Numpy修改ndarray对角线

Question

在numpy中是否有任何方法可以获得对数组对角线的引用？我希望我的数组对角线除以某个因子谢谢

Answer 1

如果X你的阵列c是因素,

X[np.diag_indices_from(X)] /= c

请参阅diag_indices_fromNumpy手册.

Answer 2

访问方形(n,n)numpy数组的对角线的快速方法是arr.flat[::n+1]:

n = 1000
c = 20
a = np.random.rand(n,n)

a[np.diag_indices_from(a)] /= c # 119 microseconds
a.flat[::n+1] /= c # 25.3 microseconds

Answer 3

该np.fill_diagonal功能是相当快的:

np.fill_diagonal(a, a.diagonal() / c)

a你的阵列在哪里,c是你的因素.在我的机器上,这种方法和@ kwgoodman的a.flat[::n+1] /= c方法一样快,在我看来更清晰(但不是那么光滑).

Answer 4

比较以上3种方法:

import numpy as np
import timeit

n = 1000
c = 20
a = np.random.rand(n,n)
a1 = a.copy()
a2 = a.copy()
a3 = a.copy()

t1 = np.zeros(1000)
t2 = np.zeros(1000)
t3 = np.zeros(1000)

for i in range(1000):
    start = timeit.default_timer()
    a1[np.diag_indices_from(a1)] /= c 
    stop = timeit.default_timer()
    t1[i] = start-stop

    start = timeit.default_timer()
    a2.flat[::n+1] /= c
    stop = timeit.default_timer()
    t2[i] = start-stop

    start = timeit.default_timer()
    np.fill_diagonal(a3,a3.diagonal() / c)
    stop = timeit.default_timer()
    t3[i] = start-stop

print([t1.mean(), t1.std()])
print([t2.mean(), t2.std()])
print([t3.mean(), t3.std()])

[-4.5693619907979154e-05, 9.3142851395411316e-06]
[-2.338075107036275e-05, 6.7119609571872443e-06]
[-2.3731951987429056e-05, 8.0455946813059586e-06]

所以你可以看到这种np.flat方法是最快但很少的.当我再运行几次时,有时候fill_diagonal方法稍快一些.但可读性明智,它可能值得使用fill_diagonal方法.