CODEWITHSUNDEEP
pythonscikit-learncross-validationgrid-search
Titus Pullo
Titus Pullo

Reputation: 3851

scikit-learn GridSearchCV with multiple repetitions

I'm trying to get the best set of parameters for an SVR model. I'd like to use the GridSearchCV over different values of C. However, from the previous test, I noticed that the split into the Training/Test set highly influences the overall performance (r2 in this instance). To address this problem, I'd like to implement a repeated 5-fold cross-validation (10 x 5CV). Is there a built-in way of performing it using GridSearchCV?

Quick solution, following the idea presented in the sci-kit official documentation:

NUM_TRIALS = 10
scores = []
for i in range(NUM_TRIALS):
     cv = KFold(n_splits=5, shuffle=True, random_state=i)
     clf = GridSearchCV(estimator=svr, param_grid=p_grid, cv=cv)
     scores.append(clf.best_score_)
print "Average Score: {0} STD: {1}".format(numpy.mean(scores), numpy.std(scores))

Upvotes: 22

Views: 20872

Answers (2)

Vivek Kumar
Vivek Kumar

Reputation: 36619

This is called as nested cross_validation. You can look at official documentation example to guide you into right direction and also have a look at my other answer here for a similar approach.

You can adapt the steps to suit your need:

svr = SVC(kernel="rbf")
c_grid = {"C": [1, 10, 100, ...  ]}

# CV Technique "LabelKFold", "LeaveOneOut", "LeaveOneLabelOut", etc.

# To be used within GridSearch (5 in your case)
inner_cv = KFold(n_splits=5, shuffle=True, random_state=i)

# To be used in outer CV (you asked for 10)
outer_cv = KFold(n_splits=10, shuffle=True, random_state=i)

# Non_nested parameter search and scoring
clf = GridSearchCV(estimator=svr, param_grid=c_grid, cv=inner_cv)
clf.fit(X_iris, y_iris)
non_nested_score = clf.best_score_

# Pass the gridSearch estimator to cross_val_score
# This will be your required 10 x 5 cvs
# 10 for outer cv and 5 for gridSearch's internal CV
clf = GridSearchCV(estimator=svr, param_grid=c_grid, cv=inner_cv)
nested_score = cross_val_score(clf, X=X_iris, y=y_iris, cv=outer_cv).mean()

Edit - Description of nested cross validation with cross_val_score() and GridSearchCV()

  1. clf = GridSearchCV(estimator, param_grid, cv= inner_cv).
  2. Pass clf, X, y, outer_cv to cross_val_score
  3. As seen in source code of cross_val_score, this X will be divided into X_outer_train, X_outer_test using outer_cv. Same for y.
  4. X_outer_test will be held back and X_outer_train will be passed on to clf for fit() (GridSearchCV in our case). Assume X_outer_train is called X_inner from here on since it is passed to inner estimator, assume y_outer_train is y_inner.
  5. X_inner will now be split into X_inner_train and X_inner_test using inner_cv in the GridSearchCV. Same for y
  6. Now the gridSearch estimator will be trained using X_inner_train and y_train_inner and scored using X_inner_test and y_inner_test.
  7. The steps 5 and 6 will be repeated for inner_cv_iters (5 in this case).
  8. The hyper-parameters for which the average score over all inner iterations (X_inner_train, X_inner_test) is best, is passed on to the clf.best_estimator_ and fitted for all data, i.e. X_outer_train.
  9. This clf (gridsearch.best_estimator_) will then be scored using X_outer_test and y_outer_test.
  10. The steps 3 to 9 will be repeated for outer_cv_iters (10 here) and array of scores will returned from cross_val_score
  11. We then use mean() to get back nested_score.

Upvotes: 36

AdamRH
AdamRH

Reputation: 321

You can supply different cross-validation generators to GridSearchCV. The default for binary or multiclass classification problems is StratifiedKFold. Otherwise, it uses KFold. But you can supply your own. In your case, it looks like you want RepeatedKFold or RepeatedStratifiedKFold.

from sklearn.model_selection import GridSearchCV, RepeatedStratifiedKFold

# Define svr here
...

# Specify cross-validation generator, in this case (10 x 5CV)
cv = RepeatedKFold(n_splits=5, n_repeats=10)
clf = GridSearchCV(estimator=svr, param_grid=p_grid, cv=cv)

# Continue as usual
clf.fit(...)

Upvotes: 32

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