I’m new in machine learning. I’m preparing my data for classification using Scikit Learn SVM. In order to select the best features I have used the following method:

SelectKBest(chi2, k=10).fit_transform(A1, A2)

Since my dataset consist of negative values, I get the following error:

ValueError                                Traceback (most recent call last)

/media/5804B87404B856AA/TFM_UC3M/test2_v.py in <module>()
----> 1 

/usr/local/lib/python2.6/dist-packages/sklearn/base.pyc in fit_transform(self, X, y,     **fit_params)
    427         else:
    428             # fit method of arity 2 (supervised transformation)

--> 429             return self.fit(X, y, **fit_params).transform(X)

/usr/local/lib/python2.6/dist-packages/sklearn/feature_selection/univariate_selection.pyc in fit(self, X, y)
    300         self._check_params(X, y)
--> 302         self.scores_, self.pvalues_ = self.score_func(X, y)
    303         self.scores_ = np.asarray(self.scores_)
    304         self.pvalues_ = np.asarray(self.pvalues_)

/usr/local/lib/python2.6/dist-  packages/sklearn/feature_selection/univariate_selection.pyc in chi2(X, y)
    190     X = atleast2d_or_csr(X)
    191     if np.any((X.data if issparse(X) else X) < 0):
--> 192         raise ValueError("Input X must be non-negative.")
    194     Y = LabelBinarizer().fit_transform(y)

ValueError: Input X must be non-negative.

Can someone tell me how can I transform my data ?

The error message Input X must be non-negative says it all: Pearson’s chi square test (goodness of fit) does not apply to negative values. It’s logical because the chi square test assumes frequencies distribution and a frequency can’t be a negative number. Consequently, sklearn.feature_selection.chi2 asserts the input is non-negative.

You are saying that your features are “min, max, mean, median and FFT of accelerometer signal”. In many cases, it may be quite safe to simply shift each feature to make it all positive, or even normalize to [0, 1] interval as suggested by EdChum.

If data transformation is for some reason not possible (e.g. a negative value is an important factor), you should pick another statistic to score your features:

Since the whole point of this procedure is to prepare the features for another method, it’s not a big deal to pick anyone, the end result usually the same or very close.