【发布时间】:2015-10-24 04:13:02
【问题描述】:
我正在学习使用 theano。我想通过计算其中每个元素的二进制 TF-IDF 来填充术语文档矩阵(一个 numpy 稀疏矩阵):
import theano
import theano.tensor as T
import numpy as np
from time import perf_counter
def tfidf_gpu(appearance_in_documents,num_documents,document_words):
start = perf_counter()
APP = T.scalar('APP',dtype='int32')
N = T.scalar('N',dtype='int32')
SF = T.scalar('S',dtype='int32')
F = (T.log(N)-T.log(APP)) / SF
TFIDF = theano.function([N,APP,SF],F)
ret = TFIDF(num_documents,appearance_in_documents,document_words)
end = perf_counter()
print("\nTFIDF_GPU ",end-start," secs.")
return ret
def tfidf_cpu(appearance_in_documents,num_documents,document_words):
start = perf_counter()
tfidf = (np.log(num_documents)-np.log(appearance_in_documents))/document_words
end = perf_counter()
print("TFIDF_CPU ",end-start," secs.\n")
return tfidf
但是numpy版本比theano实现要快很多:
Progress 1/43
TFIDF_GPU 0.05702276699594222 secs.
TFIDF_CPU 1.454801531508565e-05 secs.
Progress 2/43
TFIDF_GPU 0.023830442980397493 secs.
TFIDF_CPU 1.1073017958551645e-05 secs.
Progress 3/43
TFIDF_GPU 0.021920352999586612 secs.
TFIDF_CPU 1.0738993296399713e-05 secs.
Progress 4/43
TFIDF_GPU 0.02303648801171221 secs.
TFIDF_CPU 1.1675001587718725e-05 secs.
Progress 5/43
TFIDF_GPU 0.02359767400776036 secs.
TFIDF_CPU 1.4385004760697484e-05 secs.
....
我了解到这可能是由于开销造成的,对于小型操作可能会降低性能。
我的代码是不是很糟糕,还是应该因为开销而避免使用 GPU?
【问题讨论】:
-
据我所知,您的函数似乎只对标量输入值 (
T.scalar) 起作用。除非您要处理相当大的数组,并执行涉及多个数组元素的矢量化操作,否则使用 GPU 是没有意义的。
标签: python numpy theano tf-idf