yes, it is random forest classifier from scikit learn. Thank you. Am Fr., 8. Mai 2020 um 21:50 Uhr schrieb MRAB <pyt...@mrabarnett.plus.com>:
> On 2020-05-08 20:02, joseph pareti wrote: > > In general I prefer doing: > > > > > > X_train, X_test, y_train, y_test = train_test_split(X, y, > test_size=0.33, random_state=42) > >clf = RandomForestClassifier(n_estimators = 100, max_depth= > > None) *clf_f = clf.fit(X_train, y_train)* predicted_labels = > clf_f.predict( > > X_test) score = clf.score(X_test, y_test) score1 = > metrics.accuracy_score( > > y_test, predicted_labels) > > > > > > rather than: > > > > X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.33, > > random_state=42) clf0=RandomForestClassifier(n_estimators=100, max_depth= > > None) *clf0.fit(X_train, y_train)* y_pred =clf0.predict(X_test) score= > > metrics.accuracy_score(y_test, y_pred) > > > > > > Are the two codes really equivalent? > > > You didn't give any context and say what package you're using! > > After searching for "RandomForestClassifier", I'm guessing that you're > using scikit. > > From the documentation here: > > > https://scikit-learn.org/stable/modules/generated/sklearn.ensemble.RandomForestClassifier.html#sklearn.ensemble.RandomForestClassifier.fit > > it says: > > Returns: self : object > > so it looks like clf.fit(...) returns clf. > > That being the case, then, yes, they're equivalent. > -- > https://mail.python.org/mailman/listinfo/python-list > -- Regards, Joseph Pareti - Artificial Intelligence consultant Joseph Pareti's AI Consulting Services https://www.joepareti54-ai.com/ cell +49 1520 1600 209 cell +39 339 797 0644 -- https://mail.python.org/mailman/listinfo/python-list