import pandas as pd import numpy as np import matplotlib.pyplot as plt import matplotlib as matplot import seaborn as sns # 以内嵌的方式显示图像，不跳出新的窗口 %matplotlib inline

df = pd.read_csv('HR_comma_sep.csv', index_col=None) df.head()

df.shape

df.dtypes

df.isnull().any()

df = df.rename(columns={'satisfaction_level': 'satisfaction', 'last_evaluation': 'evaluation', 'number_project': 'projectCount', 'average_montly_hours': 'averageMonthlyHours', 'time_spend_company': 'yearsAtCompany', 'Work_accident': 'workAccident', 'promotion_last_5years': 'promotion', 'sales' : 'department', 'left' : 'turnover'})

front = df['turnover'] df.drop(labels=['turnover'], axis=1, inplace=True) # 标签数据插到首列 df.insert(0, 'turnover', front) df.head()

turnover_rate = df.turnover.value_counts() / len(df.turnover) turnover_rate

df.describe()

# 分组查看员工的特征 turnover_Summary = df.groupby('turnover') turnover_Summary.mean()

# 相关性矩阵 corr = df.corr() corr

# 通过热力图展示 sns.heatmap(corr, xticklabels=corr.columns.values, yticklabels=corr.columns.values)

# 计算离职员工和未离职员工的满意度 emp_population = df['satisfaction'][df['turnover'] == 0].mean() emp_turnover_satisfaction = df['satisfaction'][df['turnover'] == 1].mean() print( '未离职员工满意度: ' + str(emp_population)) print( '离职员工满意度: ' + str(emp_turnover_satisfaction) )

import scipy.stats as stats # 单样本t检验 stats.ttest_1samp(a=df['satisfaction'][df['turnover']==1], #离职员工样本数据 popmean=emp_population ) #未离职员工满意度均值

# 均值上相等的两份数据，分布上是不是一样 # 概率密度函数估计 fig = plt.figure(figsize=(15,4),) # 绘制样本数据的概率密度函数 ax=sns.kdeplot(df.loc[(df['turnover'] == 0),'evaluation'] , color='b',shade=True,label='no turnover') ax=sns.kdeplot(df.loc[(df['turnover'] == 1),'evaluation'] , color='r',shade=True, label='turnover') ax.set(xlabel='Employee Evaluation', ylabel='Frequency') plt.title('Employee Evaluation Distribution - Turnover V.S. No Turnover')

# 均值上相等的两份数据，分布上是不是一样 # 概率密度函数估计 fig = plt.figure(figsize=(15,4),) # 绘制样本数据的概率密度函数 ax=sns.kdeplot(df.loc[(df['turnover'] == 0),'averageMonthlyHours'] , color='b',shade=True,label='no turnover') ax=sns.kdeplot(df.loc[(df['turnover'] == 1),'averageMonthlyHours'] , color='r',shade=True, label='turnover') ax.set(xlabel='Employee averageMonthlyHours', ylabel='Frequency') plt.title('Employee averageMonthlyHours Distribution - Turnover V.S. No Turnover')

# 均值上相等的两份数据，分布上是不是一样 # 概率密度函数估计 fig = plt.figure(figsize=(15,4),) # 绘制样本数据的概率密度函数 ax=sns.kdeplot(df.loc[(df['turnover'] == 0),'satisfaction'] , color='b',shade=True,label='no turnover') ax=sns.kdeplot(df.loc[(df['turnover'] == 1),'satisfaction'] , color='r',shade=True, label='turnover') ax.set(xlabel='Employee satisfaction', ylabel='Frequency') plt.title('Employee satisfaction Distribution - Turnover V.S. No Turnover')

# 将string类型转换为整数类型 df['department'] = df['department'].astype('category').cat.codes df['salary'] = df['salary'].astype('category').cat.codes

df.head()

# 生成测试数据和训练数据 from sklearn.model_selection import train_test_split X = df.drop(['turnover'], axis=1) y = df.turnover # 由于数据中离职人员数据仅占24%，数据不平衡，不加stratify可能会造成测试数据集中没有几条离职数据 # stratify保证原始数据、训练数据、测试数据三份数据中离职员工和未离职员工比例完全一样 # 如果样本数据均衡那么没必要这么做 X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.15, random_state=123, stratify=y)

from sklearn.metrics import roc_auc_score from sklearn.metrics import classification_report from sklearn import tree from sklearn.tree import DecisionTreeClassifier # 决策树 dtree = tree.DecisionTreeClassifier( criterion='entropy', # max_depth=3, 定义树的深度，防止过拟合 min_samples_leaf=0.01 #定义叶子节点最少要包含多少个样本，防止过拟合 ) dtree = dtree.fit(X_train, y_train)

print("--决策树--") dt_roc_auc = roc_auc_score(y_test, dtree.predict(X_test)) print("决策树AUC：%2.2f" %dt_roc_auc) print(classification_report(y_test, dtree.predict(X_test)))