Python信息增益实现

Question

我目前正在使用scikit-learn对20ng数据集进行文本分类。我想计算矢量化数据集的信息增益。有人建议我可以使用sklearn中的mutual_info_classif来实现这一点。但是，这种方法非常慢，所以我试图根据这篇文章自己实现信息增益。

我得出了以下解决方案：

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

使用一个非常小的数据集，sklearn和我的实现大多数得分相等。然而，sklearn似乎考虑了频率，而我的算法显然没有考虑。例如：

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))

示例输出：

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

我在想我的实现是否有误，或者它是正确的，但与scikit-learn使用的相互信息算法有所不同。

- Roman Purgstaller

---

你找到原因了吗？ - sariii

很遗憾，我最终使用了scikit的实现。 - Roman Purgstaller

Answer 1

有点晚了，但你应该看一下Orange的实现。在他们的应用程序中，它被用作后台处理器，帮助 inform 动态模型参数构建过程。

实现本身看起来非常简单，很可能可以移植出来。首先是熵计算。

从https://github.com/biolab/orange3/blob/master/Orange/preprocess/score.py#L233开始的部分。

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))

然后是第二部分。 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

实际的评分过程发生在https://github.com/biolab/orange3/blob/master/Orange/preprocess/score.py#L218。

- Office Zombie