numpy的广播阵 [英] Numpy broadcast array

问题描述

我在numpy的以下数组：

I have the following array in NumPy:

A = array([1, 2, 3])

我怎样才能得到如下矩阵（没有明确的循环）？

How can I obtain the following matrices (without an explicit loop)?

B = [ 1 1 1
      2 2 2
      3 3 3 ]

C = [ 1 2 3
      1 2 3
      1 2 3 ]

谢谢！

推荐答案

EDIT2：该任择议定书要求的意见如何计算

The OP asks in the comments how to compute

n(i, j) = l(i, i) + l(j, j) - 2 * l(i, j)

我能想到的两种方式。我喜欢这种方式，因为它很容易概括：

I can think of two ways. I like this way because it generalizes easily:

import numpy as np

l=np.arange(9).reshape(3,3)
print(l)
# [[0 1 2]
#  [3 4 5]
#  [6 7 8]]

我们的想法是使用 np.ogrid 。这定义了两个numpy的阵列，形状（3,1）中的一个和形状中的一个（1,3）的列表：

The idea is to use np.ogrid. This defines a list of two numpy arrays, one of shape (3,1) and one of shape (1,3):

grid=np.ogrid[0:3,0:3]
print(grid)
# [array([[0],
#        [1],
#        [2]]), array([[0, 1, 2]])]

电网[0] 可作为该指数代理我，和
电网[1] 可作为该指数的代理Ĵ。

grid[0] can be used as a proxy for the index i, and grid[1] can be used as a proxy for the index j.

所以在EX pression L（I，I）+ 1（J，J）无处不在 - 2 * L（I，J），您只需替换 I - > 电网[0] 和Ĵ - > 电网[1] 和numpy的广播照顾其余的：

So everywhere in the expression l(i, i) + l(j, j) - 2 * l(i, j), you simply replace i-->grid[0], and j-->grid[1], and numpy broadcasting takes care of the rest:

n=l[grid[0],grid[0]] + l[grid[1],grid[1]] + 2*l
print(n)
# [[ 0  6 12]
#  [10 16 22]
#  [20 26 32]]

然而，在这种特殊的情况下，由于 L（I，I）和→（J，J）只是→的对角线元素，你可以这样做，而不是：

However, in this particular case, since l(i,i) and l(j,j) are just the diagonal elements of l, you could do this instead:

d=np.diag(l)
print(d)
# [0 4 8]

D [np.newaxis，：] 泵高达 D 的形状，（1,3），和
D [：，np.newaxis] D 的形状水泵达（3,1）。

d[np.newaxis,:] pumps up the shape of d to (1,3), and d[:,np.newaxis] pumps up the shape of d to (3,1).

numpy的广播泵高达 D [np.newaxis，：] 和 D [：，np.newaxis] 塑造（3,3），复制适当的值。

Numpy broadcasting pumps up d[np.newaxis,:] and d[:,np.newaxis] to shape (3,3), copying values as appropriate.

n=d[np.newaxis,:] + d[:,np.newaxis] + 2*l
print(n)
# [[ 0  6 12]
#  [10 16 22]
#  [20 26 32]]

---

EDIT1：通常你不需要形成 B 或 C 。 numpy的广播的目的是让你在 B的或 c。使用 A 。如果你告诉我们，你打算如何使用 B 或 C ，我们也许能够告诉你如何做与 A 相同，numpy的广播。

---

Usually you do not need to form B or C. The purpose of Numpy broadcasting is to allow you to use A in place of B or C. If you show us how you plan to use B or C, we might be able to show you how to do the same with A and numpy broadcasting.

（原来的答复）：

In [11]: B=A.repeat(3).reshape(3,3)

In [12]: B
Out[12]: 
array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])

In [13]: C=B.T

In [14]: C
Out[14]: 
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

或

In [25]: C=np.tile(A,(3,1))

In [26]: C
Out[26]: 
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

In [27]: B=C.T

In [28]: B
Out[28]: 
array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])

从搞鬼部门：

In [57]: np.lib.stride_tricks.as_strided(A,shape=(3,3),strides=(4,0))
Out[57]: 
array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])

In [58]: np.lib.stride_tricks.as_strided(A,shape=(3,3),strides=(0,4))
Out[58]: 
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

但需要注意的是，这些都是的意见的 A ，而不是副本（因为是上面的解决方案）。更改 B ，变造 A ：

But note that these are views of A, not copies (as were the solutions above). Changing B, alters A:

In [59]: B=np.lib.stride_tricks.as_strided(A,shape=(3,3),strides=(4,0))

In [60]: B[0,0]=100

In [61]: A
Out[61]: array([100,   2,   3])

这篇关于numpy的广播阵的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！