重新排列使用Matlab中循环数组 [英] Rearranging an array using for loop in Matlab

问题描述

我有一个1×15阵列值：

I have a 1 x 15 array of values:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

我需要使用一个for循环将它们重新组合成一个3×5矩阵：

I need to rearrange them into a 3 x 5 matrix using a for loop:

 1  2  3  4  5
 6  7  8  9 10
11 12 13 14 15

我将如何做呢？

推荐答案

我要告诉你三种方法。一，你需要有一个为循环，两个人的时候你不在：

I'm going to show you three methods. One where you need to have a for loop, and two others when you don't:

首先，创建一个矩阵，3×5，然后跟踪的指数，将通过您的数组。之后，创建一个双为循环，这将有助于您填充数组。

First, create a matrix that is 3 x 5, then keep track of an index that will go through your array. After, create a double for loop that will help you populate the array.

index = 1;
array = 1 : 15; %// Array we wish to access
matrix = zeros(3,5); %// Initialize
for m = 1 : 3
    for n = 1 : 5
         matrix(m,n) = array(index);
         index = index + 1;
    end
end

matrix =

  1     2     3     4     5
  6     7     8     9    10
 11    12    13    14    15

方法2 - 如果没有一个为循环

简单地说，使用重塑：

matrix = reshape(1:15, 5, 3).';

matrix =

  1     2     3     4     5
  6     7     8     9    10
 11    12    13    14    15

重塑将采取载体，并重组为一个矩阵，让你按列先填充矩阵。因此，我们希望把1〜5中的第一列，6至10中的第二和第11至15中的第三列。因此，我们的输出矩阵实际上是在5×3，当你看到这一点，这其实是在换位我们想要的矩阵，这就是为什么你做的版本。转置矩阵回来。

reshape will take a vector and restructure it into a matrix so that you populate the matrix by columns first. As such, we want to put 1 to 5 in the first column, 6 to 10 in the second and 11 to 15 in the third column. Therefore, our output matrix is in fact 5 x 3. When you see this, this is actually the transposed version of the matrix we want, which is why you do .' to transpose the matrix back.

您可以使用 vec2mat ，并指定你需要有5列值得你的矩阵：

You can use vec2mat, and specify that you need to have 5 columns worth for your matrix:

matrix = vec2mat(1:15, 5);

matrix =

  1     2     3     4     5
  6     7     8     9    10
 11    12    13    14    15

vec2mat 需要一个载体，如您在第二个参数中指定其重塑为多达列的矩阵。在这种情况下，我们需要5列。

vec2mat takes a vector and reshapes it into a matrix of as many columns as you specify in the second parameter. In this case, we need 5 columns.

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