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python cross validation

Trained Tuna answered on May 17, 2021 Popularity 9/10 Helpfulness 4/10

answer python cross validation

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python cross validation

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from sklearn.model_selection import cross_val_score
scores = cross_val_score(model   # Ridge(alpha=1)
                       , X_train # scaler.fit_transform(x_train)
                       , y_train # scaler.fit_transform(y_train)
                       , scoring='neg_mean_squared_error' # depends on model more at https://scikit-learn.org/stable/modules/model_evaluation.html
                       , cv=5)   # 5 fold cross validation
mean(scores) # negMSE (higher=better), adj hyper params to optimize 
model.fit(x_train, y_train)      # make sure to fit the model again after 
y_final_test_pred = model.predict(x_test) # Final predictions
mean_squared_error(y_test, y_final_test_pred) # Final MSE on 'new' data

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Contributed on May 10 2022

Trained Tuna

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# SVC: support vector classifier (one of the "built-in" classifiers in scikit-learn)
# X, y: array-like representing input and target variables
# X.shape = (N, num_of_features)
# y.shape = (N, 1) in case of classification problem

from sklearn.model_selection import cross_val_score
clf = svm.SVC(kernel='linear', C=1, random_state=42)
scores = cross_val_score(clf, X, y, cv=5) # 5-fold cross validation

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Source: scikit-learn.org

Tags: cross-validation cros

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Contributed on May 17 2021

wolf-like_hunter

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cross validation python

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from sklearn.model_selection import cross_val_score, cross_val_predict, KFold
from sklearn.model_selection import GridSearchCV, RandomizedSearchCV
from sklearn.metrics import mean_absolute_error, make_scorer

kf = KFold(n_splits=5, shuffle=True, random_state=42)
# splits = kf.split(X) # See how they are splitted, each split contains index of training and validation
# r-squared results for 5-fold cross validation score  
mae_scorer = make_scorer(mean_absolute_error)
scores = cross_val_score(your_model, X, y, cv=kf, scoring=mae_scorer)  # a list of error terms
avg_score = np.mean(scores)
# predicted_y results for 5-fold cross validation prediction
predicted_y = cross_val_predict(your_model, X, y, cv=5) # a list of predictions
avg_predicted_y = np.mean(predicted_y)

### example of ridge regression with grid search with k-fold cross validation
param_grid = {"alpha": np.arange(0.0001, 1, 10), "solver": ["sag", "lsqr"]}
ridge = Ridge()
ridge_cv = GridSearchCV(ridge, param_grid, cv=kf)
ridge_cv2 = RandomizedSearchCV(ridge, param_grid, cv=kf, n_iter=2)
ridge_cv.fit(X_train, y_train)

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Contributed on Jan 29 2023

Innocent Iguana

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