将一个 numpy 数组附加到一个列表 - 奇怪的事情

2022-01-22 00:00:00 python numpy list arrays append

问题描述

在 Raspberry Pi 上的 Raspbian 上使用 Spyder 3.1.3 中的 Python3.5.3.将两个 numpy-arrays 附加到名为list0"的列表中可以很好地与分配的 numpy 数组a"类似:

Using Python3.5.3 in Spyder 3.1.3 on Raspbian on a Raspberry Pi. Appending two numpy-arrays to a list named 'list0' works well with allocated numpy arrays 'a' like:

import numpy as np

list0 = []
a = np.array([[1,2,3],[2,3,4]])
list0.append(a)
a = np.array([[11,12,13],[12,13,14]])
list0.append(a)

print("list0 =",list0)

效果很好，作为输出提供(帖子的格式更好):

works well, giving as output (a bit better formatted for the post):

list0 = [ array([[ 1,  2,  3], [ 2,  3,  4]]), 
          array([[11, 12, 13], [12, 13, 14]]) ]

使用循环将赋值替换为a，奇怪的事情发生了:

Replacing the assignment to a using a loop, weird things happen:

import numpy as np
a = np.empty((3), int)
list0 = []
for idx in range(4):    
    for i in range(3):
        a[i] = idx*10 + i
    print("idx =",idx,"; a =",a)
    list0.append(a)
print("list0 =",list0)

第二行告诉 Python 使用的数组的形状(在我原来的例子中，它是一个三维数组).为了验证生成的名为a"的数组被打印出来.将新填充的数组 'a' 附加到 'list0' 最终显示最后一行的四倍.

The second line tells Python the shape of the array used (in my original case it is a three-dimensional array). For verification the generated arrays named 'a' are printed out. Appending the newly filled arrays 'a' to 'list0' finally shows four times the last line.

idx = 0 ; a = [ 0  1  2]
idx = 1 ; a = [10 11 12]
idx = 2 ; a = [20 21 22]
idx = 3 ; a = [30 31 32]
list0 = [ array([30, 31, 32]), array([30, 31, 32]), 
          array([30, 31, 32]), array([30, 31, 32]) ]

我想list0"只包含四倍的指向数组a"的指针，该数组只存在于一个实例/内存范围中.

I suppose that 'list0' simply contains four times a pointer to the array 'a' which only exists in one instance / memory range.

那么:如何将每个不同的数组 'a' 物理附加(复制?)到列表中?它是一个 python 错误还是只是我对某些东西的误解?当然，我应该多想pythonian ;c)

So: How can I physically append (copy?) each of the different arrays 'a' to the list? Is it a python bug or is it simply my misunderstanding of something? Certainly I should think more pythonian ;c)

谢谢你的帮助，彼得

解决方案

问题

您将相同的数组 a 附加到您的 list0 4 次.像 a 这样的数组是可变对象，这意味着，当你给它们赋值时，底层对象会发生变化.由于该数组在您的列表中出现了 4 次，因此这些更改(似乎)出现在 4 个不同的地方.

The problem

You're appending the same array a to your list0 4 times. Arrays like a are mutable objects, which means, among other things, that when you assign values to them the underlying object changes. Since the array is present in your list 4 times, those changes (seem to) show up in 4 different places.

您只需稍作改动即可修复您拥有的代码.将数组的副本附加到列表中，而不是数组本身:

You can fix the code you have with one small change. Append a copy of the array to your list, instead of the array itself:

import numpy as np
a = np.empty((3), int)
list0 = []
for idx in range(4):    
    for i in range(3):
        a[i] = idx*10 + i
    print("idx =",idx,"; a =",a)
    list0.append(a.copy())
print("list0 =",list0)

输出:

idx = 0 ; a = [0 1 2]
idx = 1 ; a = [10 11 12]
idx = 2 ; a = [20 21 22]
idx = 3 ; a = [30 31 32]
list0 = [array([0, 1, 2]), array([10, 11, 12]), array([20, 21, 22]), array([30, 31, 32])]

优化方案

Python/Numpy 提供了许多更好的方法(无论是使用更少的代码行还是运行速度更快)来初始化数组.对于像这样的一堆范围，这是一个合理的方法:

Optimized solution

Python/Numpy offer many better ways (both in terms of using fewer lines of code and running faster) to initialize arrays. For a bunch of ranges like this, here is a reasonable approach:

list0 = [np.arange(n*10, n*10+3) for n in range(4)]
print(list0)

输出:

[array([0, 1, 2]), array([10, 11, 12]), array([20, 21, 22]), array([30, 31, 32])]

您也可以考虑只使用单个二维数组来代替数组列表.单个数组通常比列表中的异构数组更易于使用.这样做的方法如下:

You might also consider just using a single 2D array in place of a list of arrays. One single array is often easier to work with than a heterogenous mix of arrays in a list. Here's how you do that:

arr0 = np.array([np.arange(n*10, n*10+3) for n in range(4)])
print(arr0)

输出:

[[ 0  1  2]
 [10 11 12]
 [20 21 22]
 [30 31 32]]

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