我有两个单独的
python函数,其中一个使用cross_val_predict返回数据集的预测值,另一个使用cross_validate返回多个错误度量值.下面显示的是用于获取度量值的方法(我已经实现了类似的方法来获取预测).
def metric_val(folds): . . . scoring = {'r_score': 'r2','abs_error': 'neg_mean_absolute_error','squared_error': 'neg_mean_squared_error'} scores = cross_validate(best_svr,X,y,scoring=scoring,cv=folds,return_train_score=True) print("****\nR2 :","",scores['test_r_score'].mean(),"| MAE :",scores['test_abs_error'].mean(),) return prediction
解决方法
有可能装备一个得分手,以便它返回预测,虽然这有点像黑客.这是怎么做的:
cross_validate()函数可以使用自定义评分函数.评分函数必须返回一个数字,但您可以在函数内部执行任何操作.由于您拥有clf和所有测试数据,只需保存clf.predict()的输出,然后返回一个虚拟值以保持得分者满意.有关详细信息,请参阅Implementing your own scoring object上的sklearn docs.
像这样:
from sklearn import svm,datasets from sklearn.model_selection import train_test_split,cross_validate,cross_val_predict # example data iris = datasets.load_iris() X,y = iris.data,iris.target clf = svm.SVC(probability=True,random_state=0)
def get_preds(clf,y): # y is required for a scorer but we won't use it with open("pred.csv","ab+") as f: # append each fold to file np.savetxt(f,clf.predict(X)) return 0 scoring = {'preds': get_preds,'accuracy': 'accuracy','recall': 'recall_macro'} # add desired scorers here k = 5 cross_validate(clf,return_train_score=True,cv = k)
重新加载get_preds(),重新整形以匹配折叠集,并在折叠中平均:
preds = np.loadtxt("pred.csv").reshape(k,len(X)) my_preds = np.mean(my_preds,axis=0).round()
与cross_val_predict()预测比较:
cv_preds = cross_val_predict(clf,cv=k) np.equal(my_preds,cv_preds).sum() # 487 out of 500
我们在makehift get_preds()方法和cross_val_predict()之间看到了几乎完美的一致.小分歧可能是由于我的平均方法与cross_val_predict不同(我只是舍入到最接近的整数类,不是非常复杂),或者它可能与sklearn
cross-validation docs中这个稍微神秘的音符有关:
Note that the result of this computation may be slightly different from those obtained using cross_val_score as the elements are grouped in different ways.