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
推荐答案
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]]
这篇关于将一个 numpy 数组附加到一个列表 - 奇怪的事情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跟版网！