Python Logistic回归模型系数

我试图对diabet进行逻辑回归，得到模型的结果。我假设每个变量都有1个系数，但结果给了我3个不同的系数列表，还有3个不同的截距。 我试过线性回归，它给每个人1

import pandas as pd
import sklearn
from sklearn.linear_model import LogisticRegression
import numpy as np
from sklearn import linear_model, preprocessing
data = pd.read_csv ('diabetestype.csv' , sep = ',')

le = preprocessing.LabelEncoder()
Age = list(data['Age']) #will take all buying to a list and transform into proper integer values 
BSf = list(data['BS Fast'])
BSp = list(data['BS pp'])
PR = list(data['Plasma R'])
PF = list(data['Plasma F'])
Hb = list(data['HbA1c'])
Type = le.fit_transform(list(data['Type']))

X = list(zip(Age, BSf,BSp,PR,PF,Hb))
y = list(Type)


x_train,x_test, y_train,y_test = sklearn.model_selection.train_test_split(X, y, test_size = 0.1)
# model = linear_model.LinearRegression()
model = LogisticRegression()
model.fit (x_train,y_train)
acc = model.score(x_test,y_test)
coef = model.coef_
inter = model.intercept_
prediction = model.predict(x_test)
for i in range (5):
    print ('predicted ', prediction[i],'variables  ', x_test[i] , 'actual', y_test[i])
print(acc)
print(coef, inter)

结果是----

predicted  1 variables (2, 9, 14, 6, 6, 10) actual 1
predicted  2 variables (33, 7, 0, 9, 8, 8) actual 2
predicted  0 variables (19, 4, 4, 3, 2, 0) actual 0
predicted  0 variables (7, 15, 9, 5, 5, 3) actual 0
predicted  0 variables (16, 4, 4, 3, 2, 0) actual 0
1.0
[[-0.02543341  0.3763792  -0.2116062  -1.36365511 -0.87416662 -1.8448327 ]
 [ 0.00940748 -1.12894486  1.50994009  1.1101098   1.23563738 -0.2574385 ]
 [ 0.01602593  0.75256566 -1.29833389  0.25354531 -0.36147076  2.1022712 ]] [ 28.79209663 -19.24933782  -9.54275881]
C:\Users\nk\anaconda3\lib\site-packages\sklearn\linear_model\_logistic.py:764: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  extra_warning_msg=_LOGISTIC_SOLVER_CONVERGENCE_MSG)

从：

系数：形状阵列（1，n个特征）或（n个类别，n个特征）

（与截距相同）

你有3门课。

在这个最小的示例中，还有3个类：

from sklearn.datasets import load_iris
from sklearn.linear_model import LogisticRegression
X, y = load_iris(return_X_y=True)
clf = LogisticRegression(random_state=0).fit(X, y)
clf.predict(X[:2, :])

clf.predict_proba(X[:2, :])


clf.score(X, y)

set(y)  # >>>{0, 1, 2} --> there are 3 classes



clf.coef_ # >>> array([[-0.41874027,  0.96699274, -2.52102832, -1.08416599],
          #            [ 0.53123044, -0.31473365, -0.20002395, -0.94866082],
          #            [-0.11249017, -0.65225909,  2.72105226,  2.03282681]])

clf.coef_.shape # >>> (3, 4)

clf.intercept_ # >>> array([  9.84028024,   2.21683511, -12.05711535])

您需要能够辨别样本是否属于哪个类。无论您测试的是哪个类，结果都将在0或1之间。

---

例如，使用第一行的

coef\uu

检查它是否属于类别1…等等。

当我尝试预测样本时，我将如何使用ax+b=c进行建模。我应该尝试所有带有变量的类吗？之后我将如何评估结果？谢谢you@N.K我不确定我是否理解这个问题。您将在使用multiclass时找到定义。否则，

样本x

属于

类别0

或

Pr（类别0 | x）

的概率为

e^（b0+b1*x）/（1+e^（b0+b1*x））