x_test,model.predict(x_test)) 有图可我不知道方程的参数 都有model了怎么能不知道 model.coef_ #array([[1.00116024]]) 斜率model.intercept...np.square(model.predict(x1) - y1)) 16.63930773735106 # 这个不错，挺小的 我不信是最好的，那就稍稍加一点截距0.01 y2 = model.coef_*x1 + model.intercept...10,10)) plt.scatter(x1,y1) plt.plot(x1,model.predict(x1),color = 'r') plt.plot(x1 , model.coef_*x1 + model.intercept
²，再看看系数 >>> print('intercept:', model.intercept_) intercept: 5.633333333333329 >>> print('slope:', model.coef...[ 8.33333333 13.73333333 19.13333333 24.53333333 29.93333333 35.33333333] 当然也可以使用下面的方法 >>> y_pred = model.intercept...of determination:', r_sq) coefficient of determination: 0.8615939258756776 >>> print('intercept:', model.intercept...of determination:', r_sq) coefficient of determination: 0.8908516262498564 >>> print('intercept:', model.intercept
model.predict(xfit[:, np.newaxis]) print "Model slope: ", model.coef_[0] print "Model intercept:", model.intercept
LogisticRegression() model.fit(x_train, y_train) print("w: ", model.coef_) print("b: ", model.intercept...load_data() model = LogisticRegression() model.fit(x_train, y_train) print("w: ", model.coef_) print("b: ", model.intercept
model.predict(x)#结果：[33.56614386 30.32827794 32.28321585] print(model.coef_) #coefficent,输出系数 0.061 print(model.intercept
linear_model.LinearRegression() model.fit(x,y) 第六步：模型评估 model_coef = model.coef_ #获取模型自变量系数并赋值给model_coef model_intercept = model.intercept
获取模型参数 拟合完成后，我们可以获取模型的参数，即斜率和截距： slope = model.coef_[0] intercept = model.intercept_ 6.
model.predict(X) , color='blue') plt.xlabel('气温') plt.ylabel('销售量') plt.show() print("截距与斜率:",model.intercept
，我们使用训练数据拟合模型： model.fit(X, y) 获取模型参数 拟合完成后，我们可以获取模型的参数，即斜率和截距： slope = model.coef_[0] intercept = model.intercept
#得到训练和测试训练集model = LinearRegression() #导入线性回归model.fit(x_train, y_train) # model.coef_ # 斜率 有三个model.intercept
train_y) # In[34]: #训练模型的系数 print('Coefficient of model :', model.coef_) #拦截模型 print('Intercept of model',model.intercept
print('\nCoefficient of model :', model.coef_) # intercept of the model print('\nIntercept of model',model.intercept
X) print('一元线性回归拟合的R平方的值：', r_sq) print('模型的均方误差值：', mean_squared_error(y, y_pred)) print('b0的值：', model.intercept
Predictions:", y_pred) print("Model coefficients:", model.coef_) # 输出斜率 print("Model intercept:", model.intercept...model.intercept_: 这是一个属性，存储了模型拟合后的截距项。
= model.predict(X_test) # 打印结果 print(f"预测值: {y_pred}") print(f"模型系数: {model.coef_}") print(f"截距: {model.intercept
>>>model.coef_ # 拟合的直线斜率 array([1.9776566]) >>>model.intercept_ # 拟合的直线截距 -0.9033107255311164
喂训练数据进去，但是需要把因变量转换成1列多行的数据 model.fit(xtrain[:,np.newaxis],ytrain) # 打印斜率 print(model.coef_) # 打印截距 print(model.intercept
label='Linear regression') plt.legend() plt.show() print("线性回归系数:", model.coef_) print("线性回归截距:", model.intercept...='Polynomial regression') plt.legend() plt.show() print("多项式回归系数:", model.coef_) print("多项式回归截距:", model.intercept
min(), X[:, 0].max(), 50), np.linspace(X[:, 1].min(), X[:, 1].max(), 50)) zz = -(model.intercept
in range(1, 6)] result model = LinearRegression().fit(train_X, train_y_ln) print('intercept:'+ str(model.intercept...数据偏移得多一点也不会对结果造成什么影响，专业一点的说法是『抗扰动能力强』 model = Ridge().fit(train_X, train_y_ln) print('intercept:'+ str(model.intercept...model = Lasso().fit(train_X, train_y_ln) print('intercept:'+ str(model.intercept_)) sns.barplot(abs(model.coef
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