Python 错误：使用序列设置数组元素。蟒蛇/麻木
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Error: Setting an array element with a sequence. Python / Numpy
提问by pceccon
I'm receiving this error when trying to assign an array to another array specific position. I was doing this before creating simple lists and doing such assignment. But Numpy is faster than simple lists and I was trying to use it now.
尝试将数组分配到另一个数组特定位置时收到此错误。在创建简单列表并进行此类分配之前，我正在这样做。但是 Numpy 比简单的列表更快，我现在正在尝试使用它。
The problem is cause I have a 2D array that stores some data and, in my code, I have, e.g., to calculate the gradient for each position value, so I create another 2D array where each position stores the gradient for its value.
问题是因为我有一个 2D 数组来存储一些数据，并且在我的代码中，我有，例如，计算每个位置值的梯度，所以我创建了另一个 2D 数组，其中每个位置存储其值的梯度。
import numpy as np
cols = 2
rows = 3
# This works
matrix_a = []
for i in range(rows):
matrix_a.append([0.0] * cols)
print matrix_a
matrix_a[0][0] = np.matrix([[0], [0]])
print matrix_a
# This doesn't work
matrix_b = np.zeros((rows, cols))
print matrix_b
matrix_b[0, 0] = np.matrix([[0], [0]])
What happens is 'cause I have a class defining a np.zeros((rows, cols))object, that stores information about some data, simplifying, e.g., images data.
发生的事情是因为我有一个定义np.zeros((rows, cols))对象的类，它存储有关某些数据的信息，简化了例如图像数据。
class Data2D(object):
def __init__(self, rows=200, cols=300):
self.cols = cols
self.rows = rows
# The 2D data structure
self.data = np.zeros((rows, cols))
In a specific method, I have to calculate the gradient for this data, which is a 2 x 2 matrix (cause of this I would like to use ndarray, and not a simple array), and, to do this, I create another instance of this object to store this new data, in which each point (pixel) should store its gradient. I was using simple lists, which works, but I though I could gain some performance with numpy.
在特定方法中，我必须计算此数据的梯度，它是一个 2 x 2 矩阵（因此我想使用ndarray，而不是简单的array），为此，我创建了另一个实例这个对象来存储这个新数据，其中每个点（像素）应该存储它的梯度。我使用的是简单的列表，这很有效，但我虽然可以使用 numpy.js 获得一些性能。
There is a way to work around this? Or a better way to do such thing? I know that I can define the array type to object, but I don't know if I lose performance doing such thing.
有办法解决这个问题吗？或者更好的方法来做这样的事情？我知道我可以将数组类型定义为object，但我不知道这样做是否会降低性能。
Thank you.
谢谢你。
采纳答案by tinybike
The trouble is that matrix_b is defaulting to a float dtype. On my machine, checking
问题是 matrix_b 默认为 float dtype。在我的机器上，检查
matrix_b.dtype
returns dtype('float64'). To create a numpy array that can hold anything, you can manually set dtype to object, which will allow you to place a matrix inside of it:
返回dtype('float64')。要创建一个可以容纳任何东西的 numpy 数组，您可以手动将 dtype 设置为 object，这将允许您在其中放置一个矩阵：
matrix_b = np.zeros((rows, cols), dtype=object)
matrix_b[0, 0] = np.matrix([[0], [0], [1]])
回答by M4rtini
You could add another dimension of size 3 to your array.
您可以向数组中添加另一个大小为 3 的维度。
import numpy as np
cols = 2
rows = 3
matrix_b = np.zeros((rows, cols, 3))
matrix_b[0, 0] = np.array([0, 0, 1])
matrix_b[0, 0] = [0, 0, 1] #This also works
Another option is to set the dtype to listand then you can set each element to a list. But this is not really recommended, as you will lost much of the speed performance of numpy by doing this.
另一种选择是将 dtype 设置为list，然后您可以将每个元素设置为一个列表。但这并不是真正推荐的，因为这样做会失去 numpy 的大部分速度性能。
matrix_b = np.zeros((rows, cols), dtype=list)
matrix_b[0, 0] = [0, 0, 1]
