436617 |
| 264 | 6.978214 | 9.177714 | 9.645041 | 4.110874 | 8.696176 | 7.142827 |
| 325 | 10.395650 | 9.728181 | 9.519735 | 11.016479 | 7.148346 | 8.632128 |
| 420 | 8.402007 | 8.569026 | 9.490015 | 3.218876 | 8.827321 | 7.239215 |
| 429 | 9.060331 | 7.467371 | 8.183118 | 3.850148 | 4.430817 | 7.824446 |
| 439 | 7.932721 | 7.437206 | 7.828038 | 4.174387 | 6.167516 | 3.951244 |
Data points considered outliers for the feature 'Detergents_Paper':
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 75 | 9.923192 | 7.036148 | 1.098612 | 8.390949 | 1.098612 | 6.882437 |
| 161 | 9.428190 | 6.291569 | 5.645447 | 6.995766 | 1.098612 | 7.711101 |
Data points considered outliers for the feature 'Delicatessen':
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| 66 | 2.197225 | 7.335634 | 8.911530 | 5.164786 | 8.151333 | 3.295837 |
| 109 | 7.248504 | 9.724899 | 10.274568 | 6.511745 | 6.728629 | 1.098612 |
| 128 | 4.941642 | 9.087834 | 8.248791 | 4.955827 | 6.967909 | 1.098612 |
| 137 | 8.034955 | 8.997147 | 9.021840 | 6.493754 | 6.580639 | 3.583519 |
| 142 | 10.519646 | 8.875147 | 9.018332 | 8.004700 | 2.995732 | 1.098612 |
| 154 | 6.432940 | 4.007333 | 4.919981 | 4.317488 | 1.945910 | 2.079442 |
| 183 | 10.514529 | 10.690808 | 9.911952 | 10.505999 | 5.476464 | 10.777768 |
| 184 | 5.789960 | 6.822197 | 8.457443 | 4.304065 | 5.811141 | 2.397895 |
| 187 | 7.798933 | 8.987447 | 9.192075 | 8.743372 | 8.148735 | 1.098612 |
| 203 | 6.368187 | 6.529419 | 7.703459 | 6.150603 | 6.860664 | 2.890372 |
| 233 | 6.871091 | 8.513988 | 8.106515 | 6.842683 | 6.013715 | 1.945910 |
| 285 | 10.602965 | 6.461468 | 8.188689 | 6.948897 | 6.077642 | 2.890372 |
| 289 | 10.663966 | 5.655992 | 6.154858 | 7.235619 | 3.465736 | 3.091042 |
| 343 | 7.431892 | 8.848509 | 10.177932 | 7.283448 | 9.646593 | 3.610918 |
Are there any data points considered outliers for more than one feature? Should these data points be removed from the dataset? If any data points were added to the outliers list to be removed, explain why.
Answer:
Yes, there are data points that were considered outliers for more than one feature. These points have the following indices: [128, 65, 154, 75, 66].
We can remove data points based on being outliers for more than a single feature. Although, we could also argue that outliers for certain features such as 'Grocery' or 'Detergents_Paper' are the only ones we need to remove since these columns biggest correlation score when fitting.
In this submission, we chosen to remove the data points that are considered outliers for multiple features as implemented in the code above.
In this section you will use principal component analysis (PCA) to draw conclusions about the underlying structure of the wholesale customer data. Since using PCA on a dataset calculates the dimensions which best maximize variance, we will find which compound combinations of features best describe customers.
Now that the data has been scaled to a more normal distribution and has had any necessary outliers removed, we can now apply PCA to the good_data to discover which dimensions about the data best maximize the variance of features involved. In addition to finding these dimensions, PCA will also report the explained variance ratio of each dimension — how much variance within the data is explained by that dimension alone. Note that a component (dimension) from PCA can be considered a new "feature" of the space, however it is a composition of the original features present in the data.
In the code block below, you will need to implement the following:
- Import
sklearn.decomposition.PCAand assign the results of fitting PCA in six dimensions withgood_datatopca. - Apply a PCA transformation of the sample log-data
log_samplesusingpca.transform, and assign the results topca_samples.
from sklearn.decomposition import PCA
# Apply PCA by fitting the good data with the same number of dimensions as features
pca = PCA(n_components=len(good_data.keys())).fit(good_data)
# Transform the sample log-data using the PCA fit above
pca_samples = pca.transform(log_samples)
# Generate PCA results plot
pca_results = rs.pca_results(good_data, pca)
# print "The cumulative variance: "
# display(pca_results['Explained Variance'].cumsum())
How much variance in the data is explained in total by the first and second principal component? What about the first four principal components? Using the visualization provided above, discuss what the first four dimensions best represent in terms of customer spending.
Hint: A positive increase in a specific dimension corresponds with an increase of the positive-weighted features and a decrease of the negative-weighted features. The rate of increase or decrease is based on the indivdual feature weights.
Answer:
Explained Variances of first and second principal components:
| Principal Component | Explained Variance |
|---|---|
| First PC | 0.4430 |
| Second PC | 0.2638 |
The cummulative variances are as follows:
| Dimension | Cumulative Variance |
|---|---|
| Dimension 1 | 0.4430 |
| Dimension 2 | 0.7068 |
| Dimension 3 | 0.8299 |
| Dimension 4 | 0.9311 |
| Dimension 5 | 0.9796 |
| Dimension 6 | 1.0000 |
-
First Principal Component:
- This component has highly positive weights for
Detergents_Paper,MilkandGrocery. It also shows negative weights forFreshandFrozen. - This patterns might represent spending on household products that are purchsed together.
- This component has highly positive weights for
-
Second Principal Component:
- This component shows a lot of variances in customers who purchase
Fresh,FrozenandDelicatessen, which is the opposite of the first component. - This component might represent spending habits of customers who aren't purchasing household items.
- This component shows a lot of variances in customers who purchase
-
Third Principal Component:
- This component shows a lot of variances in customers who purchase
FrozenandDelicatessenbut notFresh. - This component represents spending habits similar to those of component 2 (which is the opposite of the first component) however without a high variance in
Freshpurchases.
- This component shows a lot of variances in customers who purchase
-
Fourth Principal Component:
- This component has highly positive weights for
FrozenandDetergents_Paperand negative weights forFreshandDelicatessen. - This component represents spending habits similar to those of component 2 and 3 but opposite to component 2 when it comes to purchasing
Freshand opposite to 2 and 3 forDelicatessen.
- This component has highly positive weights for
Run the code below to see how the log-transformed sample data has changed after having a PCA transformation applied to it in six dimensions. Observe the numerical value for the first four dimensions of the sample points. Consider if this is consistent with your initial interpretation of the sample points.
# Display sample log-data after having a PCA transformation applied
display(pd.DataFrame(np.round(pca_samples, 4), columns = pca_results.index.values))| Dimension 1 | Dimension 2 | Dimension 3 | Dimension 4 | Dimension 5 | Dimension 6 | |
|---|---|---|---|---|---|---|
| 0 | 1.4462 | 2.7836 | 1.1083 | -0.3929 | 1.6836 | -1.5910 |
| 1 | 2.1771 | -1.8735 | -1.8437 | -1.1368 | -0.3514 | 0.1295 |
| 2 | 0.2672 | -0.0434 | -1.0736 | -0.1957 | 0.8635 | -1.3171 |
When using principal component analysis, one of the main goals is to reduce the dimensionality of the data — in effect, reducing the complexity of the problem. Dimensionality reduction comes at a cost: Fewer dimensions used implies less of the total variance in the data is being explained. Because of this, the cumulative explained variance ratio is extremely important for knowing how many dimensions are necessary for the problem. Additionally, if a signifiant amount of variance is explained by only two or three dimensions, the reduced data can be visualized afterwards.
In the code block below, you will need to implement the following:
- Assign the results of fitting PCA in two dimensions with
good_datatopca. - Apply a PCA transformation of
good_datausingpca.transform, and assign the reuslts toreduced_data. - Apply a PCA transformation of the sample log-data
log_samplesusingpca.transform, and assign the results topca_samples.
# Apply PCA by fitting the good data with only two dimensions
pca = PCA(n_components=2).fit(good_data)
# Transform the good data using the PCA fit above
reduced_data = pca.transform(good_data)
# Transform the sample log-data using the PCA fit above
pca_samples = pca.transform(log_samples)
# Create a DataFrame for the reduced data
reduced_data = pd.DataFrame(reduced_data, columns = ['Dimension 1', 'Dimension 2'])Run the code below to see how the log-transformed sample data has changed after having a PCA transformation applied to it using only two dimensions. Observe how the values for the first two dimensions remains unchanged when compared to a PCA transformation in six dimensions.
# Display sample log-data after applying PCA transformation in two dimensions
display(pd.DataFrame(np.round(pca_samples, 4), columns = ['Dimension 1', 'Dimension 2']))| Dimension 1 | Dimension 2 | |
|---|---|---|
| 0 | 1.4462 | 2.7836 |
| 1 | 2.1771 | -1.8735 |
| 2 | 0.2672 | -0.0434 |
In this section, you will choose to use either a K-Means clustering algorithm or a Gaussian Mixture Model clustering algorithm to identify the various customer segments hidden in the data. You will then recover specific data points from the clusters to understand their significance by transforming them back into their original dimension and scale.
What are the advantages to using a K-Means clustering algorithm? What are the advantages to using a Gaussian Mixture Model clustering algorithm? Given your observations about the wholesale customer data so far, which of the two algorithms will you use and why?
Answer:
K-Means Clustering:
-
Advantages
- With a large number of features, K-means can be computationally faster if K is small.
- K-Means could result in tighter clusters than hierarchical clustering.
-
Disadvantages
- Difficulty in comparing quality of the clusters produced.
- Since preset
Kvalue is required, it could be difficult to predict whichKis best. - Assumes data is globular with doesn't always occur in real-life data.
Gaussian Mixture:
-
Advantages
- Capable of "soft" classification
- Works with different distributions of the data
-
Disadvantages
- Could fail if the dimensionality of the problem is too high
Chosen Algorithim:
In this submission, the chosen algorithim is Gaussian Mixture Model because of its ability to apply "soft" classification and since we've reduced the dimensionality of the problem with PCA, GMM should be able to do the job.
If our dataset was considerably larger, we could reconsider and use K-means.
Depending on the problem, the number of clusters that you expect to be in the data may already be known. When the number of clusters is not known a priori, there is no guarantee that a given number of clusters best segments the data, since it is unclear what structure exists in the data — if any. However, we can quantify the "goodness" of a clustering by calculating each data point's silhouette coefficient. The silhouette coefficient for a data point measures how similar it is to its assigned cluster from -1 (dissimilar) to 1 (similar). Calculating the mean silhouette coefficient provides for a simple scoring method of a given clustering.
In the code block below, you will need to implement the following:
- Fit a clustering algorithm to the
reduced_dataand assign it toclusterer. - Predict the cluster for each data point in
reduced_datausingclusterer.predictand assign them topreds. - Find the cluster centers using the algorithm's respective attribute and assign them to
centers. - Predict the cluster for each sample data point in
pca_samplesand assign themsample_preds. - Import sklearn.metrics.silhouette_score and calculate the silhouette score of
reduced_dataagainstpreds.- Assign the silhouette score to
scoreand print the result.
- Assign the silhouette score to
from sklearn import mixture
from sklearn.metrics import silhouette_score
# Apply your clustering algorithm of choice to the reduced data
clusterer = mixture.GMM(n_components=2)
gmm = clusterer.fit(reduced_data)
# Predict the cluster for each data point
preds = gmm.predict(reduced_data)
# Find the cluster centers
centers = gmm.means_
# Predict the cluster for each transformed sample data point
sample_preds = gmm.predict(pca_samples)
# Calculate the mean silhouette coefficient for the number of clusters chosen
score = silhouette_score(reduced_data, preds, metric='euclidean')
# Print the score result
print "The silhouette_score is: {}".format(score)The silhouette_score is: 0.411818864386
Report the silhouette score for several cluster numbers you tried. Of these, which number of clusters has the best silhouette score?
Answer:
| Number of Clusters | Silhouette Scoree |
|---|---|
| 2 | 0.4118 |
| 3 | 0.3735 |
| 4 | 0.3333 |
| 5 | 0.2870 |
As we can see in the table above, the best number of clusters is 2.
As we increase the number of clusters, the Silhouette score gets closer to 0 which indicate overlapping clusters. Where as when using 2 clusters, we get score values closer to 1 indicating dense and well seperated clusters.
Once you've chosen the optimal number of clusters for your clustering algorithm using the scoring metric above, you can now visualize the results by executing the code block below. Note that, for experimentation purposes, you are welcome to adjust the number of clusters for your clustering algorithm to see various visualizations. The final visualization provided should, however, correspond with the optimal number of clusters.
# Display the results of the clustering from implementation
rs.cluster_results(reduced_data, preds, centers, pca_samples)
Each cluster present in the visualization above has a central point. These centers (or means) are not specifically data points from the data, but rather the averages of all the data points predicted in the respective clusters. For the problem of creating customer segments, a cluster's center point corresponds to the average customer of that segment. Since the data is currently reduced in dimension and scaled by a logarithm, we can recover the representative customer spending from these data points by applying the inverse transformations.
In the code block below, you will need to implement the following:
- Apply the inverse transform to
centersusingpca.inverse_transformand assign the new centers tolog_centers. - Apply the inverse function of
np.logtolog_centersusingnp.expand assign the true centers totrue_centers.
# Inverse transform the centers
log_centers = pca.inverse_transform(centers)
# Exponentiate the centers
true_centers = np.exp(log_centers)
# Display the true centers
segments = ['Segment {}'.format(i) for i in range(0,len(centers))]
true_centers = pd.DataFrame(np.round(true_centers), columns = data.keys())
true_centers.index = segments
print "True centers: "
display(true_centers)
# Compare the true centers to the mean and median of the data
print "Compared to the mean: "
display(true_centers - data.mean().round())
print "Compared to the median: "
display(true_centers - data.median().round())True centers:
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| Segment 0 | 4316.0 | 6347.0 | 9555.0 | 1036.0 | 3046.0 | 945.0 |
| Segment 1 | 8812.0 | 2052.0 | 2689.0 | 2058.0 | 337.0 | 712.0 |
Compared to the mean:
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| Segment 0 | -7684.0 | 551.0 | 1604.0 | -2036.0 | 165.0 | -580.0 |
| Segment 1 | -3188.0 | -3744.0 | -5262.0 | -1014.0 | -2544.0 | -813.0 |
Compared to the median:
| Fresh | Milk | Grocery | Frozen | Detergents_Paper | Delicatessen | |
|---|---|---|---|---|---|---|
| Segment 0 | -4188.0 | 2720.0 | 4799.0 | -490.0 | 2230.0 | -21.0 |
| Segment 1 | 308.0 | -1575.0 | -2067.0 | 532.0 | -479.0 | -254.0 |
Consider the total purchase cost of each product category for the representative data points above, and reference the statistical description of the dataset at the beginning of this project. What set of establishments could each of the customer segments represent?
Hint: A customer who is assigned to 'Cluster X' should best identify with the establishments represented by the feature set of 'Segment X'.
Answer:
The average customer in Segment 0 has an above average spending on Milk, Grocery and Detergents_Paper. This could represent customers who purchase a lot of household items which possibly includes Cafes or Grocery Stores.
Segment 1 shows the opposite spending habits by being far below the average in those categories mentioned above. This cluster could represent Retails.
For each sample point, which customer segment from Question 8 best represents it? Are the predictions for each sample point consistent with this?
Run the code block below to find which cluster each sample point is predicted to be.
# Display the predictions
for i, (pred, sample) in enumerate(zip(sample_preds, data_samples.values)):
print "Sample point", i, "predicted to be in Cluster", pred
print "Data for this sample: ", sample
print "-----"Sample point 0 predicted to be in Cluster 0
Data for this sample: [ 5909 23527 13699 10155 830 3636]
-----
Sample point 1 predicted to be in Cluster 0
Data for this sample: [10766 1175 2067 2096 301 167]
-----
Sample point 2 predicted to be in Cluster 1
Data for this sample: [27380 7184 12311 2809 4621 1022]
-----
Answer:
For sample point 0, the values for 'Grocery', 'Milk' are above average and mimic the Segment 0 center in those categories — so, the predicted cluster seems to be consistent with the sample.
The same behaviour is seem when inspecting Fresh and Milk purchases in sample point 2 (3rd point from above) which is also consistent with the initial findings.
Companies often run A/B tests when making small changes to their products or services. If the wholesale distributor wanted to change its delivery service from 5 days a week to 3 days a week, how would you use the structure of the data to help them decide on a group of customers to test?
Hint: Would such a change in the delivery service affect all customers equally? How could the distributor identify who it affects the most?
Answer:
The assumption is that Cafes and Diners which means they are likely to be effected the delivery service change.
Since retails make larger order and possibly have more space to stock items, they are likely to make their orders once or twice a week.
The distributor can identify which customer segment would be most affected by the delivery service change can be found by running an A/B test on the patterns of the orders for each segment.
By inspecting the number of times each of those customer segments makes an order (per week); we can assume that customers who need more than 3 deliveries per week to be affected by the service change. We can then count the number of customers affected from each segment and make a decision on whether or not to go ahead with the change.
Assume the wholesale distributor wanted to predict a new feature for each customer based on the purchasing information available. How could the wholesale distributor use the structure of the clustering data you've found to assist a supervised learning analysis?
Hint: What other input feature could the supervised learner use besides the six product features to help make a prediction?
Answer:
The segment classification (or cluster) itself could be another feature used to help with making predictions in a supervised learning model.
For example, let's consider than we fed back the cluster of each customer into the original dataset. We can then run a supervised learning model to predict if a certain customer would be interested in opting in for a new service the distributor is thinking about offering.
At the beginning of this project, it was discussed that the 'Channel' and 'Region' features would be excluded from the dataset so that the customer product categories were emphasized in the analysis. By reintroducing the 'Channel' feature to the dataset, an interesting structure emerges when considering the same PCA dimensionality reduction applied earlier on to the original dataset.
Run the code block below to see how each data point is labeled either 'HoReCa' (Hotel/Restaurant/Cafe) or 'Retail' the reduced space. In addition, you will find the sample points are circled in the plot, which will identify their labeling.
# Display the clustering results based on 'Channel' data
rs.channel_results(reduced_data, multi_feature_outliers, pca_samples[:5])
# Display the clustering results WITHOUT removing the outliers
rs.channel_results(reduced_data, [], pca_samples[:5])
How well does the clustering algorithm and number of clusters you've chosen compare to this underlying distribution of Hotel/Restaurant/Cafe customers to Retailer customers? Are there customer segments that would be classified as purely 'Retailers' or 'Hotels/Restaurants/Cafes' by this distribution? Would you consider these classifications as consistent with your previous definition of the customer segments?
Answer:
The clustering algorithim chosen has correctly identified 4 out of the 5 first samples. Also the initial intuition of these 2 customer segments has turned out to the close to accurate in terms of the kinds of customers they might be.
Since we have used the GMM learning model which uses probability instead of "hard" classification. This means that there are technically no points which are entirely segment 0 or segment 1, rather a probability of being each of those clusters.
I would consider the majority of these classifications to be inline with the real results.