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I was working through a tutorial on the titanic disaster from Kaggle and I'm getting different results depending on the details of how I use cross_validation.cross_val_score.

If I call it like:

scores = cross_validation.cross_val_score(alg, titanic[predictors], titanic["Survived"], cv=3)

print(scores.mean())

0.801346801347

I get a different set of scores than if I call it like:

kf = KFold(titanic.shape[0], n_folds=3, random_state=1)

scores = cross_validation.cross_val_score(alg, titanic[predictors], titanic["Survived"], cv=kf)

print(scores.mean())

0.785634118967

These numbers are close, but different enough to be significant. As far as I understand, both code snippets are asking for a 3-fold cross validation strategy. Can anyone explain what is going on under the hood of the second example which is leading to the slightly lower score?

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From the sklearn docs for cross_val_score's cv argument :

"For integer/None inputs, if y is binary or multiclass, StratifiedKFold used. If the estimator is a classifier or if y is neither binary nor multiclass, KFold is used."

I believe that in the first case, StratifiedKFold is being used as the default. In the second case, you are explicitly passing a KFold generator.

The difference between the two is also documented in the docs.

"KFold divides all the samples in k groups of samples, called folds (if k = n, this is equivalent to the Leave One Out strategy), of equal sizes (if possible)."

"StratifiedKFold is a variation of k-fold which returns stratified folds: each set contains approximately the same percentage of samples of each target class as the complete set."

This difference in folds is what is causing the difference in scores.

As a side note, I noticed that you are passing a random_state argument to the KFold object. However, you should note that this seed is only used if you also set KFold's shuffle parameter to True, which by default is False.

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As mentioned by @Reii Nakano if us estimator is classifier and your Y is binary StratifiedKFold will be used else KFold will be used.

Also interesting part here is you are using random state = 1 in KFold. So data in splits in case of KFold is not necessarily same as Splits in cross_val_score. Hence, your final score may differ.

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