numpy数组和Matlab矩阵不匹配[3D] [英] Numpy array and Matlab Matrix are mismatching [3D]

问题描述

以下八度代码显示了使用Octave/Matlab的示例3D矩阵

The following octave code shows a sample 3D matrix using Octave/Matlab

octave:1> A=zeros(3,3,3);
octave:2> 
octave:2> A(:,:,1)= [[1 2 3];[4 5 6];[7 8 9]];
octave:3> 
octave:3> A(:,:,2)= [[11 22 33];[44 55 66];[77 88 99]];
octave:4> 
octave:4> A(:,:,3)= [[111 222 333];[444 555 666];[777 888 999]];
octave:5> 
octave:5> 
octave:5> A
A =

ans(:,:,1) =

   1   2   3
   4   5   6
   7   8   9

ans(:,:,2) =

   11   22   33
   44   55   66
   77   88   99

ans(:,:,3) =

   111   222   333
   444   555   666
   777   888   999

octave:6> A(1,3,2)
ans =  33

而且我需要使用numpy转换相同的矩阵...不幸的是，当我尝试使用numpy中的数组访问相同的索引时，我得到的值如下所示！

And I need to convert the same matrix using numpy ... unfortunately When I'm trying to access the same index using array in numpy I get different values as shown below!!

import numpy as np
array = np.array([[[1 ,2 ,3],[4 ,5 ,6],[7 ,8 ,9]], [[11 ,22 ,33],[44 ,55 ,66],[77 ,88 ,99]], [[111 ,222 ,333],[444 ,555 ,666],[777 ,888 ,999]]])
>>> array[0,2,1]
8

我还阅读了以下文档，该文档显示了Matlab和Python numpy中矩阵实现之间的区别

Also I read the following document that shows the difference between matrix implementation in Matlab and in Python numpy Numpy for Matlab users but I didn't find a sample 3d array and the mapping of it into Matlab and vice versa!

答案是不同的，例如，使用numpy(0,2,1)访问Matlab中的element(1,3,2)与相同的索引不匹配

the answer is different for example accessing the element(1,3,2) in Matlab doesn't match the same index using numpy (0,2,1)

八度/Matlab

Octave/Matlab

八度:6> A(1,3,2)

ans = 33

Python

Python

>>>数组[0,2,1]

8

推荐答案

在numpy中构造数组的方式与在MATLAB中构造数组的方式不同.

The way your array is constructed in numpy is different than it is in MATLAB.

在您的MATLAB数组为(y, x, z)的地方，您的numpy数组为(z, y, x).您的3d numpy数组是一系列堆叠"的2d数组，因此您要为"outside-> inside"编制索引(因为缺少更好的术语).这是扩展的数组定义，因此(希望如此)更有意义:

Where your MATLAB array is (y, x, z), your numpy array is (z, y, x). Your 3d numpy array is a series of 'stacked' 2d arrays, so you're indexing "outside->inside" (for lack of a better term). Here's your array definition expanded so this (hopefully) makes a little more sense:

[[[1, 2, 3],
  [4, 5, 6],        # Z = 0
  [7 ,8 ,9]],
 [[11 ,22 ,33],
  [44 ,55 ,66],     # Z = 1
  [77 ,88 ,99]],
 [[111 ,222 ,333],
  [444 ,555 ,666],  # Z = 2
  [777 ,888 ,999]]
]

所以:

import numpy as np

A = np.array([[[1 ,2 ,3],[4 ,5 ,6],[7 ,8 ,9]], [[11 ,22 ,33],[44 ,55 ,66],[77 ,88 ,99]], [[111 ,222 ,333],[444 ,555 ,666],[777 ,888 ,999]]])
B = A[1, 0, 2]

B会按预期返回33.

如果您想要一种省力的索引数组的方法，请考虑像在MATLAB中一样生成它.

If you want a less mind-bending way to indexing your array, consider generating it as you did in MATLAB.

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