改变具有奇数索引和的元素的符号

Posted 2023-02-23

【中文标题】改变具有奇数索引和的元素的符号

【英文标题】：Change sign of elements with an odd sum of indices

【发布时间】：2018-08-19 20:37:00

【问题描述】：

arr 是一个 n 维 numpy 数组。

如何用奇数个索引改变 arr 的每个元素的符号？

例如，arr[0, 1, 2] 需要更改符号，因为它有一个索引总和 0 + 1 + 2 = 3，这是奇数。

当我将arr 转换为一个列表时，我注意到列表中的每个第二个元素都是需要更改符号的元素。

另一个例子：

Original array:

[[[ 0  1  2]
  [ 3  4  5]
  [ 6  7  8]]

 [[ 9 10 11]
  [12 13 14]
  [15 16 17]]

 [[18 19 20]
  [21 22 23]
  [24 25 26]]]

Array with signs changed:
[[[ 0  -1  2]
  [ -3  4  -5]
  [ 6  -7  8]]

 [[ -9 10 -11]
  [12 -13 14]
  [-15 16 -17]]

 [[18 -19 20]
  [-21 22 -23]
  [24 -25 26]]]

【问题讨论】：

很不清楚你在问什么。请澄清。

欢迎来到 SO。请提供 minimal reproducible example。

你的意思是你想将每个元素乘以 -1 或 +1，+-1 形成一个像棋盘一样的矩阵？

矩阵 a=i,j [1,2,3, 4,5,6, 7,8,9] 例如 a[0][0]=1 a[1][2 ]=6 如果索引的总和是奇数，你必须将它乘以 -1，所以 a[0][0]=1 将是 -1 而 a[1][2]=6 将是 -6

审稿人请注意这个问题已经被一位乐于助人的编辑澄清了。

【参考方案1】：

np.negative 比乘法快一点（因为它是 ufunc）

N = 5
arr = np.arange(N ** 3).reshape(N, N, N)
%timeit arr.ravel()[1::2] *= -1
%timeit np.negative(arr.ravel()[1::2], out = arr.ravel()[1::2])

The slowest run took 8.74 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.39 µs per loop
The slowest run took 5.57 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.12 µs per loop

N = 25
arr = np.arange(N ** 3).reshape(N, N, N)
%timeit arr.ravel()[1::2] *= -1
%timeit np.negative(arr.ravel()[1::2], out = arr.ravel()[1::2])

The slowest run took 7.03 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 10.8 µs per loop
The slowest run took 5.27 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 8.63 µs per loop

N = 101
arr = np.arange(N ** 3).reshape(N, N, N)
%timeit arr.ravel()[1::2] *= -1
%timeit np.negative(arr.ravel()[1::2], out = arr.ravel()[1::2])

1000 loops, best of 3: 663 µs per loop
1000 loops, best of 3: 512 µs per loop

【参考方案2】：

hpaulj 的建议大大简化了。

程序：

import numpy as np


def change_sign(arr):
    """Switch sign of every second element of arr in-place
    Note
    ----
    Modifies the input array (arr).
    """
    # arr.reshape(-1) makes a 1D view of arr
    #
    # [1::2] select every other element of arr,
    #   starting from the 1st element.
    #
    # *= -1 changes sign of selected elements.

    arr.reshape(-1)[1::2] *= -1
    return arr


def main():
    N = 3
    arr = np.arange(N ** 3).reshape(N, N, N)
    print("original array:")
    print(arr)
    print("change signs")
    print(change_sign(arr))


main()

结果：

original array:
[[[ 0  1  2]
  [ 3  4  5]
  [ 6  7  8]]

 [[ 9 10 11]
  [12 13 14]
  [15 16 17]]

 [[18 19 20]
  [21 22 23]
  [24 25 26]]]
change signs
[[[  0  -1   2]
  [ -3   4  -5]
  [  6  -7   8]]

 [[ -9  10 -11]
  [ 12 -13  14]
  [-15  16 -17]]

 [[ 18 -19  20]
  [-21  22 -23]
  [ 24 -25  26]]]

【讨论】：

arr.ravel()[1::2]*=-1 怎么样？否定展平视图中的所有其他元素。

@hpaulj 简单多了哈哈。

甚至np.negative(arr.ravel()[1::2], out = arr.ravel()[1::2]) 可能更快，但可读性稍差