Python 切片 3d numpy 数组
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Slicing 3d numpy arrays
提问by Patrick Rinker
Consider the following:
考虑以下：
A = np.zeros((2,3))
print(A)
[[ 0. 0. 0.]
[ 0. 0. 0.]]
This make sense to me. I'm telling numpy to make a 2x3 matrix, and that's what I get.
这对我来说很有意义。我告诉 numpy 制作一个 2x3 矩阵，这就是我得到的。
However, the following:
但是，以下内容：
B = np.zeros((2, 3, 4))
print(B)
Gives me this:
给我这个：
[[[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]]
[[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]]]
This doesn't make sense to me. Aren't I telling numpy to make a cube which has 4 2x3 matrices? I'm even more confused because although the data structure looks incorrect, the slicing works exactly as planned:
这对我来说没有意义。我不是告诉 numpy 制作一个有 4 个 2x3 矩阵的立方体吗？我更加困惑，因为虽然数据结构看起来不正确，但切片完全按计划工作：
print(B[:,:,1])
[[ 0. 0. 0.]
[ 0. 0. 0.]]
I'm missing something about how these arrays are constructed, but I'm not sure what. Can someone explain what I'm missing or not understanding?
我缺少有关如何构造这些数组的信息，但我不确定是什么。有人可以解释我缺少或不理解的内容吗？
Thanks so much!
非常感谢！
回答by adrianX
B is a 3D matrix. the indices that you specified (2x3x4) is exactly what is printed out. the outermost brackets have 2 elements, the middle brackets have 3 elements, and the innermost brackets have 4 elements.
B 是一个 3D 矩阵。您指定的索引 (2x3x4) 正是打印出来的。最外面的括号有2个元素，中间的括号有3个元素，最里面的括号有4个元素。
回答by unutbu
NumPy arrays iterate over the left-most axis first. Thus if Bhas shape
(2,3,4), then B[0]has shape (3,4) and B[1]has shape (3,4). In this sense,
you could think of Bas 2 arrays of shape (3,4). You can sort of see the two
arrays in the repr of B:
NumPy 数组首先迭代最左边的轴。因此，如果B具有形状 (2,3,4)，则B[0]具有形状 (3,4) 并B[1]具有形状 (3,4)。从这个意义上讲，您可以将其B视为 2 个形状为 (3,4) 的数组。您可以在以下的 repr 中看到两个数组B：
In [233]: B = np.arange(2*3*4).reshape((2,3,4))
array([[[ 0, 1, 2, 3],
[ 4, 5, 6, 7], <-- first (3,4) array
[ 8, 9, 10, 11]],
[[12, 13, 14, 15],
[16, 17, 18, 19], <-- second (3,4) array
[20, 21, 22, 23]]])
You can also think of Bas containing four 2x3 arrays by iterating over the last index first:
您还可以B通过首先遍历最后一个索引来将其视为包含四个 2x3 数组：
for i in range(4):
print(B[:,:,i])
# [[ 0 4 8]
# [12 16 20]]
# [[ 1 5 9]
# [13 17 21]]
# [[ 2 6 10]
# [14 18 22]]
# [[ 3 7 11]
# [15 19 23]]
but you could just as easily think of Bas three 2x4 arrays:
但是你可以很容易地想到B三个 2x4 数组：
for i in range(3):
print(B[:,i,:])
# [[ 0 1 2 3]
# [12 13 14 15]]
# [[ 4 5 6 7]
# [16 17 18 19]]
# [[ 8 9 10 11]
# [20 21 22 23]]
NumPy arrays are completely flexible this way. But as far as the reprof Bis concerned, what you seecorresponds to two (3x4) arrays since Biterates over the left-most axis first.
NumPy 数组在这种方式下是完全灵活的。但就reprofB而言，您看到的对应于两个 (3x4) 数组，因为首先B遍历最左侧的轴。
for arr in B:
print(arr)
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]]
# [[12 13 14 15]
# [16 17 18 19]
# [20 21 22 23]]
回答by Atul
I hope the below example would clarify the second part of your question where you have asked about getting a 2X3 matrics when you type print(B[:,:,1])
我希望下面的例子能澄清你的问题的第二部分，你在输入时询问获得 2X3 矩阵 print(B[:,:,1])
import numpy as np
B = [[[1,2,3,4],
[5,6,7,8],
[9,10,11,12]],
[[13,14,15,16],
[17,18,19,20],
[21,22,23,24]]]
B = np.array(B)
print(B)
print()
print(B[:,:,1])
[[[ 1 2 3 4]
[ 5 6 7 8]
[ 9 10 11 12]]
[[13 14 15 16]
[17 18 19 20]
[21 22 23 24]]]
[[ 2 6 10]
[14 18 22]]
Since the dimension of B is 2X3X4, It means you have two matrices of size 3X4as far as reprof B is concerned
由于B的尺寸2X3X4，这意味着你有大小两个矩阵3X4至于reprB的关注
Now in B[:,:,1]we are passing :, :and 1. First :indicates that we are selecting both the 3X4matrices. The second :indicates that we are selecting all the rows from both the 3X4matrices. The third parameter 1indicates that we are selecting only the second column values of all the rows from both the 3X4matrices. Hence we get
现在B[:,:,1]我们正在通过:，:并且1。First:表示我们正在选择两个3X4矩阵。第二个:表示我们正在从两个3X4矩阵中选择所有行。第三个参数1表示我们只从两个3X4矩阵中选择所有行的第二列值。因此我们得到
[[ 2 6 10]
[14 18 22]]
