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NumPy-涉及范围迭代器的向量化循环

分类: 技术问答 • 2022-06-12 18:23:45
问题描述：

没有for循环，有没有办法使这项工作可行?

Is there any way to make this work without for loops?

import import numpy as np
import matplotlib.pyplot as plt    

L = 1
N = 255
dh = 2*L/N
dh2 = dh*dh

phi_0 = 1
c = int(N/2)
r_0 = L/2

arr = np.empty((N, N))

for i in range(N):

    for j in range(N):

            arr[i, j] = phi_0 if (i - c)**2 + (j - c)**2 < r_0**2/dh2 else 0


plt.imshow(arr)

我尝试调用function(x [None ,:]，y [:, None])，其中:

I've tried calling function(x[None,:], y[:, None]), where:

function(i, j):

    return phi_0 if (i - c)**2 + (j - c)**2 < r_0**2/dh2 else 0

但是它需要列出.any或.all方法.我正在寻找专门的无功能方法(没有功能和向量化). 非常感谢！

but it requires list .any or .all methods. I'm looking for specifically functionless method (without fromfunction and vectorization). Big thanks!

答

使用开放式网格的矢量化解决方案

我们可以使用两个 open 范围/网格数组作为N来模拟与迭代器相同的行为-

Vectorized solution using open grids

We could use two open range/grid arrays for N simulating the same behavior as the iterators -

I = np.arange(N)
mask = (I[:,None] - c)**2 + (I - c)**2 < r_0**2/dh2
out = np.where(mask,phi_0,0)

对于两个循环的通用范围

对于一般情况，我们将遍历两个循环直到分别表示M和N，我们可以使用

For the generic case where we would iterate through two loops that extend till say M and N respectively, we could make use of np.ogrid to create those open grids and then use on the same lines -

I,J = np.ogrid[:M,:N]
mask = (I - c)**2 + (J - c)**2 < r_0**2/dh2

有关通用循环次数

对于一般数量的循环，只需创建与循环数一样多的变量即可.因此，对于三个循环:

For a generic number of loops, simply create as many variables as the number of loops. Hence, for three loops :

for i in range(M):
    for j in range(N):
        for k in range(P):

，我们会:

I,J,K = np.ogrid[:M,:N,:P]

，然后分别使用I,J,K而不是i,j,k进行像我们这里这样的元素操作.

, then use I,J,K instead of i,j,k respectively for element-wise operations like we have here.

在这种情况下替代最后一步

最后一步也可以通过逐元素乘法来实现，方法是将else部分设置为0s-

Last step could also be implemented with elementwise multiplication by scaling to phi_0 with mask as the else part is setting to 0s -

out = mask*phi_0

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