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.