我最近一直致力于一個項目,其中我的大部分時間花費在密集矩陣A和稀疏向量v上(見

here).在我嘗試減少計算時,我注意到A.dot(v)的運行時間不受v的零條目數的影響.

為了解釋為什么我希望在這種情況下改進運行時,讓result = A.dot.v使得j = 1的結果[j] = sum_i(A [i,j] * v [j])… v.shape [0].如果v [j] = 0,則無論值A [::,j]如何,顯然結果[j] = 0.在這種情況下,我希望numpy只設置result [j] = 0,但似乎它繼續并計算sum_i(A [i,j] * v [j])無論如何.

我繼續編寫了一個簡短的示例腳本來確認下面的這種行為.

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))

運行這個,我得到矩陣向量乘法的運行時是相同的,無論我使用密集矩陣u還是稀疏矩陣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

我想知道這是否是設計的？或者我錯過了我正在運行矩陣向量乘法的方式.就像健全性檢查一樣：我確保numpy鏈接到我的機器上的BLAS庫,并且兩個數組都是C_CONTIGUOUS(因為這顯然需要numpy來調用BLAS).

總結

以上是生活随笔為你收集整理的python 矩阵乘法 跳过nan_python – Numpy：当一些向量元素等于零时,矩阵向量乘法不会跳过计算吗？...的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。