# Assignment # 5 (demo). Solution¶

## Logistic Regression and Random Forest in the credit scoring problem¶

Same assignment as a Kaggle Kernel + solution.

In this assignment, you will build models and answer questions using data on credit scoring.

Question 1. There are 5 jurors in a courtroom. Each of them can correctly identify the guilt of the defendant with 70% probability, independent of one another. What is the probability that the jurors will jointly reach the correct verdict if the final decision is by majority vote?

1. 70.00%
2. 83.20%
3. 83.70%
4. 87.50%

Solution:

We will use the formula for $\mu$ from the article. Since the majority of votes begin with $3$, then $m = 3, ~N = 5, ~p = 0.7$. Substitute these values into the formula to get:

$$\large \mu = \sum_{i=3}^{5}{5 \choose i}0.7^i(1-0.7)^{5-i} = 83.70\%$$

Great! Let's move on to machine learning.

## Credit scoring problem setup¶

#### Problem¶

Predict whether the customer will repay their credit within 90 days. This is a binary classification problem; we will assign customers into good or bad categories based on our prediction.

#### Data description¶

Feature Variable Type Value Type Description
age Input Feature integer Customer age
DebtRatio Input Feature real Total monthly loan payments (loan, alimony, etc.) / Total monthly income percentage
NumberOfTime30-59DaysPastDueNotWorse Input Feature integer The number of cases when client has overdue 30-59 days (not worse) on other loans during the last 2 years
NumberOfTimes90DaysLate Input Feature integer Number of cases when customer had 90+dpd overdue on other credits
NumberOfTime60-89DaysPastDueNotWorse Input Feature integer Number of cased when customer has 60-89dpd (not worse) during the last 2 years
NumberOfDependents Input Feature integer The number of customer dependents
SeriousDlqin2yrs Target Variable binary:
0 or 1
Customer hasn't paid the loan debt within 90 days

Let's set up our environment:

In [1]:
# Disable warnings in Anaconda
import warnings
warnings.filterwarnings('ignore')

import numpy as np
import pandas as pd
%matplotlib inline
import matplotlib.pyplot as plt
import seaborn as sns
sns.set()

In [2]:
from matplotlib import rcParams
rcParams['figure.figsize'] = 11, 8


Let's write the function that will replace NaN values with the median for each column.

In [3]:
def fill_nan(table):
for col in table.columns:
table[col] = table[col].fillna(table[col].median())
return table


In [4]:
data = pd.read_csv('../../data/credit_scoring_sample.csv', sep =';')

Out[4]:
SeriousDlqin2yrs age NumberOfTime30-59DaysPastDueNotWorse DebtRatio NumberOfTimes90DaysLate NumberOfTime60-89DaysPastDueNotWorse MonthlyIncome NumberOfDependents
0 0 64 0 0.249908 0 0 8158.0 0.0
1 0 58 0 3870.000000 0 0 NaN 0.0
2 0 41 0 0.456127 0 0 6666.0 0.0
3 0 43 0 0.000190 0 0 10500.0 2.0
4 1 49 0 0.271820 0 0 400.0 0.0

Look at the variable types:

In [5]:
data.dtypes

Out[5]:
SeriousDlqin2yrs                          int64
age                                       int64
NumberOfTime30-59DaysPastDueNotWorse      int64
DebtRatio                               float64
NumberOfTimes90DaysLate                   int64
NumberOfTime60-89DaysPastDueNotWorse      int64
MonthlyIncome                           float64
NumberOfDependents                      float64
dtype: object

Check the class balance:

In [6]:
ax = data['SeriousDlqin2yrs'].hist(orientation='horizontal', color='red')
ax.set_xlabel("number_of_observations")
ax.set_ylabel("unique_value")
ax.set_title("Target distribution")

print('Distribution of the target:')
data['SeriousDlqin2yrs'].value_counts()/data.shape[0]

Distribution of the target:

Out[6]:
0    0.777511
1    0.222489
Name: SeriousDlqin2yrs, dtype: float64

Separate the input variable names by excluding the target:

In [7]:
independent_columns_names = [x for x in data if x != 'SeriousDlqin2yrs']
independent_columns_names

Out[7]:
['age',
'NumberOfTime30-59DaysPastDueNotWorse',
'DebtRatio',
'NumberOfTimes90DaysLate',
'NumberOfTime60-89DaysPastDueNotWorse',
'MonthlyIncome',
'NumberOfDependents']

Apply the function to replace NaN values:

In [8]:
table = fill_nan(data)


Separate the target variable and input features:

In [9]:
X = table[independent_columns_names]
y = table['SeriousDlqin2yrs']


## Bootstrapping¶

Question 2. Make an interval estimate of the average age for the customers who delayed the repayment with the confidence level equal 90%. Use the example from the article for reference. Also, use np.random.seed(0) as it was done in the article. What is the resulting interval estimate?

1. 52.59 – 52.86
2. 45.71 – 46.13
3. 45.68 – 46.17
4. 52.56 – 52.88

Solution:

In [10]:
def get_bootstrap_samples(data, n_samples):
"""Generate samples using bootstrapping."""
indices = np.random.randint(0, len(data), (n_samples, len(data)))
samples = data[indices]
return samples

def stat_intervals(stat, alpha):
"""Make an interval estimate."""
boundaries = np.percentile(stat, [100 * alpha / 2., 100 * (1 - alpha / 2.)])
return boundaries

# Save the ages of those who let a delay
churn = data[data['SeriousDlqin2yrs'] == 1]['age'].values

# Set the random seed for reproducibility
np.random.seed(0)

# Generate bootstrap samples and calculate the mean for each sample
churn_mean_scores = [np.mean(sample) for sample in get_bootstrap_samples(churn, 1000)]

# Print the interval estimate for the sample means
print("Mean interval", stat_intervals(churn_mean_scores, 0.1))

Mean interval [45.71379414 46.12700479]


## Logistic regression¶

Let's set up to use logistic regression:

In [11]:
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import GridSearchCV, StratifiedKFold


Now, we will create a LogisticRegression model and use class_weight='balanced' to make up for our unbalanced classes.

In [12]:
lr = LogisticRegression(random_state=5, class_weight='balanced')


Let's try to find the best regularization coefficient, which is the coefficient C for logistic regression. Then, we will have an optimal model that is not overfit and is a good predictor of the target variable.

In [13]:
parameters = {'C': (0.0001, 0.001, 0.01, 0.1, 1, 10)}


In order to find the optimal value of C, let's apply stratified 5-fold validation and look at the ROC AUC against different values of the parameter C. Use the StratifiedKFold function for this:

In [14]:
skf = StratifiedKFold(n_splits=5, shuffle=True, random_state=5)


One of the important metrics of model quality is the Area Under the Curve (AUC). ROC AUC varies from 0 to 1. The closer ROC AUC to 1, the better the quality of the classification model.

Question 3. Perform a Grid Search with the scoring metric "roc_auc" for the parameter C. Which value of the parameter C is optimal?

1. 0.0001
2. 0.001
3. 0.01
4. 0.1
5. 1
6. 10

Solution:

In [15]:
grid_search = GridSearchCV(lr, parameters, n_jobs=-1, scoring='roc_auc', cv=skf)
grid_search = grid_search.fit(X, y)
grid_search.best_estimator_

Out[15]:
LogisticRegression(C=0.001, class_weight='balanced', dual=False,
fit_intercept=True, intercept_scaling=1, max_iter=100,
multi_class='ovr', n_jobs=1, penalty='l2', random_state=5,
solver='liblinear', tol=0.0001, verbose=0, warm_start=False)

Question 4. Can we consider the best model stable? The model is stable if the standard deviation on validation is less than 0.5%. Save the ROC AUC value of the best model, it will be useful for the following tasks.

1. Yes
2. No

Solution:

In [16]:
grid_search.cv_results_['std_test_score'][1]

Out[16]:
0.006357471890470392

The ROC AUC value of the best model:

In [17]:
grid_search.best_score_

Out[17]:
0.795409411236062

## Feature importance¶

Question 5. Feature importance is defined by the absolute value of its corresponding coefficient. First you need to normalize all the feature values so that it will be correct to compare them. What is the most important feature for the best logistic regression model?

1. age
2. NumberOfTime30-59DaysPastDueNotWorse
3. DebtRatio
4. NumberOfTimes90DaysLate
5. NumberOfTime60-89DaysPastDueNotWorse
6. MonthlyIncome
7. NumberOfDependents

Solution:

In [18]:
from sklearn.preprocessing import StandardScaler
lr = LogisticRegression(C=0.001, random_state=5, class_weight='balanced')
scal = StandardScaler()
lr.fit(scal.fit_transform(X), y)

pd.DataFrame({'feat': independent_columns_names,
'coef': lr.coef_.flatten().tolist()}).sort_values(by='coef', ascending=False)

Out[18]:
coef feat
1 0.724004 NumberOfTime30-59DaysPastDueNotWorse
3 0.517673 NumberOfTimes90DaysLate
4 0.194732 NumberOfTime60-89DaysPastDueNotWorse
6 0.101326 NumberOfDependents
2 -0.024082 DebtRatio
5 -0.162864 MonthlyIncome
0 -0.416304 age

Question 6. Calculate how much DebtRatio affects the prediction using the softmax function. What is its value?

1. 0.38
2. -0.02
3. 0.11
4. 0.24

Solution:

In [19]:
print((np.exp(lr.coef_[0]) / np.sum(np.exp(lr.coef_[0])))[2])

0.11420536719928259


Question 7. Let's see how we can interpret the impact of our features. For this, recalculate the logistic regression with absolute values, that is without scaling. Next, modify the customer's age by adding 20 years, keeping the other features unchanged. How many times will the chance that the customer will not repay their debt increase? You can find an example of the theoretical calculation here.

1. -0.01
2. 0.70
3. 8.32
4. 0.66

Solution:

In [20]:
lr = LogisticRegression(C=0.001, random_state=5, class_weight='balanced')
lr.fit(X, y)

pd.DataFrame({'feat': independent_columns_names,
'coef': lr.coef_.flatten().tolist()}).sort_values(by='coef', ascending=False)

Out[20]:
coef feat
1 0.482349 NumberOfTime30-59DaysPastDueNotWorse
3 0.430314 NumberOfTimes90DaysLate
6 0.115356 NumberOfDependents
4 0.065958 NumberOfTime60-89DaysPastDueNotWorse
2 -0.000011 DebtRatio
5 -0.000011 MonthlyIncome
0 -0.018185 age
In [21]:
np.exp(lr.coef_[0][0]*20)

Out[21]:
0.6950957781347274

It is $\exp^{\beta\delta}$ times more likely that the customer won't repay the debt, where $\delta$ is the feature value increment. That means that if we increased the age by 20 years, the odds that the customer won't repay would increase by 0.69 times.

## Random Forest¶

Import the Random Forest classifier:

In [22]:
from sklearn.ensemble import RandomForestClassifier


Initialize Random Forest with 100 trees and balance target classes:

In [23]:
rf = RandomForestClassifier(n_estimators=100, n_jobs=-1, random_state=42,
class_weight='balanced')


We will search for the best parameters among the following values:

In [24]:
parameters = {'max_features': [1, 2, 4], 'min_samples_leaf': [3, 5, 7, 9], 'max_depth': [5,10,15]}


Also, we will use the stratified k-fold validation again. You should still have the skf variable.

Question 8. How much higher the ROC AUC of the best random forest model than that of the best logistic regression on validation?

1. 4%
2. 3%
3. 2%
4. 1%

Solution:

In [25]:
%%time
rf_grid_search = GridSearchCV(rf, parameters, n_jobs=-1, scoring='roc_auc', cv=skf, verbose=True)
rf_grid_search = rf_grid_search.fit(X, y)
print(rf_grid_search.best_score_ - grid_search.best_score_)

Fitting 5 folds for each of 36 candidates, totalling 180 fits

[Parallel(n_jobs=-1)]: Done  42 tasks      | elapsed:   58.5s
[Parallel(n_jobs=-1)]: Done 180 out of 180 | elapsed:  7.5min finished

0.039645547841116846
CPU times: user 9.23 s, sys: 316 ms, total: 9.54 s
Wall time: 7min 30s


Question 9. What feature has the weakest impact in Random Forest model?

1. age
2. NumberOfTime30-59DaysPastDueNotWorse
3. DebtRatio
4. NumberOfTimes90DaysLate
5. NumberOfTime60-89DaysPastDueNotWorse
6. MonthlyIncome
7. NumberOfDependents

Solution:

In [26]:
independent_columns_names[np.argmin(rf_grid_search.best_estimator_.feature_importances_)]

Out[26]:
'NumberOfDependents'

Rating of the feature importance:

In [27]:
pd.DataFrame({'feat': independent_columns_names,
'coef': rf_grid_search.best_estimator_.feature_importances_}).sort_values(by='coef', ascending=False)

Out[27]:
coef feat
1 0.302461 NumberOfTime30-59DaysPastDueNotWorse
3 0.278644 NumberOfTimes90DaysLate
4 0.148683 NumberOfTime60-89DaysPastDueNotWorse
0 0.115926 age
2 0.079520 DebtRatio
5 0.060429 MonthlyIncome
6 0.014337 NumberOfDependents

Question 10. What is the most significant advantage of using Logistic Regression versus Random Forest for this problem?

1. Spent less time for model fitting;
2. Fewer variables to iterate;
3. Feature interpretability;
4. Linear properties of the algorithm.

Solution:

On the one hand, the Random Forest model works better for our credit scoring problem. Its performance is 4% higher. The reason for such a result is a small number of features and the compositional property of random forests.

On the other hand, the main advantage of Logistic Regression is that we can interpret the feature impact on the model outcome.

## Bagging¶

Import modules and set up the parameters for bagging:

In [28]:
from sklearn.ensemble import BaggingClassifier
from sklearn.model_selection import cross_val_score, RandomizedSearchCV

parameters = {'max_features': [2, 3, 4], 'max_samples': [0.5, 0.7, 0.9],
'base_estimator__C': [0.0001, 0.001, 0.01, 1, 10, 100]}


Question 11. Fit a bagging classifier with random_state=42. For the base classifiers, use 100 logistic regressors and use RandomizedSearchCV instead of GridSearchCV. It will take a lot of time to iterate over all 54 variants, so set the maximum number of iterations for RandomizedSearchCV to 20. Don't forget to set the parameters cv and random_state=1. What is the best ROC AUC you achieve?

1. 80.75%
2. 80.12%
3. 79.62%
4. 76.50%

Solution:

In [29]:
bg = BaggingClassifier(LogisticRegression(class_weight='balanced'),
n_estimators=100, n_jobs=-1, random_state=42)
r_grid_search = RandomizedSearchCV(bg, parameters, n_jobs=-1,
scoring='roc_auc', cv=skf, n_iter=20, random_state=1,
verbose=True)
r_grid_search = r_grid_search.fit(X, y)

Fitting 5 folds for each of 20 candidates, totalling 100 fits

[Parallel(n_jobs=-1)]: Done  42 tasks      | elapsed:  7.0min
[Parallel(n_jobs=-1)]: Done 100 out of 100 | elapsed: 16.9min finished

In [30]:
r_grid_search.best_score_

Out[30]:
0.8076172570918905
In [31]:
r_grid_search.best_estimator_

Out[31]:
BaggingClassifier(base_estimator=LogisticRegression(C=0.001, class_weight='balanced', dual=False,
fit_intercept=True, intercept_scaling=1, max_iter=100,
multi_class='ovr', n_jobs=1, penalty='l2', random_state=None,
solver='liblinear', tol=0.0001, verbose=0, warm_start=False),
bootstrap=True, bootstrap_features=False, max_features=2,
max_samples=0.7, n_estimators=100, n_jobs=-1, oob_score=False,
random_state=42, verbose=0, warm_start=False)

Question 12. Give an interpretation of the best parameters for bagging. Why are these values of max_features and max_samples the best?

1. For bagging it's important to use as few features as possible;
2. Bagging works better on small samples;
3. Less correlation between single models;
4. The higher the number of features, the lower the loss of information.