基于生存分析模型的用户流失预测

HsuHeinrich

发布于 2023-05-25 17:00:43

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基于生存分析模型的用户流失预测

小O：有没有什么很好的办法在预测用户流失的同时，提供一些建议帮助我们运营呢？小H：这简单，如果我可以告诉你什么样的人群容易流失、什么时间点容易流失、用户的可能存活多节可以吗？小O：这太可以了～

生存模型就能很好的地解决上面的问题，生存分析(Survival analysis)是指根据历史数据对人的生存时间进行分析和推断，研究生存情况与众多影响因素间的关系。本文参考自python数据分析案例-利用生存分析Kaplan-Meier法与COX比例风险回归模型进行客户流失分析与剩余价值预测[1]。

数据探索

导入相关库

# pip install -i https://pypi.doubanio.com/simple/ lifelines  # 采用镜像安装

import pandas as pd                
import numpy as np                 
import matplotlib.pyplot as plt 
import seaborn as sns
from lifelines import NelsonAalenFitter, CoxPHFitter, KaplanMeierFitter   
from lifelines.statistics import logrank_test, multivariate_logrank_test 
from sklearn.model_selection import train_test_split, GridSearchCV
from lifelines.utils.sklearn_adapter import sklearn_adapter
from sklearn import metrics
import time
import math
import toad
import itertools
from lifelines.plotting import add_at_risk_counts
from lifelines.statistics import pairwise_logrank_test
from sklearn import preprocessing
from lifelines.calibration import survival_probability_calibration
from sklearn.metrics import brier_score_loss
from lifelines.utils import median_survival_times, qth_survival_times

plt.rcParams['font.sans-serif'] = ['SimHei'] 
plt.rcParams['axes.unicode_minus'] = False  
import warnings
warnings.filterwarnings("ignore")

数据准备

# 读取数据
raw_data = pd.read_csv('Customer-Churn.csv')
raw_data.head()

image-20230206152049606

数据预处理

df = raw_data.copy()
# 删除tenure=0（This model does not allow for non-positive durations）
df = df[df['tenure']>0]
# 转换成连续型变量
df['TotalCharges'] = pd.to_numeric(df.TotalCharges, errors='coerce')
# 填充缺失值
df['TotalCharges'].fillna(0, inplace=True)
# 转换成类别变量
df['SeniorCitizen'] = df['SeniorCitizen'].astype("str")
# 将是否流失转换为01分布
df['Churn'] = df['Churn'].map({'No':0,'Yes':1})

生存分析探索

查看整体生存曲线

fig, ax = plt.subplots(nrows=2, ncols=1, figsize=(12,10))  

# 整体生存曲线
kmf = KaplanMeierFitter()
kmf.fit(df.tenure,  # 代表生存时长
        event_observed=df.Churn  # 代表事件的终点
        
        )
kmf.plot_survival_function(ax=ax[0], label='all')
ax[0].grid()
# 整体生存变化率
naf = NelsonAalenFitter()
naf.fit(df.tenure, event_observed=df.Churn)
naf.plot_hazard(ax=ax[1], bandwidth=20, label='all')
ax[1].grid()
plt.show()

output_8_0

截至观察期（最大70个月），整体还有60%的用户未流失。不存在半衰期，即当用户流失达到50%的时间节点
0-10个月用户流失加快，50-60个月的用户流失速率也有所提升

# 缩短时间查看前20个月
t = np.linspace(0, 20, 21)
fig, ax = plt.subplots(nrows=2, ncols=1, figsize=(12,10))  

# 整体生存曲线
kmf = KaplanMeierFitter()
kmf.fit(df.tenure,  # 代表生存时长
        event_observed=df.Churn,  # 代表事件的终点
        timeline=t
        )
kmf.plot_survival_function(ax=ax[0], label='all')
# 整体生存变化率
naf = NelsonAalenFitter()
naf.fit(df.tenure, event_observed=df.Churn, timeline=t)
naf.plot_hazard(ax=ax[1], bandwidth=20, label='all')
plt.show()

output_10_0

第一个月是用户流失重灾时间
前10个月中，前5个月流失速率加快
查看单变量生存曲线差异

# 连续变量分箱
combiner = toad.transform.Combiner()  
combiner.fit(df[['MonthlyCharges','TotalCharges']], df['Churn'], method='chi', # 卡方分箱
                min_samples=0.05) 
df_t = combiner.transform(df)

# 筛选变量
obj_list = df_t.select_dtypes(include="object").columns[1:].to_list()
obj_list += ['MonthlyCharges','TotalCharges']

# 计算行列数
num_plots = len(obj_list)
num_cols = math.ceil(np.sqrt(num_plots))
num_rows = math.ceil(num_plots/num_cols)
cr = [d for d in itertools.product(range(num_rows),range(num_cols))]
# 绘制生存曲线
fig, ax = plt.subplots(nrows=num_rows, ncols=num_cols, figsize=(48,36)) 

for ind,feature in enumerate(obj_list):
    for i in df_t[feature].unique():
        # KaplanMeier检验
        kmf=KaplanMeierFitter()  
        df_tmp = df_t.loc[df_t[feature] == i]
        kmf.fit(df_tmp.tenure,  # 代表生存时长
                event_observed=df_tmp.Churn,  # 代表事件的终点
                label=i)
        # 绘制生存曲线
        kmf.plot_survival_function(ci_show=False, 
                                   ax = ax[cr[ind][0]][cr[ind][1]])
    # 对数秩检验，得到p值
    p_value = multivariate_logrank_test(event_durations = df_t.tenure,  # 代表生存时长
                                        groups=df_t[feature],  # 代表检验的组别
                                        event_observed=df_t.Churn  # 代表事件的终点
                                       ).p_value
    p_value_text = ['p-value < 0.001' if p_value < 0.001 else 'p-value = %.4F'%p_value][0]
    ax[cr[ind][0]][cr[ind][1]].set_title("{}\n{}".format(feature, p_value_text))

output_13_0

验证多分类单变量生存曲线差异

# 查看多分类变量的组间差异：以PaymentMethod为例
fig, ax = plt.subplots(figsize=(10,6)) 
feature = 'PaymentMethod'
for i in df_t[feature].unique():
    # KaplanMeier检验
    kmf=KaplanMeierFitter()  
    df_tmp = df_t.loc[df_t[feature] == i]
    kmf.fit(df_tmp.tenure,  # 代表生存时长
            event_observed=df_tmp.Churn,  # 代表事件的终点
            label=i)
    # 绘制生存曲线
    kmf.plot_survival_function(ci_show=False, 
                               ax = ax)
# 对数秩检验，得到p值
p_value = multivariate_logrank_test(event_durations = df_t.tenure,  # 代表生存时长
                                    groups=df_t[feature],  # 代表检验的组别
                                    event_observed=df_t.Churn  # 代表事件的终点
                                   ).p_value
p_value_text = ['p-value < 0.001' if p_value < 0.001 else 'p-value = %.4F'%p_value][0]
ax.set_title("{}\n{}".format(feature, p_value_text))
plt.show()

pairwise_logrank_test(
        event_durations = df_t.tenure,
        groups=df_t.PaymentMethod,
        event_observed=df_t.Churn 
)

image-20221221185003629

生存曲线表明PaymentMethod存在组间显著差异
检验表明除了Bank transfer (automatic)与Credit card (automatic)不存在显著差异，其余组间均显著差异
交互变量的生存曲线

# 以性别gender为例，查看与其他变量交互后，gender的流失情况是否存在显著差异
for feature in obj_list:
    for i in df_t[feature].unique():
        df_tmp = df_t.loc[df_t[feature] == i]
        p_value = multivariate_logrank_test(event_durations = df_tmp.tenure,
                                            groups=df_tmp.gender,
                                            event_observed=df_tmp.Churn 
                                           ).p_value
        if p_value <= 0.05:
            p_value_text = ['p-value < 0.001' if p_value < 0.001 else 'p-value = %.4F'%p_value][0]
            print('gender : {}={}, logrank test {}'.format(feature, i, p_value_text))

gender : PaymentMethod=Credit card (automatic), logrank test p-value = 0.0288
gender : MonthlyCharges=3, logrank test p-value = 0.0360
gender : MonthlyCharges=4, logrank test p-value = 0.0181

当 PaymentMethod=Credit card (automatic)时，性别的流失情况有显著差异。同样的，当MonthlyCharges=3或者=4时，性别的流失情况有显著差异

# 查看生存曲线差异：不同性别在PaymentMethod=Credit card (automatic)时的流失差异
fig,ax = plt.subplots(figsize=(12,8))

# 绘制Male：PaymentMethod=Credit card (automatic)生存曲线
data_male = df_t.loc[(df_t.gender=='Male') & (df_t.PaymentMethod == 'Credit card (automatic)')]
kmf_male = KaplanMeierFitter().fit(data_male.tenure, 
                                   data_male.Churn, 
                                   label='Male : PaymentMethod=Credit card (automatic)')
ax = kmf_male.plot_survival_function(ax=ax)
# 绘制Female：PaymentMethod=Credit card (automatic)生存曲线
data_female = df_t.loc[(df_t.gender=='Female') & (df_t.PaymentMethod == 'Credit card (automatic)')]
kmf_female = KaplanMeierFitter().fit(data_female.tenure, 
                                     data_female.Churn, 
                                     label='Female : PaymentMethod=Credit card (automatic)')
ax = kmf_female.plot_survival_function(ax=ax)
# 调整位置
plt.grid()
plt.tight_layout()

output_20_0

在Credit card (automatic)客户中，Female更容易流失

特征工程

# 特征工程
df_model = pd.get_dummies(df.iloc[:,1:], drop_first=True) # 剔除客户ID，并对类别变量进行哑变量处理(剔除首列防止虚拟变量陷阱)
# 拆分训练集
train, test = train_test_split(df_model, test_size=0.2)

数据建模

模型拟合

# 模型拟合
cph = CoxPHFitter(penalizer = 0.01)
cph.fit(train, duration_col='tenure', event_col='Churn')
cph.print_summary(decimals=1)

model	lifelines.CoxPHFitter
duration col	'tenure'
event col	'Churn'
penalizer	0.01
l1 ratio	0
baseline estimation	breslow
number of observations	5625
number of events observed	1500
partial log-likelihood	-10138.3
time fit was run	2022-09-16 15:52:24 UTC

image-20221221185924375

Concordance=0.9，模型拟合不错。Concordance Index是AUC的推广，值越接近1效果越好
系数为正则表示该变量促进流失，系数为负表示该变量防止流失
显著影响的为p<0.05的变量。例如TotalCharges、Partner_Yes等
exp(coef)即为相对危险度。对于连续变量TotalCharges，总费用增加一单位，流失概率是原来的1.00倍（基本没变。）；对于分类变量Contract。表明One year和Two year是对照组（Month-to-month）的0.30和0.09倍

模型优化

# 剔除不显著变量
drop_col = list(cph.summary['p'][cph.summary['p'].values>0.05].index)
train_s = train.drop(drop_col, axis=1)
test_s = test.drop(drop_col, axis=1)

# 重新拟合
cph = CoxPHFitter(penalizer = 0.01)
cph.fit(train_s, duration_col='tenure', event_col='Churn')
cph.print_summary(decimals=1)

image-20221221190131160

特征重要性

# 特征重要性
fig,ax = plt.subplots(figsize=(12,9))
cph.plot(ax=ax)
plt.show()

output_27_0

加速流失的变量有：InternetService_Fiber optic、PaymentMethod_Mailed check等
防止流失的变量有：Contract_Two year、Contract_One year等

风险观察

# 累积风险曲线
fig,ax = plt.subplots(nrows=2, ncols=1, figsize=(10,12))
cph.baseline_cumulative_hazard_.plot(ax=ax[0])
# 各时间节点的风险曲线
cph.baseline_hazard_.plot(ax=ax[1])
plt.show()

output_29_0

在60个月以后流失风险明显提高

模型评估

# 模型校准性
fig,ax = plt.subplots(figsize=(12,9))
survival_probability_calibration(cph, test_s, t0=50, ax=ax) 
ax.grid()

ICI =  0.05854302858419381
E50 =  0.021989969295193035

output_32_1

x轴为预测的流失概率，y轴为观测的流失概率
以50个月为例，模型与基准值（对角线）偏离较大，且一直高估了用户的流失情况
建议样本均衡处理，剔除具有相关性的特征等

# 使用brier score观测校准距离：Brier分数对于一组预测值越低，预测校准越好
loss_dict = {} 
for i in range(1,73): 
    score = brier_score_loss( 
        test_s.Churn, 
        1 - np.array(cph.predict_survival_function(test_s).loc[i]),   
        pos_label=1 ) 
    loss_dict[i] = [score] 
    
loss_df = pd.DataFrame(loss_dict).T 
 
fig, ax = plt.subplots(figsize=(12,9))
ax.plot(loss_df.index, loss_df) 
ax.set(xlabel='Prediction Time', ylabel='Calibration Loss', title='Cox PH Model Calibration Loss / Time') 
ax.grid()

output_34_0

模型在10月-20月的预测效果较好

模型应用

预测剩余价值

# 筛选未流失用户
churn0 = df_model.query("Churn == 0")
# 预测中位数生存时间
churn0_median_survive = cph.predict_median(churn0).rename('median_survive')
# 计算剩余价值=月消费*（预测中位数生存时间）-已存续时间
values = pd.concat([churn0_median_survive, churn0[['MonthlyCharges','tenure']]], axis=1)
values['RemainTenure'] = values['median_survive'] - values['tenure']
values['RemainingValue'] = values['MonthlyCharges'] * values['RemainTenure']
# 计算RemainingValue负值与inf的占比
inf_rate = values[values['RemainingValue']==float('inf')].shape[0]/values.shape[0]
neg_rate = values[values['RemainingValue']<0].shape[0]/values.shape[0]
print('{:*^60}'.format('inf_rate&neg_rate:'), '\n', 'inf_rate=%.2f' % inf_rate, 'neg_rate=%.2f' % neg_rate)
values.tail()

*********************inf_rate&neg_rate:********************* 
 inf_rate=0.32 neg_rate=0.04

image-20230206152201057

如果用户生存曲线不超过0.5，预测的中位生存时间是inf，可以采用cph.predict_percentile(churn0,p=0.6)计算分为数存活时间
预测的最大存活时间为tenure的最大值，即无法预测到观测截面时间后的生存情况。因此也可以将inf定义为最大值
一些用户会在流失前被预测为流失，因此存在剩余生存时间为负。可以通过校准纠偏

# 预测未流失客户的生存曲线
unconditioned_sf = cph.predict_survival_function(churn0)
# 校准
conditioned_sf = unconditioned_sf.apply(lambda c: (c / c.loc[df.loc[c.name, 'tenure']]).clip(upper=1))
# 绘制校准曲线
fig, ax = plt.subplots(figsize=(12,9))
subject = 7038  
tenures = df_model['tenure'].loc[subject,] # 查询index的位置，非自然位置
unconditioned_sf[subject].plot(ls="--", color="#A60628", label="unconditioned") 
conditioned_sf[subject].plot(color="#A60628", label="conditioned on $T>{:.0f}$".format(tenures)) 
ax.legend()
ax.grid()

output_40_0

# 根据校准曲线纠偏
churn0_median_survive_new = median_survival_times(conditioned_sf) 
values_new = pd.concat([churn0[['tenure','MonthlyCharges']], churn0_median_survive_new.T], axis=1)
values_new.rename(columns={0.5:'median_survive'}, inplace=True)
values_new['RemainTenure'] = values_new['median_survive'] - values_new['tenure']
values_new['RemainingValue'] = values_new['MonthlyCharges'] * values_new['RemainTenure']
# 计算RemainingValue负值与inf的占比
inf_rate = values_new[values_new['RemainingValue']==float('inf')].shape[0]/values_new.shape[0]
neg_rate = values_new[values_new['RemainingValue']<0].shape[0]/values_new.shape[0]
print('{:*^60}'.format('inf_rate&neg_rate:'), '\n', 'inf_rate=%.2f' % inf_rate, 'neg_rate=%.2f' % neg_rate)
values_new.tail()

*********************inf_rate&neg_rate:********************* 
 inf_rate=0.33 neg_rate=0.00

image-20230206152222706

剩余价值提升

# 将之前的校准写成函数
def predict_median_func(churn0):
    unconditioned_sf = cph.predict_survival_function(churn0)
    conditioned_sf = unconditioned_sf.apply(lambda c: (c / c.loc[df.loc[c.name, 'tenure']]).clip(upper=1))
    churn0_median_survive_new = median_survival_times(conditioned_sf).T
    return churn0_median_survive_new

# 将月签合同提升至一年合同，若原先是两年合同则维持不变
tmp =df_model.loc[values_new.index,].copy()
tmp['Contract_One year'] = tmp.apply(lambda x: 1 if x['Contract_Two year'] == 0 else 0, axis=1)
values_new['contract_1y_tenure_pred'] = predict_median_func(tmp)
# 将月签合同或者一年合同提升至两年合同
tmp =df_model.loc[values_new.index,].copy()
tmp['Contract_Two year'] = 1
tmp['Contract_One year'] = 0
values_new['contract_2y_tenure_pred'] = predict_median_func(tmp)

# 提升在线安全服务
tmp =df_model.loc[values_new.index,].copy()
tmp['OnlineSecurity_Yes'] = 1
values_new['OnlineSecurity_tenure_pred'] = predict_median_func(tmp)

# 计算剩余价值
values_new['Contract_1y Diff'] = (values_new['contract_1y_tenure_pred'] - values_new['median_survive']) * values_new['MonthlyCharges'] 
values_new['Contract_2y Diff'] = (values_new['contract_2y_tenure_pred'] - values_new['median_survive']) * values_new['MonthlyCharges']
values_new['OnlineSecurity Diff'] = (values_new['OnlineSecurity_tenure_pred'] - values_new['median_survive']) * values_new['MonthlyCharges'] 
# 预览结果
values_new.head().T

image-20230206152247133

# 客户基本信息
df.loc[0,]

customerID                7590-VHVEG
gender                        Female
SeniorCitizen                      0
Partner                          Yes
Dependents                        No
tenure                             1
PhoneService                      No
MultipleLines       No phone service
InternetService                  DSL
OnlineSecurity                    No
OnlineBackup                     Yes
DeviceProtection                  No
TechSupport                       No
StreamingTV                       No
StreamingMovies                   No
Contract              Month-to-month
PaperlessBilling                 Yes
PaymentMethod       Electronic check
MonthlyCharges                 29.85
TotalCharges                   29.85
Churn                              0
Name: 0, dtype: object

0号客户基本信息：使用时长（tenure）1个月，没有OnlineSecurity，Contract为月签。月费为29.85
0号客户基本预测信息：预测27个月，剩余价值805.95
0号客户价值提升信息：
- 更换1年合同后，预测43个月，剩余价值较月签合同提升了447.75
- 更换2年合同后，预测65个月，剩余价值较月签合同提升了1104.45
- 添加OnlineSecurity后，预测31个月，剩余价值较月签合同提升了89.55