我有以下代碼：
import numpy as np
epsilon = np.array([[0. , 0.00172667, 0.00071437, 0.00091779, 0.00154501],
[0.00128983, 0. , 0.00028139, 0.00215905, 0.00094862],
[0.00035811, 0.00018714, 0. , 0.00029365, 0.00036993],
[0.00035631, 0.00112175, 0.00022906, 0. , 0.00291149],
[0.00021527, 0.00017653, 0.00010341, 0.00104458, 0. ]])
Sii = np.array([19998169., 14998140., 9997923., 7798321., 2797958.])
n = len(Sii)
epsilonijSjj = np.zeros((n,n))
for i in range(n):
for j in range(n):
epsilonijSjj[i,j] = epsilon[i][j]*Sii[j]
print (epsilonijSjj)
如何避免雙重 for 回圈并以快速 Pythonic 方式撰寫代碼？
先感謝您
uj5u.com熱心網友回復：
Numpy 允許您直接將 2 個陣列相乘。
因此，與其定義一個0基于陣列并用另一個陣列的更改元素填充它，您可以簡單地創建另一個陣列的副本并直接應用乘法，如下所示：
import numpy as np
epsilon = np.array([[0. , 0.00172667, 0.00071437, 0.00091779, 0.00154501],
[0.00128983, 0. , 0.00028139, 0.00215905, 0.00094862],
[0.00035811, 0.00018714, 0. , 0.00029365, 0.00036993],
[0.00035631, 0.00112175, 0.00022906, 0. , 0.00291149],
[0.00021527, 0.00017653, 0.00010341, 0.00104458, 0. ]])
Sii = np.array([19998169., 14998140., 9997923., 7798321., 2797958.])
epsilonijSjj = epsilon.copy()
epsilonijSjj *= Sii
print(epsilonijSjj)
輸出：
[[ 0. 25896.8383938 7142.21625351 7157.22103059
4322.87308958]
[25794.23832127 0. 2813.31555297 16836.96495505
2654.19891796]
[ 7161.54430059 2806.7519196 0. 2289.97696165
1035.04860294]
[ 7125.54759639 16824.163545 2290.12424238 0.
8146.22673742]
[ 4305.00584063 2647.6216542 1033.88521743 8145.97015018
0. ]]
或者，只需這樣做，這會更快，因為它不需要創建陣列的副本：
import numpy as np
epsilon = np.array([[0. , 0.00172667, 0.00071437, 0.00091779, 0.00154501],
[0.00128983, 0. , 0.00028139, 0.00215905, 0.00094862],
[0.00035811, 0.00018714, 0. , 0.00029365, 0.00036993],
[0.00035631, 0.00112175, 0.00022906, 0. , 0.00291149],
[0.00021527, 0.00017653, 0.00010341, 0.00104458, 0. ]])
Sii = np.array([19998169., 14998140., 9997923., 7798321., 2797958.])
epsilonijSjj = epsilon * Sii
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