模型分数()与r2_score之间的差异 、 我正在训练一个LinearRegression()分类器,并试图估计它的预测应计性。from sklearn.metrics importr2_scoreregr_rf = LinearRegressionregr_rf.predict(df[features])print score score2 =r2_score( 浏览7提问于2017-08-06得票数10 ...
我们可以用测试集test_x数据产生预测的房价pred_test_y,下面我们用sklearn中的r2_score来验证测试集的准确率 8.准确率计算 from sklearn.metrics import r2_score pred_acc = r2_score(test_y, pred_test_y) print('pred_acc',pred_acc) 9.为了更直观我们在进行可视化一下。 #绘图 import matplotlib....
AI代码解释 from sklearn.metrics import classification_report, confusion_matrix, r2_score, recall_score, accuracy_score In 35: 代码语言:txt AI代码解释 # 精度、R2 print("多分类预测建模的精度acc为: ",accuracy_score(test_labels,y_predict)) print("多分类预测建模的R方为: ",r2_score(test_labels...
R2 = metrics.r2_score(predicted_stock_price, real_stock_price) print('均方误差: %.5f' % MSE) print('均方根误差: %.5f' % RMSE) print('平均绝对误差: %.5f' % MAE) print('R2: %.5f' % R2) # test_set = data.loc[len(data) - n_timestamp * 2:len(data), 'close].values #...
import warningswarnings.filterwarnings("ignore")import numpy as npnp.random.seed(1337)from keras.models import Sequentialfrom keras.layers import Densefrom sklearn.metrics import r2_scoreimport matplotlib.pyplot as plt Using TensorFlow backend.
import numpy as np np.random.seed(1337) from keras.models import Sequential from keras.layers import Dense from sklearn.metrics import r2_score import matplotlib.pyplot as plt # 创建数据集 # 在[-1,1]的区间内等间隔创建200个样本数 X = np.linspace(-1, 1, 200) np.random.shuffle(X) # ...
R2=metrics.r2_score(TestY,TestPrediction) RMSE=metrics.mean_squared_error(TestY,TestPrediction)**0.5 print('Pearson correlation coefficient is {0}, and RMSE is {1}.'.format(Pearsonr[0],RMSE)) return (Pearsonr[0],R2,RMSE) # Save key parameters. ...
R2=metrics.r2_score(TestY,TestPrediction) RMSE=metrics.mean_squared_error(TestY,TestPrediction)**0.5 print('Pearson correlation coefficient is {0}, and RMSE is {1}.'.format(Pearsonr[0],RMSE)) return (Pearsonr[0],R2,RMSE) # Save key parameters. ...
Note that besides the MSE and MAE scores, you could also use the R2 score or the regression score function. Here, you should go for a score of 1.0, which is the best. However, the score can also be negative!from sklearn.metrics import r2_score r2_score(Y[test], y_pred) ...
Note that besides the MSE and MAE scores, you could also use the R2 score or the regression score function. Here, you should go for a score of 1.0, which is the best. However, the score can also be negative!from sklearn.metrics import r2_score r2_score(Y[test], y_pred) ...