Python邏輯回歸模型應用舉例

取UCI公共測試數據庫中澳大利亞信貸批准數據集作為本例數據集，
其擁有14個特征，1個分類標簽y（1--同意貸款，0--不同意貸款）共計690個申請者記錄

1、數據獲取

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、訓練樣本與測試樣本划分

#訓練用的特征數據用x表示，預測變量用y表示   測試樣本分別記為x1,y1
#以前600數據為訓練數據，后90個為測試數據

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、邏輯回歸分析

#導入邏輯回歸模塊（LR）
from sklearn.linear_model import LogisticRegression as LR

#利用LR創建邏輯回歸對象lr
lr = LR(max_iter=3000)

#調用lr中的fit()方法進行訓練
lr.fit(x,y)

LogisticRegression(max_iter=3000)

這里遇到一個問題：TOP: TOTAL NO. of ITERATIONS REACHED LIMIT......extra_warning_msg=_LOGISTIC_SOLVER_CONVERGENCE_MSG

解決辦法
意思是達到限制的迭代總數，只需要增加迭代次數（最大值）或縮放數據就可以。
將代碼改為（增加迭代次數）：

最大迭代次數默認值為1000，把它改為3000即可
lr = LR(max_iter=3000)

#調用lr中的score（）方法返回模型准確率
r = lr.score(x,y)  #模型准確率(針對訓練數據)
r

0.875

#調用lr中的predict()方法，對測試樣本x1進行預測，獲取預測結果
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)

#預測准確率
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)   #創建邏輯回歸模型類
lr.fit(x, y) #訓練數據
r=lr.score(x, y); # 模型准確率（針對訓練數據）
print('模型准確率（針對訓練數據）:',r)
R=lr.predict(x1)
Z=R-y1
Rs=len(Z[Z==0])/len(Z)
print('預測結果為：',R)
print('預測准確率為：',Rs)

模型准確率（針對訓練數據）: 0.875
預測結果為： [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]
預測准確率為： 0.8666666666666667