pairs of consecutive numbers

Hello,

I have a vector A=[1 2 5] i would like to have something that stores the pair of consecutive numbers eg B= [1 2] and something that stores the values unused eg [C= 5]

i would like this working for A=[5 2 1], A=[5 1 2] and all the permutations of the elements .

Thanks

Here is some piece of code that I have tried but I have no idea how to get the element unused, 'C' in my example above

A= [6 3 4];
p=[find(diff(A)==1); find(diff(A) == -1)];
q=[p;p+1];
A(q)

[Andrei Bobrov]

Please set your result for [1 2 3 4 5 4 3].

Answer 1

[Guillaume]

Well, you're almost there. When the diff is 1 or -1, this means that the number at the same index and the next one are consecutive. The easiest way to find the non-consecutive numbers is to take the set difference of all the indices with the pair indices.

A = [6 2 3 2 4 1 3 2 1 0 5]
pairindices = find(ismember(diff(A), [-1, 1])) + [0;1]  %assumes A is a row vector
nonconsindices = setdiff(1:numel(A), pairindices)

Another method, which uses the undocumented feature that strfind works with numeric vectors:

nonconsindices = strfind([0, abs(diff(A)) == 1, 0], [0, 0])

[Lenin Cruz]

Thanks for your reply Guillaume. The problem that I'm facing now is that I would like to find the pairs of consecutive numbers(array elements) even when they are not next to each other in the array e.g

A= [2 6 3];

I would like my output to be:

pairindices= [2;3]

But my code outputs an empty matrix when I set an Array 'A' like the one above.

Any advice on how to do that?

Thanks a lot.

[Guillaume]

That's completely different from what you asked originally, but easily worked around. Sort A beforehand:

A = [2 6 20 3 15 9 10 8 14 12]
[sorted, origindex] = sort(A);
pairindices = origindex(find(abs(diff(sorted)) == 1) + [0; 1])
nonconsindices = setdiff(1:numel(A), pairindices)

[Lenin Cruz]

I am thinking about reorganizing my matrix A above, is there any way to reorganize the elements of an array from the shortest to the greatest number?

[Lenin Cruz]

Sorry about the misunderstanding, Thanks a lot for your help.

[Ancalagon8]

Does this also work with a timetable? (timestamp in first collumn and values in the second).

[Rik]

@Ancalagon8 did you try? What happened? Did you expect that result?

[Ancalagon8]

Yes I tried only with my value column

pairindices = find(ismember(diff(A), [-1, 1])) + [0;1] 
nonconsindices = setdiff(1:numel(A), pairindices)

Receiving error : Matrix dimensions must agree.

While

nonconsindices = strfind([0, abs(diff(A)) == 1, 0], [0, 0])

gave me: Error using horzcat: Dimensions of arrays being concatenated are not consistent.

[Bruno Luong]

Guys, esepecially @Ancalagon8 you post your question/comment at the wrong place.

If you want to get any useful help you should provide the data or a representatve example of it (A).

[Ancalagon8]

i tried @Guillaume code from above.I attach A collumn.

[sorted, origindex] = sort(A);
pairindices = origindex(find(abs(diff(sorted)) == 1) + [0; 1])
nonconsindices = setdiff(1:numel(A), pairindices)

[Bruno Luong]

"I attach A collumn."

@Guillaume works for row vector, so just traspose it

load('A.mat');
A = A.';
[sorted, origindex] = sort(A);
pairindices = origindex(find(abs(diff(sorted)) == 1) + [0; 1])

pairindices = 1×2

365 2

nonconsindices = setdiff(1:numel(A), pairindices)

nonconsindices = 1×363

1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

[Ancalagon8]

Thanks for your reply! I am trying to find consecutive numbers from a row but I also have another row with the date and time information. Is it possible to combine the above code with timestamp? I mean after having found the consecutive numbers e.g from row position 14 to 19, 33 to 36 etc to assign these number to dates (14-Jan to 19-Jan, 02-Febr to 05-Febr). Any ideas how to implement the timestamp information?

[Stephen23]

"from row position 14 to 19, 33 to 36 etc to assign these number to dates (14-Jan to 19-Jan, 02-Febr to 05-Febr)"

V = [14;19;33;36];
D = datetime(2022,1,V)

D = 4×1 datetime array
   14-Jan-2022
   19-Jan-2022
   02-Feb-2022
   05-Feb-2022

[Ancalagon8]

@Stephen23 thank you very much!

Answer 2

[Rik]

I have expanded your example array a bit to more clearly show the indeded effect. I have replicated the effect of your code, please confirm that is what you meant.

A=[3 1 2 0 3 4 5];

L= abs(diff(A))==1;
L=[L false;false L];
L_unused= sum(L,1)==0 ;

B=[A(L(1,:));A(L(2,:))];
C=A(L_unused);

[Lenin Cruz]

Thanks for your reply. Yes, that is similar to what my code does, but let's say that now

A=[5 3 4 1];

I would expect my matrix B to be: B=[3 4; 4 5]

but I am just getting B=[3;4]

Any advice on how to do that? Thanks

[Rik]

The solution is the same as with the other method: sort A beforehand.

Answer 3

[Bruno Luong]

A=[5 3 4 1];

As=sort(A);
b=diff(As)==1;
pairs=As(b)'+[0 1];

notpair = setdiff(A,pairs(:))

notpair = 1