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Msen Rezaee February 2016
### How to extract vectors of consecutive numbers?

Suppose that I have a Q vector which is defined as `Q = [1 2 3 4 5 8 9 10 15];`

and I would like to find a way to extract different vectors of consecutive numbers and also a vector for the rest of the elements. So my result would be like:

```
q1 = [1 2 3 4 5];
q2 = [8 9 10 ];
q3 = [15];
```

You can do this using `diff`

, `cumsum`

and `accumarray`

:

```
q = accumarray(cumsum([1, diff(Q)~=1])', Q', [], @(x){x})
```

which returns:

```
{[1,2,3,4,5];
[8,9,10];
[15]}
```

i.e. `q{1}`

gives you `[1,2,3,4,5]`

etc which is a far cleaner solution to having separately named vectors. But if you really really wanted to have them, and you know exactly how many groups you will get out, you can do it as follows:

```
[q1,q2,q3] = q{:};
```

Explanation:

`accumarray`

will apply an aggregation function (4th input) to elements of a vector (2nd input) based on groupings specified in another vector (1st input).

To use the notation in the docs:

```
sub = cumsum([1, diff(Q)~=1])';
val = Q';
fun = @(x){x};
```

Note that `sub`

needs to start from `1`

. The idea is to use `diff`

to find elements that are consecutive (i.e. where `Q(i+1) - Q(i) == 1`

) which is vectorized using the `diff`

function. By specifying `diff(Q)~=1`

we can find the breaks between groups of consecutive numbers (concatenating the `1`

at the beginning to force a break at the start). `cumsum`

then just converts these breaks into vector of in the right form for `sub`

i.e.

```
sub = [1 1 1 1 1 2 2 2 3]
```

The aggregation function we specify is just cell concatenation.

Asked in February 2016

Viewed 1,108 times

Voted 6

Answered 1 times

Viewed 1,108 times

Voted 6

Answered 1 times