# Predicting customer churn with scikit-learn

#### by Eric Chiang

Learn More Tweet### Customer Churn

"Churn Rate" is a business term describing the rate at which customers leave or cease paying for a product or service. It's a critical figure in many businesses, as it's often the case that acquiring *new* customers is a lot more costly than retaining *existing* ones (in some cases, 5 to 20 times more expensive).

Understanding what keeps customers engaged, therefore, is incredibly valuable, as it is a logical foundation from which to develop retention strategies and roll out operational practices aimed to keep customers from walking out the door. Consequently, there's growing interest among companies to develop better churn-detection techniques, leading many to look to data mining and machine learning for new and creative approaches.

Predicting churn is particularly important for businesses w/ subscription models such as cell phone, cable, or merchant credit card processing plans. But modeling churn has wide reaching applications in many domains. For example, casinos have used predictive models to predict ideal room conditions for keeping patrons at the blackjack table and when to reward unlucky gamblers with front row seats to Celine Dion. Similarly, airlines may offer first class upgrades to complaining customers. The list goes on.

This is a post about modeling customer churn using Python.

#### Wait, don't go!

So what are some of ops strategies that companies employ to prevent churn? Well, reducing churn, it turns out, often requires non-trivial resources. Specialized retention teams are common in many industries and exist expressly to call down lists of at-risk customers to plead for their continued business.

Organizing and running such teams is tough. From an ops perspective, cross-geographic teams must be well organized and trained to respond to a huge spectrum of customer complaints. Customers must be accurately targeted based on churn-risk, and retention treatments must be well-conceived and correspond reasonably to match expected customer value to ensure the economics make sense. Spending \$1,000 on someone who wasn't about to leave can get expensive pretty quickly.

The good news is that we live in the data age and have some pretty great tools at our disposal to help answer these questions. John Forman of MailChimp summarizes this well:

"I work in the e-mail marketing industry for a website called MailChimp.com. We help customers send e-mail newsletters to their audience, and every time someone uses the term 'e-mail blast,' a little part of me dies.

"Why? Because e-mail addresses are no longer black boxes that you lob 'blasts' at like flash grenades. No, in e-mail marketing (as with many other forms of online engagement, including tweets, Facebook posts, and Pinterest campaigns), a business receives feedback on how their audience is engaging at the individual level through click tracking, online purchases, social sharing, and so on. This data is not noise. It characterizes your audience. But to the uninitiated, it might as well be Greek. Or Esperanto."

_{- John Foreman, DataSmart}

Within this frame of mind, efficiently dealing with turnover is an exercise of distinguishing who is likely to churn from who is not using the data at our disposal. The remainder of this post will explore a simple case study to show how Python and its scientific libraries can be used to predict churn and how you might deploy such a solution within operations to guide a retention team.

### The Dataset

The data set I'll be using is a longstanding telecom customer data set which you can download here.

The data is straightforward. Each row represents a subscribing telephone customer. Each column contains customer attributes such as phone number, call minutes used during different times of day, charges incurred for services, lifetime account duration, and whether or not the customer is still a customer.

```
from __future__ import division
import pandas as pd
import numpy as np
churn_df = pd.read_csv('churn.csv')
col_names = churn_df.columns.tolist()
print "Column names:"
print col_names
to_show = col_names[:6] + col_names[-6:]
print "\nSample data:"
churn_df[to_show].head(6)
```

I'll be keeping the statistical model pretty simple for this blog so the feature space is almost unchanged from what you see above. The following code simply drops irrelevant columns and converts strings to boolean values (since models don't handle "yes" and "no" very well). The rest of the numeric columns are left untouched.

```
# Isolate target data
churn_result = churn_df['Churn?']
y = np.where(churn_result == 'True.',1,0)
# We don't need these columns
to_drop = ['State','Area Code','Phone','Churn?']
churn_feat_space = churn_df.drop(to_drop,axis=1)
# 'yes'/'no' has to be converted to boolean values
# NumPy converts these from boolean to 1. and 0. later
yes_no_cols = ["Int'l Plan","VMail Plan"]
churn_feat_space[yes_no_cols] = churn_feat_space[yes_no_cols] == 'yes'
# Pull out features for future use
features = churn_feat_space.columns
X = churn_feat_space.as_matrix().astype(np.float)
# This is important
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
X = scaler.fit_transform(X)
print "Feature space holds %d observations and %d features" % X.shape
print "Unique target labels:", np.unique(y)
```

One slight side note. Many predictors care about the relative size of different features even though those scales might be arbitrary. For instance: the number of points a basketball team scores per game will naturally be a couple orders of magnitude larger than their win percentage. But this doesn't mean that the latter is 100 times less significant. `StandardScaler`

fixes this by normalizing each feature to a range of around 1.0 to -1.0 thereby preventing models from misbehaving. Well, at least for that reason.

Great, I now have a feature space '`X`

' and a set of target values '`y`

'. On to the predictions!

### How good is your model?

Express, test, cycle. A machine learning pipeline should be anything but static. There are always new features to design, new data to use, new classifiers to consider each with unique parameters to tune. And for every change it's critical to be able to ask, *"Is the new version better than the last?"* So how do I do that?

As a good start, cross validation will be used throughout this blog. Cross validation attempts to avoid overfitting (training on and predicting the same datapoint) while still producing a prediction for each observation dataset. This is accomplished by systematically hiding different subsets of the data while training a set of models. After training, each model predicts on the subset that had been hidden to it, emulating multiple train-test splits. When done correctly, every observation will have a 'fair' corresponding prediction.

Here's what that looks like using `scikit-learn`

libraries.

```
from sklearn.cross_validation import KFold
def run_cv(X,y,clf_class,**kwargs):
# Construct a kfolds object
kf = KFold(len(y),n_folds=5,shuffle=True)
y_pred = y.copy()
# Iterate through folds
for train_index, test_index in kf:
X_train, X_test = X[train_index], X[test_index]
y_train = y[train_index]
# Initialize a classifier with key word arguments
clf = clf_class(**kwargs)
clf.fit(X_train,y_train)
y_pred[test_index] = clf.predict(X_test)
return y_pred
```

I've decided to compare three fairly unique algorithms support vector machines, random forest, and k-nearest-neighbors. Nothing fancy here, just passing each to cross validation and determining how often the classifier predicted the correct class.

```
from sklearn.svm import SVC
from sklearn.ensemble import RandomForestClassifier as RF
from sklearn.neighbors import KNeighborsClassifier as KNN
def accuracy(y_true,y_pred):
# NumPy interprets True and False as 1. and 0.
return np.mean(y_true == y_pred)
print "Support vector machines:"
print "%.3f" % accuracy(y, run_cv(X,y,SVC))
print "Random forest:"
print "%.3f" % accuracy(y, run_cv(X,y,RF))
print "K-nearest-neighbors:"
print "%.3f" % accuracy(y, run_cv(X,y,KNN))
```

Random forest won, right?

### Precision and recall

Measurements aren't golden formulas which always spit out high numbers for good models and low numbers for bad ones. Inherently they convey something sentiment about a model's performance, and it's the job of the human designer to determine each number's validity. The problem with accuracy is that outcomes aren't necessarily equal. If my classifier predicted a customer would churn and they didn't, that's not the best but it's forgivable. However, if my classifier predicted a customer would return, I didn't act, and then they churned... that's really bad.

I'll be using another built in `scikit-learn`

function to construction a confusion matrix. A confusion matrix is a way of visualizing predictions made by a classifier and is just a table showing the distribution of predictions for a specific class. The x-axis indicates the true class of each observation (if a customer churned or not) while the y-axis corresponds to the class predicted by the model (if my classifier said a customer would churned or not).

```
from sklearn.metrics import confusion_matrix
y = np.array(y)
class_names = np.unique(y)
confusion_matrices = [
( "Support Vector Machines", confusion_matrix(y,run_cv(X,y,SVC)) ),
( "Random Forest", confusion_matrix(y,run_cv(X,y,RF)) ),
( "K-Nearest-Neighbors", confusion_matrix(y,run_cv(X,y,KNN)) ),
]
# Pyplot code not included to reduce clutter
from churn_display import draw_confusion_matrices
%matplotlib inline
draw_confusion_matrices(confusion_matrices,class_names)
```

An important question to ask might be, *When an individual churns, how often does my classifier predict that correctly?* This measurement is called "recall" and a quick look at these diagrams can demonstrate that random forest is clearly best for this criteria. Out of all the churn cases (outcome "`1`

") random forest correctly retrieved 330 out of 482. This translates to a churn "recall" of about \(68\%\) (\(330 / 482 \approx 2/3\)), far better than support vector machines (\(\approx 50\%\)) or k-nearest-neighbors (\(\approx 35\%\)).

Another question of importance is "precision" or, *When a classifier predicts an individual will churn, how often does that individual actually churn?* The differences in semantic are small from the previous question, but it makes quite a different. Random forest again out preforms the other two at about \(93\%\) precision (330 out of 356) with support vector machines a little behind at about \(87\%\) (235 out of 269). K-nearest-neighbors lags at about \(80\%\).

While, just like accuracy, precision and recall still rank random forest above SVC and KNN, this won't always be true. When different measurements do return a different pecking order, understanding the values and trade-offs of each rating should effect how you proceed.

### Thinking in Probabilities

Decision making often favors probability over simple classifications. There's plainly more information in statements like *"there's a 20% chance of rain tomorrow"* and *"about 55% of test takers pass the California bar exam"* than just saying *"it shouldn't rain tomorrow"* or *"you'll probably pass."* Probability predictions for churn also allow us to gauge a customers expected value, and their expected loss. Who do you want to reach out to first, the client with a 80% churn risk who pays \$20,000 annually, or the client who's worth \$100,000 a year with a 40% risk? How much should you spend on each client?

While I'm moving a bit away from my expertise, being able to ask that question requires producing predictions a little differently. However, `scikit-learn`

makes moving to probabilities easy; my three models have `predict_proba()`

built right into their class objects. This is the same cross validation code with only a few lines changed.

```
def run_prob_cv(X, y, clf_class, **kwargs):
kf = KFold(len(y), n_folds=5, shuffle=True)
y_prob = np.zeros((len(y),2))
for train_index, test_index in kf:
X_train, X_test = X[train_index], X[test_index]
y_train = y[train_index]
clf = clf_class(**kwargs)
clf.fit(X_train,y_train)
# Predict probabilities, not classes
y_prob[test_index] = clf.predict_proba(X_test)
return y_prob
```

### How good is good?

Determining how good a predictor which gives *probabilities* rather than *classes* is a bit more difficult. If I predict there's a 20% likelihood of rain tomorrow I don't get to live out all the possible outcomes of the universe. It either rains or it doesn't.

What helps is that the predictors aren't making one prediction, they're making 3000+. So for every time I predict an event to occur 20% of the time I can see how often those events actually happen. Here's I use `pandas`

to help me compare the predictions made by random forest against the actual outcomes.

```
import warnings
warnings.filterwarnings('ignore')
# Use 10 estimators so predictions are all multiples of 0.1
pred_prob = run_prob_cv(X, y, RF, n_estimators=10)
pred_churn = pred_prob[:,1]
is_churn = y == 1
# Number of times a predicted probability is assigned to an observation
counts = pd.value_counts(pred_churn)
# calculate true probabilities
true_prob = {}
for prob in counts.index:
true_prob[prob] = np.mean(is_churn[pred_churn == prob])
true_prob = pd.Series(true_prob)
# pandas-fu
counts = pd.concat([counts,true_prob], axis=1).reset_index()
counts.columns = ['pred_prob', 'count', 'true_prob']
counts
```

We can see that random forests predicted that 75 individuals would have a 0.9 probability of churn and in actuality that group had a ~0.97 rate.

### Calibration and Discrimination

Using the `DataFrame`

above I can draw a pretty simple graph to help visualize probability measurements. The x axis represents the churn probabilities which random forest assigned to a group of individuals. The y axis is the actual rate of churn within that group, and each point is scaled relative to the size of the group.

```
from ggplot import *
%matplotlib inline
baseline = np.mean(is_churn)
ggplot(counts,aes(x='pred_prob',y='true_prob',size='count')) + \
geom_point(color='blue') + \
stat_function(fun = lambda x: x, color='red') + \
stat_function(fun = lambda x: baseline, color='green') + \
xlim(-0.05, 1.05) + ylim(-0.05,1.05) + \
ggtitle("Random Forest") + \
xlab("Predicted probability") + ylab("Relative frequency of outcome")
```

You may have also noticed the two lines I drew with `stat_function()`

.

The red line represents a perfect prediction for a given group, or when the churn probability forecasted equals the outcome frequency. The green line shows the baseline probability of churn. For this dataset it's about 0.15.

Calibration is a relatively simple measurement and can be summed up as so: *Events predicted to happen 60% of the time should happen 60% of the time.* For all individuals I predict to have a churn risk of between 30 and 40%, the true churn rate for that group should be about 35%. For the graph above think of it as, *How close are my predictions to the red line?*

Discrimination measures *How far are my predictions away from the green line?* Why is that important?

Well, if I assign a churn probability of 15% to every individual I'll have near perfect calibration due to averages, but I'll be lacking any real insight. Discrimination gives a model a better score if it's able to isolate groups which are further from the base set.

`Scikit-learn`

doesn't come with these measurements, meaning I've had to implement them myself. For everyone's sake I've kept the math and source code out of this blog. Equations are replicated from Yang, Yates, and Smith (1991) and the code I wrote in the `churn_measurements`

import below can be found on GitHub here.

`from churn_measurements import calibration, discrimination`

Let's see how my three models fair on these measurements.

```
def print_measurements(pred_prob):
churn_prob, is_churn = pred_prob[:,1], y == 1
print " %-20s %.4f" % ("Calibration Error", calibration(churn_prob, is_churn))
print " %-20s %.4f" % ("Discrimination", discrimination(churn_prob,is_churn))
print "Note -- Lower calibration is better, higher discrimination is better"
print "Support vector machines:"
print_measurements(run_prob_cv(X,y,SVC,probability=True))
print "Random forests:"
print_measurements(run_prob_cv(X,y,RF,n_estimators=18))
print "K-nearest-neighbors:"
print_measurements(run_prob_cv(X,y,KNN))
```

Unlike the classification comparisons earlier, random forest isn't as clearly the front-runner here. While it's good at differentiating between high and low probability churn events, it has trouble assigning an accurate probability estimate to those events. For example the group which random forest predicts to have a 30% churn rate actually had a true churn rate of 14%. Clearly there's more work to be done, but I leave that to you as a challenge.

### Putting the model to use with Yhat

Time to upload a model to the cloud! In order to show some cool functionality, I'm going to go ahead and create a test set from the original churn data using `test_train_split()`

from sklearn. From there, I fit a support vector classifier for my deployment.

```
from sklearn.cross_validation import train_test_split
train_index,test_index = train_test_split(churn_df.index)
clf = SVC(probability=True)
clf.fit(X[train_index],y[train_index])
test_churn_df = churn_df.ix[test_index]
test_churn_df.to_csv("test_churn.csv")
```

The model I'm going to deploy using `yhat`

replicates the pipeline of this blog with a few modifications. Because I already defined variables within my global scope such as `yes_no_cols`

, `features`

, and `scaler`

I can just use them without having to specify them further.

On the methodological side I've added a few calculations. First the customer worth has been added (the sum of the total charges to that individual). Combining this value with probability of churn creates a very important measurement: the expected loss of revenue from that customer. This is where an accurate prediction model plays an important role as it's impossible to produce these values with only classifications.

```
from yhat import Yhat,YhatModel,preprocess
class ChurnModel(YhatModel):
# Type casts incoming data as a dataframe
@preprocess(in_type=pd.DataFrame,out_type=pd.DataFrame)
def execute(self,data):
# Collect customer meta data
response = data[['Area Code','Phone']]
charges = ['Day Charge','Eve Charge','Night Charge','Intl Charge']
response['customer_worth'] = data[charges].sum(axis=1)
# Convert yes no columns to bool
data[yes_no_cols] = data[yes_no_cols] == 'yes'
# Create feature space
X = data[features].as_matrix().astype(float)
X = scaler.transform(X)
# Make prediction
churn_prob = clf.predict_proba(X)
response['churn_prob'] = churn_prob[:,1]
# Calculate expected loss by churn
response['expected_loss'] = response['churn_prob'] * response['customer_worth']
response = response.sort('expected_loss', ascending=False)
# Return response DataFrame
return response
yh = Yhat(
"e[at]yhathq.com",
"MY API KEY",
"http://cloud.yhathq.com/"
)
response = yh.deploy("PythonChurnModel",ChurnModel,globals())
```

### Yhat batch mode

There comes a point when data science tools need to stop being scripts on an EC2 instance and start solving problems. In this case, empowering a retention team by warning them about customers likely to churn.

Our *M.O.* here at Yhat is finding ways to make data science applicable and practical as quickly as possible. Because my `execute()`

function takes and returns a `DataFrame`

, Yhat allows me to invoke my routine through a batch-scoring mode. The concept is simple. Upload a csv file from anywhere, the model pipeline is executed and the user gets a csv file back. This means that my method for scoring customers for churn-risk can be utilized by anyone else at my company regardless of their technical know-how, understanding of machine learning, or technical dependencies like `Python`

or `R`

.

Logging into cloud.yhathq.com and selecting my model gives me the following screen. The csv file I uploaded was the training data I created in the last section--in practice, this would be a new file exported from our CRM or customer database in the same format as the one I used to train the model.

Wait a second, download the resulting file at the bottom of the page when it's ready, and open it in Excel.

And there you go. Over 800 customers ranked and analyzed via drag and drop.