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python - 快速信息增益计算

转载 作者:太空狗 更新时间:2023-10-29 20:22:55 26 4
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我需要为 文本分类 的 >10k 文档中的 >100k 特征计算信息增益分数。下面的代码工作正常,但完整数据集的速度非常慢 - 在笔记本电脑上需要一个多小时。数据集是 20newsgroup,我正在使用 scikit-learn, chi2 scikit 中提供的功能运行速度非常快。

知道如何更快地计算此类数据集的信息增益吗?

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, []

编辑:

我合并了内部函数并运行 cProfiler 如下(在限制为 ~15k 特征和~1k 文档的数据集上):

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 前 20 名的结果:

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花在了 _get_row_slice 上。我不完全确定第一行,看起来它涵盖了我提供给 cProfile.runctx 的整个 block ,虽然我不知道为什么第一行 totime 之间有这么大的差距=60.27 和第二个 tottime=1.362。差价花在了哪里?是否可以在 cProfile 中检查它?

基本上,看起来问题出在稀疏矩阵运算(切片、获取元素)——解决方案可能是使用矩阵代数计算信息增益(如 chi2 is implemented in scikit )。但我不知道如何用矩阵运算来表达这个计算...有人有想法吗??

最佳答案

一年过去了,不知道还有没有用。但是现在我恰好面临着同样的文本分类任务。我已经使用 nonzero() 重写了您的代码为稀疏矩阵提供的函数。然后我就扫描nz,统计对应的y_value,计算熵。

以下代码仅需几秒即可运行 news20 数据集(使用 libsvm 稀疏矩阵格式加载)。

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)

关于python - 快速信息增益计算,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25462407/

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