我在下面撰寫了使用 for 回圈的代碼。我想問是否有辦法在第二個 for 回圈中對操作進行矢量化，因為我打算使用更大的矩陣。
import numpy as np
num = 5
A = np.array([[1,2,3,4,5], [4,5,6,4,5], [7,8,9,4,5], [10,11,12,4,5], [13,14,15,4,5]])
sm_factor = np.array([0.1 ,0.1, 0.1, 0.1, 0.1])
d2m = np.zeros((num, num))
d2m[0, 0] = 2
d2m[0, 1] = -5
d2m[0, 2] = 4
d2m[0, 3] = -1
for k in range(1, num-1):
d2m[k, k-1] = 1
d2m[k, k] = -2
d2m[k, k 1] = 1
d2m[num-1, num-4] = -1
d2m[num-1, num-3] = 4
d2m[num-1, num-2] = -5
d2m[num-1, num-1] = 2
x_smf = 0
for i in range(len(sm_factor)):
x_smf = x_smf sm_factor[i] * (d2m @ (A[i, :]).T).T @ (d2m @ (A[i, :]).T)
x_smf
# 324.0
uj5u.com熱心網友回復：
您可以使用sps.diags創建稀疏三對角矩陣來避免d2m矩陣創建和x_smf向量計算的回圈，您可以將其轉換為陣列以便能夠編輯第一行和最后一行。您的代碼將如下所示（請注意，第10 行的結果已使用scipy.sparse.dia_matrix.toarray方法轉換為密集的 ndarray ）：diags
import numpy as np
import scipy.sparse as sps
# Dense tridiagonal matrix
d2m = sps.diags([1, -2, 1], [-1, 0, 1], shape=(num, num)).toarray() # cast to array
# First line boundary conditions
d2m[0, 0] = 2
d2m[0, 1] = -5
d2m[0, 2] = 4
d2m[0, 3] = -1
# Last line boundary conditions
d2m[num-1, num-4] = -1
d2m[num-1, num-3] = 4
d2m[num-1, num-2] = -5
d2m[num-1, num-1] = 2
Valdi_Bo提出的解決方案使您能夠洗掉第二個 FOR 回圈：
x_smf = np.sum(sm_factor * np.square(d2m @ A.T).sum(axis=0))
但是，我想引起您的注意，x_smf矩陣是稀疏的，并且將其存盤為密集的 ndarray 對計算時間和記憶體存盤都是不利的。我建議您不要轉換為密集的 ndarray，而是轉換為稀疏矩陣格式。例如lil_matrix，它是串列稀疏矩陣格式的串列，使用 tolil() 方法而不是 toarray()：
# Sparse tridiagonal matrix
d2m_s = sps.diags([1, -2, 1], [-1, 0, 1], shape=(num, num)).tolil() # cast to lil
這是一個腳本，它在更大的情況下比較了三個實作num=4000（對于num=5所有 give 324）。對于這種大小，我已經看到了使用稀疏矩陣的好處，這是整個腳本（第一行是對num不同 from的代碼的概括5）：
from time import time
import numpy as np
import scipy.sparse as sps
num = 4000
A = np.concatenate([np.arange(1, (num-2)*num 1).reshape(num, num-2), np.repeat([[4, 5]], num, axis=0)], axis=1)
sm_factor = 0.1*np.ones(num)
########## DENSE matrix FOR loop ##########
d2m = sps.diags([1, -2, 1], [-1, 0, 1], shape=(num, num)).toarray() # cast to array
# First line boundary conditions
d2m[0, 0] = 2
d2m[0, 1] = -5
d2m[0, 2] = 4
d2m[0, 3] = -1
# Last line boundary conditions
d2m[num-1, num-4] = -1
d2m[num-1, num-3] = 4
d2m[num-1, num-2] = -5
d2m[num-1, num-1] = 2
# FOR loop version
t_start = time()
x_smf = 0
for i in range(len(sm_factor)):
x_smf = x_smf sm_factor[i] * (d2m @ (A[i, :]).T).T @ (d2m @ (A[i, :]).T)
print(f'FOR loop version time: {time()-t_start}s')
print(f'FOR loop version value: {x_smf}\n')
########## DENSE matrix VECTORIZED ##########
t_start = time()
x_smf_v = np.sum(sm_factor * np.square(d2m @ A.T).sum(axis=0))
print(f'VECTORIZED version time: {time()-t_start}s')
print(f'VECTORIZED version value: {x_smf_v}\n')
########## SPARSE matrix VECTORIZED ##########
d2m_s = sps.diags([1, -2, 1], [-1, 0, 1], shape=(num, num)).tolil() # cast to lil
# First line boundary conditions
d2m_s[0, 0] = 2
d2m_s[0, 1] = -5
d2m_s[0, 2] = 4
d2m_s[0, 3] = -1
# Last line boundary conditions
d2m_s[num-1, num-4] = -1
d2m_s[num-1, num-3] = 4
d2m_s[num-1, num-2] = -5
d2m_s[num-1, num-1] = 2
t_start = time()
x_smf_s = np.sum(sm_factor * np.square(d2m_s @ A.T).sum(axis=0))
print(f'SPARSE VECTORIZED version time: {time()-t_start}s')
print(f'SPARSE VECTORIZED version value: {x_smf_s}\n')
這是我在運行代碼時得到的：
FOR loop version time: 25.878241777420044s
FOR loop version value: 3.752317536763356e 17
VECTORIZED version time: 1.0873610973358154s
VECTORIZED version value: 3.752317536763356e 17
SPARSE VECTORIZED version time: 0.37279224395751953s
SPARSE VECTORIZED version value: 3.752317536763356e 17
正如您所看到的，使用稀疏矩陣可以讓您在計算時間上贏得另一個因素 3，并且不需要您調整之后的代碼。測驗稀疏矩陣的各種 scipy 實作（tocsc()、tocsr()、todok() 等）也是一個很好的策略，有些可能更適合您的情況。
uj5u.com熱心網友回復：
在對回圈的中間結果進行一些研究和列印輸出之后，我找到了解決方案：
x_smf = np.sum(sm_factor * np.square(d2m @ A.T).sum(axis=0))
結果是：
324.0
順便說一句：dm2的創建可以縮短為：
d2m = np.zeros((num, num), dtype='int')
d2m[0, :4] = [ 2, -5, 4, -1]
for k in range(1, num-1):
d2m[k, k-1:k 2] = [ 1, -2, 1]
d2m[-1, -4:] = [-1, 4, -5, 2]
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