# 10-fold cross-validation with K=5 for KNN (the n_neighbors parameter)
knn = KNeighborsClassifier(n_neighbors=5)
scores = cross_val_score(knn, X, y, cv=10, scoring='accuracy')
# use average accuracy as an estimate of out-of-sample accuracy
print(scores.mean())
# Output
0.9666666666666668