Python Information gain implementation

问题

I am currently using scikit-learn for text classification on the 20ng dataset. I want to calculate the information gain for a vectorized dataset. It has been suggested to me that this can be accomplished, using mutual_info_classif from sklearn. However, this method is really slow, so I was trying to implement information gain myself based on this post.

I came up with the following solution:

from scipy.stats import entropy
import numpy as np

def information_gain(X, y):

    def _entropy(labels):
        counts = np.bincount(labels)
        return entropy(counts, base=None)

    def _ig(x, y):
        # indices where x is set/not set
        x_set = np.nonzero(x)[1]
        x_not_set = np.delete(np.arange(x.shape[1]), x_set)

        h_x_set = _entropy(y[x_set])
        h_x_not_set = _entropy(y[x_not_set])

        return entropy_full - (((len(x_set) / f_size) * h_x_set)
                             + ((len(x_not_set) / f_size) * h_x_not_set))

    entropy_full = _entropy(y)

    f_size = float(X.shape[0])

    scores = np.array([_ig(x, y) for x in X.T])
    return scores

Using a very small dataset, most scores from sklearn and my implementation are equal. However, sklearn seems to take frequencies into account, which my algorithm clearly doesn't. For example

categories = ['talk.religion.misc', 'comp.graphics', 'sci.space']
newsgroups_train = fetch_20newsgroups(subset='train',
                                      categories=categories)

X, y = newsgroups_train.data, newsgroups_train.target
cv = CountVectorizer(max_df=0.95, min_df=2,
                                     max_features=100,
                                     stop_words='english')
X_vec = cv.fit_transform(X)

t0 = time()
res_sk = mutual_info_classif(X_vec, y, discrete_features=True)
print("Time passed for sklearn method: %3f" % (time()-t0))
t0 = time()
res_ig = information_gain(X_vec, y)
print("Time passed for ig: %3f" % (time()-t0))

for name, res_mi, res_ig in zip(cv.get_feature_names(), res_sk, res_ig):
    print("%s: mi=%f, ig=%f" % (name, res_mi, res_ig))

sample output:

center: mi=0.011824, ig=0.003548
christian: mi=0.128629, ig=0.127122
color: mi=0.028413, ig=0.026397    
com: mi=0.041184, ig=0.030458
computer: mi=0.020590, ig=0.012327
cs: mi=0.007291, ig=0.001574
data: mi=0.020734, ig=0.008986
did: mi=0.035613, ig=0.024604
different: mi=0.011432, ig=0.005492
distribution: mi=0.007175, ig=0.004675
does: mi=0.019564, ig=0.006162
don: mi=0.024000, ig=0.017605
earth: mi=0.039409, ig=0.032981
edu: mi=0.023659, ig=0.008442
file: mi=0.048056, ig=0.045746
files: mi=0.041367, ig=0.037860
ftp: mi=0.031302, ig=0.026949
gif: mi=0.028128, ig=0.023744
god: mi=0.122525, ig=0.113637
good: mi=0.016181, ig=0.008511
gov: mi=0.053547, ig=0.048207

So I was wondering if my implementation is wrong, or it is correct, but a different variation of the mutual information algorithm scikit-learn uses.

回答1:

A little late with my answer but you should look at Orange's implementation. Within their app it is used as a behind-the-scenes processor to help inform the dynamic model parameter building process.

The implementation itself looks fairly straightforward and could most likely be ported out. The entropy calculation first

The sections starting at https://github.com/biolab/orange3/blob/master/Orange/preprocess/score.py#L233

def _entropy(dist):
    """Entropy of class-distribution matrix"""
    p = dist / np.sum(dist, axis=0)
    pc = np.clip(p, 1e-15, 1)
    return np.sum(np.sum(- p * np.log2(pc), axis=0) * np.sum(dist, axis=0) / np.sum(dist))

Then the second portion. https://github.com/biolab/orange3/blob/master/Orange/preprocess/score.py#L305

class GainRatio(ClassificationScorer):
    """
    Information gain ratio is the ratio between information gain and
    the entropy of the feature's
    value distribution. The score was introduced in [Quinlan1986]_
    to alleviate overestimation for multi-valued features. See `Wikipedia entry on gain ratio
    <http://en.wikipedia.org/wiki/Information_gain_ratio>`_.
    .. [Quinlan1986] J R Quinlan: Induction of Decision Trees, Machine Learning, 1986.
    """
    def from_contingency(self, cont, nan_adjustment):
        h_class = _entropy(np.sum(cont, axis=1))
        h_residual = _entropy(np.compress(np.sum(cont, axis=0), cont, axis=1))
        h_attribute = _entropy(np.sum(cont, axis=0))
        if h_attribute == 0:
            h_attribute = 1
        return nan_adjustment * (h_class - h_residual) / h_attribute

The actual scoring process happens at https://github.com/biolab/orange3/blob/master/Orange/preprocess/score.py#L218

来源：https://stackoverflow.com/questions/47241525/python-information-gain-implementation

标签

python

machine-learning

scikit-learn

information-gain