Python逻辑回归模型应用举例
取UCI公共测试数据库中澳大利亚信贷批准数据集作为本例数据集，
其拥有14个特征，1个分类标签y（1--同意贷款，0--不同意贷款）共计690个申请者记录
1、数据获取
import pandas as pd
data = pd.read_excel('credit.xlsx')
data
| x1 | x2 | x3 | x4 | x5 | x6 | x7 | x8 | x9 | x10 | x11 | x12 | x13 | x14 | d | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | 22.08 | 11.460 | 2 | 4 | 4 | 1.585 | 0 | 0 | 0 | 1 | 2 | 100 | 1213 | 0 |
| 1 | 0 | 22.67 | 7.000 | 2 | 8 | 4 | 0.165 | 0 | 0 | 0 | 0 | 2 | 160 | 1 | 0 |
| 2 | 0 | 29.58 | 1.750 | 1 | 4 | 4 | 1.250 | 0 | 0 | 0 | 1 | 2 | 280 | 1 | 0 |
| 3 | 0 | 21.67 | 11.500 | 1 | 5 | 3 | 0.000 | 1 | 1 | 11 | 1 | 2 | 0 | 1 | 1 |
| 4 | 1 | 20.17 | 8.170 | 2 | 6 | 4 | 1.960 | 1 | 1 | 14 | 0 | 2 | 60 | 159 | 1 |
| ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... |
| 685 | 1 | 31.57 | 10.500 | 2 | 14 | 4 | 6.500 | 1 | 0 | 0 | 0 | 2 | 0 | 1 | 1 |
| 686 | 1 | 20.67 | 0.415 | 2 | 8 | 4 | 0.125 | 0 | 0 | 0 | 0 | 2 | 0 | 45 | 0 |
| 687 | 0 | 18.83 | 9.540 | 2 | 6 | 4 | 0.085 | 1 | 0 | 0 | 0 | 2 | 100 | 1 | 1 |
| 688 | 0 | 27.42 | 14.500 | 2 | 14 | 8 | 3.085 | 1 | 1 | 1 | 0 | 2 | 120 | 12 | 1 |
| 689 | 1 | 41.00 | 0.040 | 2 | 10 | 4 | 0.040 | 0 | 1 | 1 | 0 | 1 | 560 | 1 | 1 |
690 rows × 15 columns
2、训练样本与测试样本划分
#训练用的特征数据用x表示，预测变量用y表示 测试样本分别记为x1,y1
#以前600数据为训练数据，后90个为测试数据
x = data.iloc[:600,:14].values
x
array([[1.000e+00, 2.208e+01, 1.146e+01, ..., 2.000e+00, 1.000e+02,
1.213e+03],
[0.000e+00, 2.267e+01, 7.000e+00, ..., 2.000e+00, 1.600e+02,
1.000e+00],
[0.000e+00, 2.958e+01, 1.750e+00, ..., 2.000e+00, 2.800e+02,
1.000e+00],
...,
[1.000e+00, 3.492e+01, 2.500e+00, ..., 2.000e+00, 2.390e+02,
2.010e+02],
[1.000e+00, 2.408e+01, 8.750e-01, ..., 2.000e+00, 2.540e+02,
1.951e+03],
[1.000e+00, 3.733e+01, 6.500e+00, ..., 2.000e+00, 9.300e+01,
1.000e+00]])
y = data.iloc[:600,14].values
y
array([0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0,
0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0,
0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1,
0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1,
1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0,
0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0,
1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0,
0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0,
0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1,
1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1,
1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1,
1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1,
1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0,
1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0,
1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0,
1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1,
0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0,
1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1,
0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0,
0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0,
1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1,
0, 0, 1, 1, 0, 1], dtype=int64)
x1 = data.iloc[600:,:14].values
x1
array([[0.000e+00, 2.075e+01, 9.540e+00, ..., 2.000e+00, 2.000e+02,
1.001e+03],
[1.000e+00, 3.667e+01, 3.250e+00, ..., 2.000e+00, 1.020e+02,
6.400e+02],
[1.000e+00, 2.258e+01, 1.004e+01, ..., 2.000e+00, 6.000e+01,
3.970e+02],
...,
[0.000e+00, 1.883e+01, 9.540e+00, ..., 2.000e+00, 1.000e+02,
1.000e+00],
[0.000e+00, 2.742e+01, 1.450e+01, ..., 2.000e+00, 1.200e+02,
1.200e+01],
[1.000e+00, 4.100e+01, 4.000e-02, ..., 1.000e+00, 5.600e+02,
1.000e+00]])
y1 = data.iloc[600:,14].values
y1
array([0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1,
1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0,
0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0,
0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1,
1, 1], dtype=int64)
3、逻辑回归分析
#导入逻辑回归模块（LR）
from sklearn.linear_model import LogisticRegression as LR
#利用LR创建逻辑回归对象lr
lr = LR(max_iter=3000)
#调用lr中的fit()方法进行训练
lr.fit(x,y)
LogisticRegression(max_iter=3000)
这里遇到一个问题：TOP: TOTAL NO. of ITERATIONS REACHED LIMIT......extra_warning_msg=_LOGISTIC_SOLVER_CONVERGENCE_MSG
解决办法
意思是达到限制的迭代总数，只需要增加迭代次数（最大值）或缩放数据就可以。
将代码改为（增加迭代次数）：
最大迭代次数默认值为1000，把它改为3000即可
lr = LR(max_iter=3000)
#调用lr中的score（）方法返回模型准确率
r = lr.score(x,y) #模型准确率(针对训练数据)
r
0.875
#调用lr中的predict()方法，对测试样本x1进行预测，获取预测结果
R = lr.predict(x1)
R
array([0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1,
1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0,
1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0,
0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1,
1, 0], dtype=int64)
#预测准确率
Z = R-y1
Rs=len(Z[Z==0])/len(Z)
Rs
0.8666666666666667
import pandas as pd
data = pd.read_excel('credit.xlsx')
x = data.iloc[:600,:14].values
y = data.iloc[:600,14].values
x1= data.iloc[600:,:14].values
y1= data.iloc[600:,14].values
from sklearn.linear_model import LogisticRegression as LR
lr = LR(max_iter=3000) #创建逻辑回归模型类
lr.fit(x, y) #训练数据
r=lr.score(x, y); # 模型准确率（针对训练数据）
print('模型准确率（针对训练数据）:',r)
R=lr.predict(x1)
Z=R-y1
Rs=len(Z[Z==0])/len(Z)
print('预测结果为：',R)
print('预测准确率为：',Rs)
模型准确率（针对训练数据）: 0.875
预测结果为： [0 1 1 1 1 0 0 1 0 1 1 0 1 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 0 0 0 0 1 0 0 1 0
0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1 1 0 1
0 0 0 0 0 1 0 1 1 0 1 1 0 1 1 0]
预测准确率为： 0.8666666666666667
