我有一个非常稀疏的矩阵,比如说5000x3000,双精度浮点数。这个矩阵的80%是零。我需要计算每一行的和。所有这些都在python/cython中实现。我想加快这个过程。因为我需要计算这个和几百万次,所以我认为如果我对非零元素进行索引,并且只对它们求和,会更快。结果变得比原来的“蛮力”全零求和慢得多。
下面是一个最小的例子:
#cython: language_level=2
import numpy as np
cimport numpy as np
import time
cdef int Ncells = 5000, KCells = 400, Ne= 350
cdef double x0=0.1, x1=20., x2=1.4, x3=2.8, p=0.2
# Setting up weight
all_weights = np.zeros( (Ncells,KCells) )
all_weights[ :Ne, :Ne ] = x0
all_weights[ :Ne, Ne: ] = x1
all_weights[Ne: , :Ne ] = x2
all_weights[Ne: , Ne: ] = x3
all_weights = all_weights * (np.random.rand(Ncells,KCells) < p)
# Making a memory view
cdef np.float64_t[:,:] my_weights = all_weights
# make an index of non zero weights
x,y = np.where( np.array(my_weights) > 0.)
#np_pawid = np.column_stack( (x ,y ) )
np_pawid = np.column_stack( (x ,y ) ).astype(int)
cdef np.int_t[:,:] pawid = np_pawid
# Making vector for column sum
summEE = np.zeros(KCells)
# Memory view
cdef np.float64_t [:] my_summEE = summEE
cdef int cc,dd,i
# brute-force summing
ntm = time.time()
for cc in range(KCells):
my_summEE[cc] = 0
for dd in range(Ncells):
my_summEE[cc] += my_weights[dd,cc]
stm = time.time()
print "BRUTE-FORCE summation : %f s"%(stm-ntm)
my_summEE[:] = 0
# summing only non zero indices
ntm = time.time()
for dd,cc in pawid:
my_summEE[cc] += my_weights[dd,cc]
stm = time.time()
print "INDEX summation : %f s"%(stm-ntm)
my_summEE[:] = 0
# summing only non zero indices unpacked by zip
ntm = time.time()
for dd,cc in zip(pawid[:,0],pawid[:,1]):
my_summEE[cc] += my_weights[dd,cc]
stm = time.time()
print "ZIPPED INDEX summation : %f s"%(stm-ntm)
my_summEE[:] = 0
# summing only non zero indices unpacked by zip
ntm = time.time()
for i in range(pawid.shape[0]):
dd = pawid[i,0]
cc = pawid[i,1]
my_summEE[cc] += my_weights[dd,cc]
stm = time.time()
print "INDEXING over INDEX summation: %f s"%(stm-ntm)
# Numpy brute-froce summing
ntm = time.time()
sumwee = np.sum(all_weights,axis=0)
stm = time.time()
print "NUMPY BRUTE-FORCE summation : %f s"%(stm-ntm)
#>
print
print "Number of brute-froce summs :",my_weights.shape[0]*my_weights.shape[1]
print "Number of indexing summs :",pawid.shape[0]
#<我在Raspberry Pi 3上运行了它,但在PC上似乎也是一样的结果。
BRUTE-FORCE summation : 0.381014 s
INDEX summation : 18.479018 s
ZIPPED INDEX summation : 3.615952 s
INDEXING over INDEX summation: 0.450131 s
NUMPY BRUTE-FORCE summation : 0.013017 s
Number of brute-froce summs : 2000000
Number of indexing summs : 400820
NUMPY BRUTE-FORCE in Python : 0.029143 s有人能解释一下为什么cython代码比numpy慢3-4倍吗?为什么索引将求和次数从2000000减少到400820,速度要慢45倍?这没有任何意义。
发布于 2019-01-29 03:39:49
,
@cython.wraparound(False)和@cython.boundscheck(False)作为装饰器添加到函数定义中。请注意它们的实际用途--仅在cdefed numpy数组或类型化内存视图上关闭这些功能,而不适用于其他任何类型(因此,不要将它们应用于所有地方作为货物狂热类型)。查看问题所在的好方法是运行cython -a <filename>并查看带注释的html文件。带有黄色的区域可能没有优化,您可以展开这些行来查看底层的C代码。显然,在这方面只需要担心频繁调用的函数和循环-事实上,设置Numpy数组的代码包含Python调用是意料之中的,而不是问题。
以下是一些测量结果:
正如你所写的
BRUTE-FORCE summation : 0.008625 s
INDEX summation : 0.713661 s
ZIPPED INDEX summation : 0.127343 s
INDEXING over INDEX summation: 0.002154 s
NUMPY BRUTE-FORCE summation : 0.001461 s在函数中
BRUTE-FORCE summation : 0.007706 s
INDEX summation : 0.681892 s
ZIPPED INDEX summation : 0.123176 s
INDEXING over INDEX summation: 0.002069 s
NUMPY BRUTE-FORCE summation : 0.001429 s在关闭了boundscheck和wraparound的函数中:
BRUTE-FORCE summation : 0.005208 s
INDEX summation : 0.672948 s
ZIPPED INDEX summation : 0.124641 s
INDEXING over INDEX summation: 0.002006 s
NUMPY BRUTE-FORCE summation : 0.001467 s我的建议确实有帮助,但不是太戏剧性。我的差异并不像您看到的那样显著(即使对于未更改的代码也是如此)。Numpy仍然赢了--猜一猜:
我怀疑它在整个数组上的multithreading.
https://stackoverflow.com/questions/54408389
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