例如，我有一个向量
A = [1 2 3 4 5 6 7 8]
我想使用windowsize=4和stepsize=2对它进行“重塑”以使其矩阵化，使得最终的矩阵为
b = [ 1 3 5;
2 4 6;
3 5 7;
4 6 8 ]
答案 0 :(得分：4)
您可以设置索引矩阵，然后只需索引到 <!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Title</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/js/bootstrap.min.js"></script>
</head>
<body>
<div class="container">
<div class="table-responsive">
<table class="table">
<thead>
<tr>
<th>1</th>
<th>Header 2</th>
<th>Header 3</th>
<th>Header 4</th>
<th>Header 5</th>
<th>Header 6</th>
<th>Header 7</th>
<th>Header 8</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>2</td>
<td>3</td>
<td>4</td>
<td>5</td>
<td>longtextlongtextlongtext</td>
<td>6</td>
<td>7</td>
</tr>
</tbody>
</table>
</div>
</div>
</body>
</html> ...
A注意；在这种情况下，A = [1 2 3 4 5 6 7 8];
windowsize = 4;
stepsize = 2;
% Implicit expansion to create a matrix of indices
idx = bsxfun( @plus, (1:windowsize).', 0:stepsize:(numel(A)-windowsize) );
b = A(idx);
和idx是相同的，但是您需要在最终索引步骤中假设b不仅仅是实际示例中的连续整数。