快速信息增益计算 [英] Fast Information Gain computation
问题描述
我需要为文本分类计算超过1万个文档中超过10万个功能的信息增益得分.下面的代码可以正常工作，但是整个数据集非常慢-在笔记本电脑上需要一个多小时的时间.数据集是20newsgroup，我正在使用scikit-learn， chi2 (该功能在scikit中提供)非常快速.
I need to compute Information Gain scores for >100k features in >10k documents for text classification. Code below works fine but for the full dataset is very slow - takes more than an hour on a laptop. Dataset is 20newsgroup and I am using scikit-learn, chi2 function which is provided in scikit works extremely fast.
有什么想法可以更快地为此类数据集计算信息增益吗?
Any idea how to compute Information Gain faster for such dataset?
def information_gain(x, y):
def _entropy(values):
counts = np.bincount(values)
probs = counts[np.nonzero(counts)] / float(len(values))
return - np.sum(probs * np.log(probs))
def _information_gain(feature, y):
feature_set_indices = np.nonzero(feature)[1]
feature_not_set_indices = [i for i in feature_range if i not in feature_set_indices]
entropy_x_set = _entropy(y[feature_set_indices])
entropy_x_not_set = _entropy(y[feature_not_set_indices])
return entropy_before - (((len(feature_set_indices) / float(feature_size)) * entropy_x_set)
+ ((len(feature_not_set_indices) / float(feature_size)) * entropy_x_not_set))
feature_size = x.shape[0]
feature_range = range(0, feature_size)
entropy_before = _entropy(y)
information_gain_scores = []
for feature in x.T:
information_gain_scores.append(_information_gain(feature, y))
return information_gain_scores, []
我合并了内部函数，并按以下方式运行cProfiler(在仅限于〜15,000个要素和〜1k个文档的数据集中):
I merged the internal functions and ran cProfiler as below (on a dataset limited to ~15k features and ~1k documents):
cProfile.runctx(
"""for feature in x.T:
feature_set_indices = np.nonzero(feature)[1]
feature_not_set_indices = [i for i in feature_range if i not in feature_set_indices]
values = y[feature_set_indices]
counts = np.bincount(values)
probs = counts[np.nonzero(counts)] / float(len(values))
entropy_x_set = - np.sum(probs * np.log(probs))
values = y[feature_not_set_indices]
counts = np.bincount(values)
probs = counts[np.nonzero(counts)] / float(len(values))
entropy_x_not_set = - np.sum(probs * np.log(probs))
result = entropy_before - (((len(feature_set_indices) / float(feature_size)) * entropy_x_set)
+ ((len(feature_not_set_indices) / float(feature_size)) * entropy_x_not_set))
information_gain_scores.append(result)""",
globals(), locals())
tottime的结果前20名:
ncalls tottime percall cumtime percall filename:lineno(function)
1 60.27 60.27 65.48 65.48 <string>:1(<module>)
16171 1.362 0 2.801 0 csr.py:313(_get_row_slice)
16171 0.523 0 0.892 0 coo.py:201(_check)
16173 0.394 0 0.89 0 compressed.py:101(check_format)
210235 0.297 0 0.297 0 {numpy.core.multiarray.array}
16173 0.287 0 0.331 0 compressed.py:631(prune)
16171 0.197 0 1.529 0 compressed.py:534(tocoo)
16173 0.165 0 1.263 0 compressed.py:20(__init__)
16171 0.139 0 1.669 0 base.py:415(nonzero)
16171 0.124 0 1.201 0 coo.py:111(__init__)
32342 0.123 0 0.123 0 {method 'max' of 'numpy.ndarray' objects}
48513 0.117 0 0.218 0 sputils.py:93(isintlike)
32342 0.114 0 0.114 0 {method 'sum' of 'numpy.ndarray' objects}
16171 0.106 0 3.081 0 csr.py:186(__getitem__)
32342 0.105 0 0.105 0 {numpy.lib._compiled_base.bincount}
32344 0.09 0 0.094 0 base.py:59(set_shape)
210227 0.088 0 0.088 0 {isinstance}
48513 0.081 0 1.777 0 fromnumeric.py:1129(nonzero)
32342 0.078 0 0.078 0 {method 'min' of 'numpy.ndarray' objects}
97032 0.066 0 0.153 0 numeric.py:167(asarray)
看起来大部分时间都用在_get_row_slice中.我不确定第一行，它看起来覆盖了我提供给cProfile.runctx的整个块，尽管我不知道为什么第一行totime=60.27和第二行tottime=1.362之间有这么大的差距.差额花在哪里了?是否可以在cProfile中对其进行检查?
Looks that most of the time is spent in _get_row_slice. I am not entirely sure about the first row, looks it covers the whole block I provided to cProfile.runctx, though I don't know why there is such a big gap between first line totime=60.27 and second one tottime=1.362. Where was the difference spent in? Is it possible to check it in cProfile?
基本上，看起来问题出在稀疏矩阵运算(切片，获取元素)上-解决方案可能是使用矩阵代数(例如
Basically, looks the problem is with sparse matrix operations (slicing, getting elements) -- the solution probably would be to calculate Information Gain using matrix algebra (like chi2 is implemented in scikit). But I have no idea how to express this calculation in terms of matrices operations... Anyone has an idea??
推荐答案
自一年过去以来，尚不知道它是否仍然有帮助.但是现在我恰好面临着文本分类的相同任务.我已经使用 nonzero()函数.然后，我只扫描nz，计算相应的y_value并计算熵.
Don't know whether it still helps since a year has passed. But now I happen to be faced with the same task for text classification. I've rewritten your code using the nonzero() function provided for sparse matrix. Then I just scan nz, count the corresponding y_value and calculate the entropy.
以下代码只需几秒钟即可运行news20数据集(使用libsvm稀疏矩阵格式加载).
The following code only needs seconds to run news20 dataset (loaded in using libsvm sparse matrix format).
def information_gain(X, y):
def _calIg():
entropy_x_set = 0
entropy_x_not_set = 0
for c in classCnt:
probs = classCnt[c] / float(featureTot)
entropy_x_set = entropy_x_set - probs * np.log(probs)
probs = (classTotCnt[c] - classCnt[c]) / float(tot - featureTot)
entropy_x_not_set = entropy_x_not_set - probs * np.log(probs)
for c in classTotCnt:
if c not in classCnt:
probs = classTotCnt[c] / float(tot - featureTot)
entropy_x_not_set = entropy_x_not_set - probs * np.log(probs)
return entropy_before - ((featureTot / float(tot)) * entropy_x_set
+ ((tot - featureTot) / float(tot)) * entropy_x_not_set)
tot = X.shape[0]
classTotCnt = {}
entropy_before = 0
for i in y:
if i not in classTotCnt:
classTotCnt[i] = 1
else:
classTotCnt[i] = classTotCnt[i] + 1
for c in classTotCnt:
probs = classTotCnt[c] / float(tot)
entropy_before = entropy_before - probs * np.log(probs)
nz = X.T.nonzero()
pre = 0
classCnt = {}
featureTot = 0
information_gain = []
for i in range(0, len(nz[0])):
if (i != 0 and nz[0][i] != pre):
for notappear in range(pre+1, nz[0][i]):
information_gain.append(0)
ig = _calIg()
information_gain.append(ig)
pre = nz[0][i]
classCnt = {}
featureTot = 0
featureTot = featureTot + 1
yclass = y[nz[1][i]]
if yclass not in classCnt:
classCnt[yclass] = 1
else:
classCnt[yclass] = classCnt[yclass] + 1
ig = _calIg()
information_gain.append(ig)
return np.asarray(information_gain)
这篇关于快速信息增益计算的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
