Waze Project¶

Prediction of customer churn

This notebook will show how I would build a machine learning model to predict user churn at the company Waze. To get the best results, two tree-based models are compared: random forest and XGBoost.

Course 6 End-of-Course Project: Build a machine learning model¶

In this activity, you will practice using tree-based modeling techniques to predict on a binary target class.

The purpose of this model is to find factors that drive user churn.

The goal of this model is to predict whether or not a Waze user is retained or churned.

Build a machine learning model¶

Imports and data loading¶

Import packages and libraries needed to build and evaluate random forest and XGBoost classification models.

In [9]:
# Import packages for data manipulation
import pandas as pd

# Import packages for data visualization
import seaborn as sns
import matplotlib.pyplot as plt

# This lets us see all of the columns, preventing Juptyer from redacting them.


# Import packages for data modeling
import sklearn.preprocessing as pp

# This is the function that helps plot feature importance
import sklearn.metrics as m


# This module lets us save our models once we fit them.
import pickle

Read and inspect the dataset

In [13]:
# Import dataset
df0 = pd.read_csv('waze_dataset.csv')
In [14]:
# Inspect the first five rows
df0.head()
Out[14]:
ID label sessions drives total_sessions n_days_after_onboarding total_navigations_fav1 total_navigations_fav2 driven_km_drives duration_minutes_drives activity_days driving_days device
0 0 retained 283 226 296.748273 2276 208 0 2628.845068 1985.775061 28 19 Android
1 1 retained 133 107 326.896596 1225 19 64 13715.920550 3160.472914 13 11 iPhone
2 2 retained 114 95 135.522926 2651 0 0 3059.148818 1610.735904 14 8 Android
3 3 retained 49 40 67.589221 15 322 7 913.591123 587.196542 7 3 iPhone
4 4 retained 84 68 168.247020 1562 166 5 3950.202008 1219.555924 27 18 Android

Basic Feature engineering¶

In [15]:
# Copy the df0 dataframe
df0.isna().sum()
df = df0.dropna().drop(columns='ID').copy() # drop missing labels
df = pd.get_dummies(df, drop_first=True) # create dummy variables

Call info() on the new dataframe so the existing columns can be easily referenced.

In [16]:
df.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 14299 entries, 0 to 14998
Data columns (total 12 columns):
 #   Column                   Non-Null Count  Dtype  
---  ------                   --------------  -----  
 0   sessions                 14299 non-null  int64  
 1   drives                   14299 non-null  int64  
 2   total_sessions           14299 non-null  float64
 3   n_days_after_onboarding  14299 non-null  int64  
 4   total_navigations_fav1   14299 non-null  int64  
 5   total_navigations_fav2   14299 non-null  int64  
 6   driven_km_drives         14299 non-null  float64
 7   duration_minutes_drives  14299 non-null  float64
 8   activity_days            14299 non-null  int64  
 9   driving_days             14299 non-null  int64  
 10  label_retained           14299 non-null  uint8  
 11  device_iPhone            14299 non-null  uint8  
dtypes: float64(3), int64(7), uint8(2)
memory usage: 1.2 MB

km_per_driving_day¶

  1. Create a feature representing the mean number of kilometers driven on each driving day in the last month for each user. Add this feature as a column to df.

  2. Get descriptive statistics for this new feature

In [18]:
# 1. Create `km_per_driving_day` feature
df['km_per_driving_day'] = df.driven_km_drives / df.driving_days
df.loc[df.driving_days == 0,'km_per_driving_day'] = 0 # fix cases of missing values
# 2. Get descriptive stats
df.km_per_driving_day.describe()
Out[18]:
count    14299.000000
mean       581.942399
std       1038.254509
min          0.000000
25%        136.168003
50%        273.301012
75%        558.018761
max      15420.234110
Name: km_per_driving_day, dtype: float64

percent_sessions_in_last_month¶

  1. Create a new column percent_sessions_in_last_month that represents the percentage of each user's total sessions that were logged in their last month of use.

  2. Get descriptive statistics for this new feature

In [19]:
# 1. Create `percent_sessions_in_last_month` feature
df['percent_sessions_in_last_month'] = df.sessions / round( df.total_sessions )

# 1. Get descriptive stats
df.percent_sessions_in_last_month.describe()
df.loc[round(df.total_sessions)==0,'percent_sessions_in_last_month'] = 0
df.percent_sessions_in_last_month.isna().sum() # missing values!

#sns.histplot(df.percent_sessions_in_last_month)
#df[df.percent_sessions_in_last_month>1]
#df.loc[df.sessions == round(df.total_sessions)]
Out[19]:
0

professional_driver¶

Create a new, binary feature called professional_driver that is a 1 for users who had 60 or more drives **and** drove on 15+ days in the last month.

In [293]:
# Create `professional_driver` feature
df['professional_driver'] = ((df.drives >= 60) & (df.driving_days >= 15))

total_sessions_per_day¶

Now, create a new column that represents the mean number of sessions per day since onboarding.

In [21]:
# Create `total_sessions_per_day` feature
df['total_sessions_per_day'] = df.total_sessions / df.n_days_after_onboarding  
df.total_sessions_per_day.describe()
Out[21]:
count    14299.000000
mean         0.338207
std          1.319814
min          0.000298
25%          0.050818
50%          0.100457
75%          0.215210
max         39.763874
Name: total_sessions_per_day, dtype: float64

As with other features, get descriptive statistics for this new feature.

In [295]:
# Get descriptive stats
df.total_sessions_per_day.describe()
Out[295]:
count    14299.000000
mean         0.338207
std          1.319814
min          0.000298
25%          0.050818
50%          0.100457
75%          0.215210
max         39.763874
Name: total_sessions_per_day, dtype: float64

km_per_hour¶

Create a column representing the mean kilometers per hour driven in the last month.

In [35]:
# Create `km_per_hour` feature
import numpy as np
df['duration_hours_drives'] = round( df.duration_minutes_drives / 60.0, 1 )
df['km_per_hour'] = df.driven_km_drives / df.duration_hours_drives
df.loc[df.km_per_hour == np.Inf,'km_per_hour'] = 0
df.km_per_hour.describe()
df[ df.km_per_hour > 120 ]
sns.histplot(df.km_per_hour[df.km_per_hour< 200])


# !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
# Major change to the data -- drop unlikely obs.
# !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
# Something is wrong -- it makes no sense for there to be people who travel > 200km per hour!
# Perhaps they have the app open on a plane?! Or in meters rather than km??
look = df.km_per_hour >= 200
df = df[df.km_per_hour <= 150] # Drop these rows!!!
df.shape
Out[35]:
(9000, 20)

km_per_drive¶

Create a column representing the mean number of kilometers per drive made in the last month for each user. Then, print descriptive statistics for the feature.

In [36]:
# Create `km_per_drive` feature
df['km_per_drive'] = df.driven_km_drives / df.drives
df.loc[df.drives==0,'km_per_drive']=0 
df.km_per_drive.describe()
Out[36]:
count     9000.000000
mean       224.622611
std        618.644479
min          0.000000
25%         30.208270
50%         68.953665
75%        168.655354
max      15777.426560
Name: km_per_drive, dtype: float64

percent_of_sessions_to_favorite¶

Finally, create a new column that represents the percentage of total sessions that were used to navigate to one of the users' favorite places. Then, print descriptive statistics for the new column.

This is a proxy representation for the percent of overall drives that are to a favorite place. Since total drives since onboarding are not contained in this dataset, total sessions must serve as a reasonable approximation.

People whose drives to non-favorite places make up a higher percentage of their total drives might be less likely to churn, since they're making more drives to less familiar places.

In [37]:
# Create `percent_of_sessions_to_favorite` feature
df['percent_of_sessions_to_favorite'] = ( df.total_navigations_fav1 + df.total_navigations_fav2 ) / df.sessions
df.loc[df.sessions==0,'percent_of_sessions_to_favorite'] = 0
# Get descriptive stats
df.percent_of_sessions_to_favorite.describe()
Out[37]:
count    9000.000000
mean        8.103972
std        28.185086
min         0.000000
25%         0.489608
50%         1.790895
75%         5.425987
max       799.000000
Name: percent_of_sessions_to_favorite, dtype: float64

Task 3. Drop missing values¶

Because you know from previous EDA that there is no evidence of a non-random cause of the 700 missing values in the label column, and because these observations comprise less than 5% of the data, use the dropna() method to drop the rows that are missing this data.

In [38]:
# Drop rows with missing values
# Already done.

Task 4. Outliers¶

You know from previous EDA that many of these columns have outliers. However, tree-based models are resilient to outliers, so there is no need to make any imputations.

Task 5. Variable encoding¶

Dummying features¶

In order to use device as an X variable, you will need to convert it to binary, since this variable is categorical.

In [39]:
# Create new `device2` variable
df.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 9000 entries, 0 to 14998
Data columns (total 20 columns):
 #   Column                           Non-Null Count  Dtype  
---  ------                           --------------  -----  
 0   sessions                         9000 non-null   int64  
 1   drives                           9000 non-null   int64  
 2   total_sessions                   9000 non-null   float64
 3   n_days_after_onboarding          9000 non-null   int64  
 4   total_navigations_fav1           9000 non-null   int64  
 5   total_navigations_fav2           9000 non-null   int64  
 6   driven_km_drives                 9000 non-null   float64
 7   duration_minutes_drives          9000 non-null   float64
 8   activity_days                    9000 non-null   int64  
 9   driving_days                     9000 non-null   int64  
 10  label_retained                   9000 non-null   uint8  
 11  device_iPhone                    9000 non-null   uint8  
 12  km_per_driving_day               9000 non-null   float64
 13  percent_sessions_in_last_month   9000 non-null   float64
 14  total_sessions_per_day           9000 non-null   float64
 15  duration_hours_drives            9000 non-null   float64
 16  km_per_hour                      9000 non-null   float64
 17  km_per_drive                     9000 non-null   float64
 18  percent_of_sessions_to_favorite  9000 non-null   float64
 19  churned                          9000 non-null   int32  
dtypes: float64(10), int32(1), int64(7), uint8(2)
memory usage: 1.3 MB

Target encoding¶

The target variable is also categorical, since a user is labeled as either "churned" or "retained." Change the data type of the label column to be binary. This change is needed to train the models.

Assign a 0 for all retained users.

Assign a 1 for all churned users.

Save this variable as label2 so as not to overwrite the original label variable.

Note: There are many ways to do this. Consider using np.where() as you did earlier in this notebook.

In [40]:
# Create binary `label2` column
df['churned'] = 0 + 1*(df.label_retained == 0)

Task 6. Feature selection¶

Tree-based models can handle multicollinearity, so the only feature that can be cut is ID, since it doesn't contain any information relevant to churn.

Note, however, that device won't be used simply because it's a copy of device2.

Drop ID from the df dataframe.

In [41]:
# Drop `ID` column
# done

Task 7. Evaluation metric¶

Before modeling, you must decide on an evaluation metric. This will depend on the class balance of the target variable and the use case of the model.

First, examine the class balance of your target variable.

In [42]:
# Get class balance of 'label' col
df.churned.value_counts(normalize=True)
Out[42]:
0    0.816
1    0.184
Name: churned, dtype: float64

Approximately 18% of the users in this dataset churned. This is an unbalanced dataset, but not extremely so. It can be modeled without any class rebalancing.

Now, consider which evaluation metric is best. Remember, accuracy might not be the best gauge of performance because a model can have high accuracy on an imbalanced dataset and still fail to predict the minority class.

It was already determined that the risks involved in making a false positive prediction are minimal. No one stands to get hurt, lose money, or suffer any other significant consequence if they are predicted to churn. Therefore, select the model based on the recall score.

Task 8. Modeling workflow and model selection process¶

The final modeling dataset contains 14,299 samples. This is towards the lower end of what might be considered sufficient to conduct a robust model selection process, but still doable.

  1. Split the data into train/validation/test sets (60/20/20)

Task 9. Split the data¶

Ready to model. Split the data into features/target variable and training/validation/test sets.

In [43]:
# 1. Isolate X variables
import sklearn.model_selection as ms

df1 = df.drop(columns='label_retained')
df1.head()
X = df1.drop(columns='churned')

# 2. Isolate y variable
y = df1.churned

# 3. Split into train and test sets
X_train, X_test, y_train, y_test = ms.train_test_split(X,y, random_state=42, train_size=.80, stratify=y)

# 4. Split into train and validate sets
X_trn, X_val, y_trn, y_val = ms.train_test_split(X_train,y_train, random_state=42, train_size=.75, stratify=y_train)

Verify the number of samples in the partitioned data.

In [44]:
print(X_train.shape)
print(X_trn.shape)
print(X_val.shape)
print(X_test.shape)
(7200, 18)
(5400, 18)
(1800, 18)
(1800, 18)

This aligns with expectations.

Task 10. Modeling¶

Random forest¶

Using GridSearchCV to tune a random forest model.

In [45]:
# 1. Instantiate the random forest classifier
import sklearn.tree as tree
import sklearn.ensemble as en

rf = en.RandomForestClassifier()

# 2. Create a dictionary of hyperparameters to tune
cv_params = {#'max_depth':[2,4],
             'max_features':[2,4,6],
             'min_samples_leaf':[5,10],
             'n_estimators':[50,100,500]
            }

# 3. Define a dictionary of scoring metrics to capture
scoring = ['f1','recall','precision','accuracy']

# 4. Instantiate the GridSearchCV object
gs = ms.GridSearchCV(estimator=rf,param_grid=cv_params,scoring=scoring,refit='recall',cv=5)

Now fit the model to the training data.

In [46]:
%%time
f0 = gs.fit(X_trn,y_trn)
CPU times: total: 5min 1s
Wall time: 5min 1s

Examine the best average score across all the validation folds.

In [47]:
# Examine best score
f0.best_estimator_
Out[47]:
RandomForestClassifier(max_features=6, min_samples_leaf=5, n_estimators=50)
In a Jupyter environment, please rerun this cell to show the HTML representation or trust the notebook.
On GitHub, the HTML representation is unable to render, please try loading this page with nbviewer.org.
RandomForestClassifier(max_features=6, min_samples_leaf=5, n_estimators=50)

Examine the best combination of hyperparameters.

In [48]:
# Examine best hyperparameter combo
# RandomForestClassifier(max_features=6, min_samples_leaf=5, n_estimators=50)

Use the make_results() function to output all of the scores of your model. Note that the function accepts three arguments.

HINT
To learn more about how this function accesses the cross-validation results, refer to the [`GridSearchCV` scikit-learn documentation](https://scikit-learn.org/stable/modules/generated/sklearn.model_selection.GridSearchCV.html?highlight=gridsearchcv#sklearn.model_selection.GridSearchCV) for the `cv_results_` attribute.
In [49]:
def make_results(model_name:str, model_object, metric:str,X_val=X_val):
    '''
    Arguments:
        model_name (string): what you want the model to be called in the output table
        model_object: a fit GridSearchCV object
        metric (string): precision, recall, f1, or accuracy

    Returns a pandas df with the F1, recall, precision, and accuracy scores
    for the model with the best mean 'metric' score across all validation folds.
    '''

  # Create dictionary that maps input metric to actual metric name in GridSearchCV
    model = model_object.best_estimator_
    y_pred = model.predict(X_val)
    bs = model_object.best_score_
  # Get all the results from the CV and put them in a df
    acc = m.accuracy_score(y_val,y_pred)
    prec = m.precision_score(y_val,y_pred)
    rec = m.recall_score(y_val,y_pred)
    f1 = m.f1_score(y_val,y_pred)
    d = {'model':model_name,'Recall':rec,'Prec':prec,'F1':f1,'Acc':acc}
    sdf = pd.DataFrame(d.items()).transpose()
    sdf.columns = list( sdf.iloc[0] ) 
    sdf = sdf.iloc[1:,:]
    
  # Isolate the row of the df with the max(metric) score
    return sdf

  # Extract Accuracy, precision, recall, and f1 score from that row

  # Create table of results

results_rf = make_results("Random Forest",f0,'recall')
results_rf
Out[49]:
model Recall Prec F1 Acc
1 Random Forest 0.10574 0.4375 0.170316 0.810556

Pass the GridSearch object to the make_results() function.

In [51]:
!python -m pip install xgboost
Looking in indexes: https://pypi.org/simple, https://pypi.ngc.nvidia.com
Collecting xgboost
  Downloading xgboost-2.0.0-py3-none-win_amd64.whl (99.7 MB)
     ---------------------------------------- 99.7/99.7 MB 3.7 MB/s eta 0:00:00
Requirement already satisfied: scipy in c:\users\jon\anaconda3\envs\bertopic\lib\site-packages (from xgboost) (1.10.0)
Requirement already satisfied: numpy in c:\users\jon\anaconda3\envs\bertopic\lib\site-packages (from xgboost) (1.23.5)
Installing collected packages: xgboost
Successfully installed xgboost-2.0.0
WARNING: Ignoring invalid distribution -atplotlib (c:\users\jon\anaconda3\envs\bertopic\lib\site-packages)
WARNING: Ignoring invalid distribution -atplotlib (c:\users\jon\anaconda3\envs\bertopic\lib\site-packages)
WARNING: Ignoring invalid distribution -atplotlib (c:\users\jon\anaconda3\envs\bertopic\lib\site-packages)
WARNING: Ignoring invalid distribution -atplotlib (c:\users\jon\anaconda3\envs\bertopic\lib\site-packages)
WARNING: Ignoring invalid distribution -atplotlib (c:\users\jon\anaconda3\envs\bertopic\lib\site-packages)
WARNING: Ignoring invalid distribution -atplotlib (c:\users\jon\anaconda3\envs\bertopic\lib\site-packages)
WARNING: Ignoring invalid distribution -atplotlib (c:\users\jon\anaconda3\envs\bertopic\lib\site-packages)

Asside from the accuracy, the scores aren't that good. However, recall that when you built the logistic regression model in the last course the recall was ~0.09, which means that this model has 33% better recall and about the same accuracy, and it was trained on less data.

If you want, feel free to try retuning your hyperparameters to try to get a better score. You might be able to marginally improve the model.

XGBoost¶

Try to improve scores using an XGBoost model.

  1. Instantiate the XGBoost classifier xgb and set objective='binary:logistic'. Also set the random state.

  2. Create a dictionary cv_params of the following hyperparameters and their corresponding values to tune:

    • max_depth
    • min_child_weight
    • learning_rate
    • n_estimators
  3. Define a dictionary scoring of scoring metrics for grid search to capture (precision, recall, F1 score, and accuracy).

  4. Instantiate the GridSearchCV object xgb_cv. Pass to it as arguments:

    • estimator=xgb
    • param_grid=cv_params
    • scoring=scoring
    • cv: define the number of cross-validation folds you want (cv=_)
    • refit: indicate which evaluation metric you want to use to select the model (refit='recall')
In [53]:
# 1. Instantiate the XGBoost classifier
from xgboost import XGBClassifier
xgb = XGBClassifier()

# 2. Create a dictionary of hyperparameters to tune
param_grid = {
 'max_depth':[2,6,8,10,12,14],
 'min_child_weight':[.5,1,2,10],
 'learning_rate':[.1,.2,.3],
 'n_estimators':[100,200,400]   
}
# 3. Define a dictionary of scoring metrics to capture
# same as before!
print( scoring)

# 4. Instantiate the GridSearchCV object
gscv = ms.GridSearchCV(estimator=xgb, param_grid=param_grid, scoring=scoring, refit='recall', cv=5 )
['f1', 'recall', 'precision', 'accuracy']

Now fit the model to the X_train and y_train data.

Note this cell might take several minutes to run.

In [54]:
%%time
f1 = gscv.fit(X_trn,y_trn)
CPU times: total: 3h 11min 2s
Wall time: 19min 32s

Get the best score from this model.

In [55]:
# Examine best score
f1.best_score_
Out[55]:
0.24046495101771487

And the best parameters.

In [56]:
# Examine best parameters
f1.best_params_
Out[56]:
{'learning_rate': 0.3,
 'max_depth': 8,
 'min_child_weight': 10,
 'n_estimators': 400}

Use the make_results() function to output all of the scores of your model. Note that the function accepts three arguments.

In [57]:
# Call 'make_results()' on the GridSearch object
results_xgb = make_results("XGBoost",f1,'recall')
results_xgb
Out[57]:
model Recall Prec F1 Acc
1 XGBoost 0.214502 0.420118 0.284 0.801111

This model fit the data even better than the random forest model. The recall score is nearly double the recall score from the logistic regression model from the previous course, and it's almost 50% better than the random forest model's recall score, while maintaining a similar accuracy and precision score.

Task 11. Model selection¶

Now, use the best random forest model and the best XGBoost model to predict on the validation data. Whichever performs better will be selected as the champion model.

Random forest¶

In [58]:
# Use random forest model to predict on validation data
y_pred = f1.predict(X_val)

Use the get_test_scores() function to generate a table of scores from the predictions on the validation data.

In [59]:
def get_test_scores(model_name:str, preds, y_test_data):
    '''
    Generate a table of test scores.

    In:
        model_name (string): Your choice: how the model will be named in the output table
        preds: numpy array of test predictions
        y_test_data: numpy array of y_test data

    Out:
        table: a pandas df of precision, recall, f1, and accuracy scores for your model
    '''
    accuracy = m.accuracy_score(y_test_data, preds)
    precision = m.precision_score(y_test_data, preds)
    recall = m.recall_score(y_test_data, preds)
    f1 = m.f1_score(y_test_data, preds)

    table = pd.DataFrame({'model': [model_name],
                          'precision': [precision],
                          'recall': [recall],
                          'F1': [f1],
                          'accuracy': [accuracy]
                          })

    return table
In [60]:
# Get validation scores for RF model
rs0 = get_test_scores('XGBoost', y_pred, y_val)

# Append to the results table
results_all = pd.concat([results_rf, results_xgb])
results_all
Out[60]:
model Recall Prec F1 Acc
1 Random Forest 0.10574 0.4375 0.170316 0.810556
1 XGBoost 0.214502 0.420118 0.284 0.801111

Notice that the scores went down from the training scores across all metrics, but only by very little. This means that the model did not overfit the training data.

Task 12. Use champion model to predict on test data¶

Now, use the champion model to predict on the test dataset. This is to give a final indication of how you should expect the model to perform on new future data, should you decide to use the model.

In [61]:
# Use XGBoost model to predict on test data
y_pred_test = f1.predict(X_test)
results_champ = make_results("XGBoost champion",f1,'recall',X_val=X_test)
#results_champ = make_results("RF champion",f0,'recall',X_val=X_test)
results_champ 
Out[61]:
model Recall Prec F1 Acc
1 XGBoost champion 0.102719 0.179894 0.130769 0.748889

The recall was exactly the same as it was on the validation data, but the precision declined notably, which caused all of the other scores to drop slightly. Nonetheless, this is stil within the acceptable range for performance discrepancy between validation and test scores.

Task 13. Confusion matrix¶

Plot a confusion matrix of the champion model's predictions on the test data.

In [62]:
# Generate array of values for confusion matrix
cm = m.confusion_matrix(y_true=y_test,y_pred=f1.predict(X_test))
cm
# Plot confusion matrix
p = m.ConfusionMatrixDisplay(cm, display_labels=f1.classes_)
p.plot(values_format='')
Out[62]:
<sklearn.metrics._plot.confusion_matrix.ConfusionMatrixDisplay at 0x1562176c1f0>

The model predicted three times as many false negatives than it did false positives, and it correctly identified only 16.6% of the users who actually churned.

Task 14. Feature importance¶

Use the plot_importance function to inspect the most important features of your final model.

In [63]:
from xgboost import plot_importance
plot_importance(f1.best_estimator_)
Out[63]:
<AxesSubplot:title={'center':'Feature importance'}, xlabel='F score', ylabel='Features'>

The XGBoost model made more use of many of the features than did the logistic regression model from the previous course, which weighted a single feature (activity_days) very heavily in its final prediction.

If anything, this underscores the importance of feature engineering. Notice that engineered features accounted for six of the top 10 features (and three of the top five). Feature engineering is often one of the best and easiest ways to boost model performance.

Also, note that the important features in one model might not be the same as the important features in another model. That's why you shouldn't discount features as unimportant without thoroughly examining them and understanding their relationship with the dependent variable, if possible. These discrepancies between features selected by models are typically caused by complex feature interactions.

Remember, sometimes your data simply will not be predictive of your chosen target. This is common. Machine learning is a powerful tool, but it is not magic. If your data does not contain predictive signal, even the most complex algorithm will not be able to deliver consistent and accurate predictions. Do not be afraid to draw this conclusion.

Even if you cannot use the model to make strong predictions, was the work done in vain? What insights can you report back to stakeholders?

Conclusion¶

  1. Would you recommend using this model for churn prediction? Why or why not?

No - the recall from the model is not strong enough to justify making business decisions with. However, it may be useful as a starting point for further analysis.

  1. What tradeoff was made by splitting the data into training, validation, and test sets as opposed to just training and test sets?

Splitting the model into train, validation and test meant that there was less information overall but it gives us more confidence that the chosen model will work as expected on unseen data.

  1. What is the benefit of using a logistic regression model over an ensemble of tree-based models (like random forest or XGBoost) for classification tasks?

The tree based models give us a sense of variable importance, but the logistic regression gives us the strength and direction of the independent variables.

  1. What is the benefit of using an ensemble of tree-based models like random forest or XGBoost over a logistic regression model for classification tasks?

Tree based models are easier to work with because there are relatively few constraints and less variable cleaning is required. For example, a random forest model is a series of decision trees that use a subset of explanatory variables, the results of which are aggregated to determine an outcome.

  1. What could you do to improve this model?

One way to improve the model is to remove observations where the distance travelled per hours is excessively high. It seems that there is problematic data. However, I would need to determine the root cause of this problem and determine if those observations are systematically different from the rest of the data.

Additional features may prove to be advantageous.

  1. What additional features would you like to have to help improve the model?

Geography, time of trip and other information could be useful.

BONUS¶

The following content is not required, but demonstrates further steps that you might take to tailor your model to your use case.

Identify an optimal decision threshold¶

The default decision threshold for most implementations of classification algorithms—including scikit-learn's—is 0.5. This means that, in the case of the Waze models, if they predicted that a given user had a 50% probability or greater of churning, then that user was assigned a predicted value of 1—the user was predicted to churn.

With imbalanced datasets where the response class is a minority, this threshold might not be ideal. You learned that a precision-recall curve can help to visualize the trade-off between your model's precision and recall.

Here's the precision-recall curve for the XGBoost champion model on the test data.

In [64]:
# Plot precision-recall curve
# classifier, X_test, y_test, name="LinearSVC", plot_chance_level=True
m.PrecisionRecallDisplay.from_estimator(f1.best_estimator_, X_test, y_test)
Out[64]:
<sklearn.metrics._plot.precision_recall_curve.PrecisionRecallDisplay at 0x1561f8edee0>

As recall increases, precision decreases. But what if you determined that false positives aren't much of a problem? For example, in the case of this Waze project, a false positive could just mean that a user who will not actually churn gets an email and a banner notification on their phone. It's very low risk.

So, what if instead of using the default 0.5 decision threshold of the model, you used a lower threshold?

Here's an example where the threshold is set to 0.4:

In [65]:
# Get predicted probabilities on the test data

The predict_proba() method returns a 2-D array of probabilities where each row represents a user. The first number in the row is the probability of belonging to the negative class, the second number in the row is the probability of belonging to the positive class. (Notice that the two numbers in each row are complimentary to each other and sum to one.)

You can generate new predictions based on this array of probabilities by changing the decision threshold for what is considered a positive response. For example, the following code converts the predicted probabilities to {0, 1} predictions with a threshold of 0.4. In other words, any users who have a value ≥ 0.4 in the second column will get assigned a prediction of 1, indicating that they churned.

In [66]:
# Create a list of just the second column values (probability of target)
prob_churned = f1.predict_proba(X_trn)[:,1] >= 0.5

# Create an array of new predictions that assigns a 1 to any value >= 0.4
pd.DataFrame(prob_churned).value_counts()
Out[66]:
False    4406
True      994
dtype: int64
In [67]:
# Get evaluation metrics for when the threshold is 0.4

Compare these numbers with the results from earlier.

In [68]:
 

Recall and F1 score increased significantly, while precision and accuracy decreased.

So, using the precision-recall curve as a guide, suppose you knew that you'd be satisfied if the model had a recall score of 0.5 and you were willing to accept the ~30% precision score that comes with it. In other words, you'd be happy if the model successfully identified half of the people who will actually churn, even if it means that when the model says someone will churn, it's only correct about 30% of the time.

What threshold will yield this result? There are a number of ways to determine this. Here's one way that uses a function to accomplish this.

In [69]:
def threshold_finder(y_test_data, probabilities, desired_recall):
    '''
    Find the threshold that most closely yields a desired recall score.

    Inputs:
        y_test_data: Array of true y values
        probabilities: The results of the `predict_proba()` model method
        desired_recall: The recall that you want the model to have

    Outputs:
        threshold: The threshold that most closely yields the desired recall
        recall: The exact recall score associated with `threshold`
    '''
    probs = [x[1] for x in probabilities]  # Isolate second column of `probabilities`
    thresholds = np.arange(0, 1, 0.001)    # Set a grid of 1,000 thresholds to test

    scores = []
    for threshold in thresholds:
        # Create a new array of {0, 1} predictions based on new threshold
        preds = np.array([1 if x >= threshold else 0 for x in probs])
        # Calculate recall score for that threshold
        recall = m.recall_score(y_test_data, preds)
        # Append the threshold and its corresponding recall score as a tuple to `scores`
        scores.append((threshold, recall))

    distances = []
    for idx, score in enumerate(scores):
        # Calculate how close each actual score is to the desired score
        distance = abs(score[1] - desired_recall)
        # Append the (index#, distance) tuple to `distances`
        distances.append((idx, distance))

    # Sort `distances` by the second value in each of its tuples (least to greatest)
    sorted_distances = sorted(distances, key=lambda x: x[1], reverse=False)
    # Identify the tuple with the actual recall closest to desired recall
    best = sorted_distances[0]
    # Isolate the index of the threshold with the closest recall score
    best_idx = best[0]
    # Retrieve the threshold and actual recall score closest to desired recall
    threshold, recall = scores[best_idx]

    return threshold, recall

Now, test the function to find the threshold that results in a recall score closest to 0.5.

In [70]:
# Get the predicted probabilities from the champion model
pp = f1.best_estimator_.predict_proba(X_test)
# Call the function
threshold_finder(y_test, pp, desired_recall=.50)
Out[70]:
(0.08, 0.4984894259818731)

Setting a threshold of 0.124 will result in a recall of 0.503.

To verify, you can repeat the steps performed earlier to get the other evaluation metrics for when the model has a threshold of 0.124 (or 0.08 in subsample!). Based on the precision-recall curve, a 0.5 recall score should have a precision of ~0.3.

In [77]:
# Create an array of new predictions that assigns a 1 to any value >= 0.124
probs = [x[1] for x in pp]
new_preds = np.array([1 if x >= 0.08 else 0 for x in probs])

# Get evaluation metrics for when the threshold is 0.124
get_test_scores('XGB, threshold = 0.124', new_preds, y_test)
Out[77]:
model precision recall F1 accuracy
0 XGB, threshold = 0.124 0.287958 0.498489 0.365044 0.681111

It worked!