问题描述:

I have a vector `x = [1,3,5,6,7]`

and I want to produce a matrix `y`

in which rows y(k) = x(k:k+2). So the resulting matrix in this case would be

`1 3 5`

3 5 6

5 6 7

How can I achieve this **without** using a loop? Is there a clever way of doing it with indexing?

This is the top non-zero square of a Hankel matrix. Just use `hankel`

:

```
>> X = hankel(x)
X =
1 3 5 6 7
3 5 6 7 0
5 6 7 0 0
6 7 0 0 0
7 0 0 0 0
>> X = X(1:3,1:3)
X =
1 3 5
3 5 6
5 6 7
```

Generalized, `hankel`

output specified exactly:

```
w = floor(numel(x)/2);
X = hankel(x(1:end-w),x(w+1:end))
```

A slightly contrived way using `meshgrid`

:

```
k = (length(x) + 1) / 2;
[a b] = meshgrid(1:k, 0:k-1);
y = x(a+b);
```

Or the compact equivalent using `bsxfun`

```
y = x(bsxfun(@plus, (1:k)', 0:k-1));
```

Or a *really* silly one-liner:

```
y = x(interp2([1 3], [1;3], [1 3; 3 5], 1:3, (1:3)'));
```

You can do this the following way without direct loop:

```
cell2mat(arrayfun(@(k) x(k:k+2), 1:numel(x) - 2, 'UniformOutput', false)')
ans =
1 3 5
3 5 6
5 6 7
```

Though, `arrayfun`

actually loops over elements `1:numel(x) - 2`

. So its a bit of cheating, i guess.

Using convolutions:

```
n = numel(x)-2; %// x is a row vector with arbitrary length
result = conv2(x,rot90(eye(n)));
result = result(:,n:end-n+1);
```