我想加速以下双for循环：
n = 10000
A = np.zeros((n, n))
twopi = 2.0 * np.pi
for i in range(0, n):
for j in range(0, n):
A[i,j] = twopi * i*j/n
我该怎么做？我的想法是：计算一个包含从0到n-1的所有元素的向量v：
v = np.arange(n)
然后构造两个矩阵I和J，一个具有等于v的所有行，另一个具有与v相等的所有列。如果我没有错的话，那么我可以用代替双for循环
A = twopi * I*J/n
这是正确的吗？如何构建I和J？有更好的方法吗？
类似这样的东西：
from numpy import outer, pi, arange
n = 10
i = arange(n)
A = 2 * pi / n * outer(i, i)
outer产品满足您的要求。
编辑：测试性能的简单方法如下：
from time import time
n = 10000
t = time()
for _ in range(10):
i = arange(n)
A = 2 * pi / n * outer(i, i)
t = time() - t
print("Time used [s]", t)
对我来说，这10次重复需要5秒，range和arange之间没有显著差异。
答案
你确实可以使用numpy的广播能力。
import numpy as np
n = 10000
twopi = 2.0 * np.pi / n
A = np.arange(n) * np.arange(n).reshape(-1, 1) * twopi
第二个CCD_ 16的重塑推动了产品的传播。
这与V博士调用np.outer()非常相似，即使从时间性能的角度来看也是如此。
基准
配置：
import numpy as np
n = 10000
twopi = 2 * np.pi / n
次数：
# List comprehension
A = np.array([
[twopi*i*j for j in range(n)]
for i in range(n)])
# timeit > 10.2 s ± 129 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# Outer
i = np.arange(n)
A = twopi * np.outer(i, i)
# timeit > 234 ms ± 6.87 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# Product
A = np.arange(n) * np.arange(n).reshape(-1, 1) * twopi
# timeit > 208 ms ± 41.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
如前所述，外部方法和产品方法非常相似。但列表理解并不是因为它违背了numpy的目的，numpy是通过分布式子任务(通常还有低级优化(来向量化数组计算。
您可以使用列表综合：
A = np.array([
[twopi*i*j/n for j in range(n)]
for i in range(n)])
默认情况下，range从零开始，因此range(0,n)而不是range(n)是多余的。
In [118]: matrix = np.array([np.arange(n) for _ in range(n)])
In [119]: matrix
Out[119]:
array([[ 0, 1, 2, ..., 9997, 9998, 9999],
[ 0, 1, 2, ..., 9997, 9998, 9999],
[ 0, 1, 2, ..., 9997, 9998, 9999],
...,
[ 0, 1, 2, ..., 9997, 9998, 9999],
[ 0, 1, 2, ..., 9997, 9998, 9999],
[ 0, 1, 2, ..., 9997, 9998, 9999]])
In [120]: matrix [0,1]
Out[120]: 1
In [121]: for j in range(n): matrix[:,j] *= j
In [122]: matrix
Out[122]:
array([[ 0, 1, 4, ..., 99940009, 99960004, 99980001],
[ 0, 1, 4, ..., 99940009, 99960004, 99980001],
[ 0, 1, 4, ..., 99940009, 99960004, 99980001],
...,
[ 0, 1, 4, ..., 99940009, 99960004, 99980001],
[ 0, 1, 4, ..., 99940009, 99960004, 99980001],
[ 0, 1, 4, ..., 99940009, 99960004, 99980001]])
In [123]: matrix = matrix/n
In [124]: matrix = matrix * np.pi*2
In [125]: matrix
Out[125]:
array([[0.00000000e+00, 6.28318531e-04, 2.51327412e-03, ...,
6.27941596e+04, 6.28067228e+04, 6.28192873e+04],
[0.00000000e+00, 6.28318531e-04, 2.51327412e-03, ...,
6.27941596e+04, 6.28067228e+04, 6.28192873e+04],
[0.00000000e+00, 6.28318531e-04, 2.51327412e-03, ...,
6.27941596e+04, 6.28067228e+04, 6.28192873e+04],
...,
[0.00000000e+00, 6.28318531e-04, 2.51327412e-03, ...,
6.27941596e+04, 6.28067228e+04, 6.28192873e+04],
[0.00000000e+00, 6.28318531e-04, 2.51327412e-03, ...,
6.27941596e+04, 6.28067228e+04, 6.28192873e+04],
[0.00000000e+00, 6.28318531e-04, 2.51327412e-03, ...,
6.27941596e+04, 6.28067228e+04, 6.28192873e+04]])
与脚本相同：
matrix = np.array([np.arange(n) for _ in range(n)])
for j in range(n): matrix[:,j] *= j
matrix = matrix/n
matrix = matrix * np.pi*2