01- 机器学习经典流程 (中国人寿保费项目) (项目一)

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删除特征: data = data.drop(['region', 'sex'], axis=1)
特征数据调整: data.apply( )

# 体重指数离散化转换体重两种情况标准、肥胖
def convert(df,bmi):
    df['bmi'] = 'fat' if df['bmi'] >= bmi else 'standard'
    return df
data = data.apply(convert, axis = 1, args=(30,))  # args 是传参

将非数字类型特征转换为数值: data = pd.get_dummies(data)
设定目标值和特征值:

X = data.drop('charges', axis=1) # 训练数据
y = data['charges'] # 目标值

中国人寿保费项目

数据存储: https://blog.csdn.net/March_A/article/details/128985290

1. 导入数据

import numpy as np
import pandas as pd
data = pd.read_excel('./中国人寿.xlsx')
data.head(10)

2. 查看数据特征

import seaborn as sns
# 性别对保费影响
sns.kdeplot(data['charges'],shade = True,hue = data['sex'])

# 地区对保费影响
sns.kdeplot(data['charges'],shade = True,hue = data['region'])

# 吸烟对保费影响
sns.kdeplot(data['charges'],shade = True,hue = data['smoker'])

# 孩子数量对保费影响
sns.kdeplot(data['charges'],shade = True,hue = data['children'],palette='Set1')

不同性别对保费影响不大不同性别的保费的概率分布曲线基本重合因此这个特征无足轻重可以删除, 地区同理
吸烟与否对保费的概率分布曲线差别很大整体来说不吸烟更加健康那么保费就低这个特征很重要
家庭孩子数量对保费有一定影响

3. 删除不重要数据

data = data.drop(['region', 'sex'], axis=1)

4 特征属性调整

# 体重指数离散化转换体重两种情况标准、肥胖
def convert(df,bmi):
    df['bmi'] = 'fat' if df['bmi'] >= bmi else 'standard'
    return df
data = data.apply(convert, axis = 1, args=(30,))  # args 是传参
data.head()

5 非数值转换为数据类型 (去字符串)

# 特征提取离散型数据转换为数值型数据
data = pd.get_dummies(data)
data.head()

6 设定目标值和特征值

# 特征和目标值抽取
X = data.drop('charges', axis=1) # 训练数据
y = data['charges'] # 目标值
X.head()

7 数据拆分

# 数据拆分
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)

8 数据升维

# 特征升维
from sklearn.preprocessing import PolynomialFeatures
poly = PolynomialFeatures(degree= 2, include_bias = False)
X_train_poly = poly.fit_transform(X_train)
X_test_poly = poly.fit_transform(X_test)

9 模型训练

# 模型训练
from sklearn.linear_model import LinearRegression
model_1 = LinearRegression()
model_1.fit(X_train_poly, y_train)

# 使用弹性网络训练
from sklearn.linear_model import ElasticNet
model_2 = ElasticNet(alpha = 0.3,l1_ratio = 0.5,max_iter = 50000)
model_2.fit(X_train_poly,y_train)

10 模型评估

model_1.score(X_train_poly, y_train)  # 0.8624083431844988
model_2.score(X_test_poly, model_2.predict(X_test_poly))   # 1.0

# 模型评估
from sklearn.metrics import mean_squared_error,mean_squared_log_error
print('训练数据均方误差',
      np.sqrt(mean_squared_error(y_train,model_1.predict(X_train_poly))))
print('测试数据均方误差',
      np.sqrt(mean_squared_error(y_test,model_1.predict(X_test_poly))))

print('训练数据对数误差',
      np.sqrt(mean_squared_log_error(y_train,model_1.predict(X_train_poly))))
print('测试数据对数误差',
      np.sqrt(mean_squared_log_error(y_test,model_1.predict(X_test_poly))))

结论