Cross validation and pipeline in sci-kit learn

Question

For a machine learning project, i'm trying to predict a categorical outcome variable using features extracted from text.

Using cross validation, i split my X and Y into a test set and training set. The training set is trained using a pipeline. However, when i compute the performance using X from my test set my performance is 0.0. This is while there are no features extracted from X_test yet.

Is it possible to split the dataset within the pipeline?

My code:

X, Y = read_data('development2.csv')

X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size=0.33, random_state=42)

train_pipeline = Pipeline([('vect', CountVectorizer()), #ngram_range=(1,2), analyzer='word'
                 ('tfidf', TfidfTransformer(use_idf=False)),
                 ('clf', OneVsRestClassifier(SVC(kernel='linear', probability=True))),
                 ])

train_pipeline.fit(X_train, Y_train)

predicted = train_pipeline.predict(X_test)

print accuracy_score(Y_test, predicted)

The traceback when using SVC:

File     "/Users/Robbert/Documents/pipeline.py", line     62, in <module>
train_pipeline.fit(X_train, Y_train)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/sklearn/pipeline.py", line 130, in fit
self.steps[-1][-1].fit(Xt, y, **fit_params)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/sklearn/svm/base.py", line 138, in fit
y = self._validate_targets(y)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/sklearn/svm/base.py", line 441, in _validate_targets
y_ = column_or_1d(y, warn=True)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/sklearn/utils/validation.py", line 319, in column_or_1d
raise ValueError("bad input shape {0}".format(shape))
ValueError: bad input shape (670, 5)

Answer 1

I solved the problem.

The target variable (Y) did not have the appropriate format. The variables were stored like this: [[0 0 0 0 1],[0 0 1 0 0]] . I converted this to a different array format like this: [5, 3] .

This did the trick for me.

Thanks for all answers.