我最近一直致力于一个项目,其中我的大部分时间花费在密集矩阵A和稀疏向量v上(见

here).在我尝试减少计算时,我注意到A.dot(v)的运行时间不受v的零条目数的影响.

为了解释为什么我希望在这种情况下改进运行时,让result = A.dot.v使得j = 1的结果[j] = sum_i(A [i,j] * v [j])… v.shape [0].如果v [j] = 0,则无论值A [::,j]如何,显然结果[j] = 0.在这种情况下,我希望numpy只设置result [j] = 0,但似乎它继续并计算sum_i(A [i,j] * v [j])无论如何.

我继续编写了一个简短的示例脚本来确认下面的这种行为.

import time

import numpy as np

np.__config__.show() #make sure BLAS/LAPACK is being used

np.random.seed(seed = 0)

n_rows, n_cols = 1e5, 1e3

#initialize matrix and vector

A = np.random.rand(n_rows, n_cols)

u = np.random.rand(n_cols)

u = np.require(u, dtype=A.dtype, requirements = ['C'])

#time

start_time = time.time()

A.dot(u)

print "time with %d non-zero entries: %1.5f seconds" % (sum(u==0.0), (time.time() - start_time))

#set all but one entry of u to zero

v = u

set_to_zero = np.random.choice(np.array(range(0, u.shape[0])), size = (u.shape[0]-2), replace=False)

v[set_to_zero] = 0.0

start_time = time.time()

A.dot(v)

print "time with %d non-zero entries: %1.5f seconds" % (sum(v==0.0), (time.time() - start_time))

#what I would really expect it to take

non_zero_index = np.squeeze(v != 0.0)

A_effective = A[::,non_zero_index]

v_effective = v[non_zero_index]

start_time = time.time()

A_effective.dot(v_effective)

print "expected time with %d non-zero entries: %1.5f seconds" % (sum(v==0.0), (time.time() - start_time))

运行这个,我得到矩阵向量乘法的运行时是相同的,无论我使用密集矩阵u还是稀疏矩阵v：

time with 0 non-zero entries: 0.04279 seconds

time with 999 non-zero entries: 0.04050 seconds

expected time with 999 non-zero entries: 0.00466 seconds

我想知道这是否是设计的？或者我错过了我正在运行矩阵向量乘法的方式.就像健全性检查一样：我确保numpy链接到我的机器上的BLAS库,并且两个数组都是C_CONTIGUOUS(因为这显然需要numpy来调用BLAS).