Finding consecutive values in a vector

Hi. I'm trying to find the consecutive values sorted in a vector and count the total number of the consecutive values. Here is what I have:

A = input('Enter a vector: ');
diff(find(diff([nan A nan])~=1));
% By entering the vector [1 2 3 4 4 4 3 2 1], the output would be 
% ans =
%    4     1     1     1     1     1

The reason I'm stuck is that my code only works for increment vectors([1 2 3 4]), but not for identical([4 4 4]) or decrement vectors([4 3 2 1]). Any suggestions or hints will be of great help to me.

[Peter Wang]

I've tried both

diff(find(diff([nan A nan])~=-1|0|1))

and

diff(find(diff([nan A nan])~=-1&0&1))

However neither of them worked.

Answer 1

[Matt J]

Here's a way to do it using group1s from the File Exchange

https://www.mathworks.com/matlabcentral/fileexchange/78008-group1s

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

D=diff([0,A])==1;
G=group1s(D);
G(~G)=max(G)+1:max(G)+nnz(~G);

s=splitapply(@numel,G,G);
clear result; 
  result=s(unique(G,'stable'))

result =

     4     1     1     1     1     1     1     5

[Peter Wang]

Hi Matt,

Thank you for replying. I'm trying to write a code to find consecutive values are sorted, neither increasing nor decreasing, and I want to write one code that satisfies both conditions, and that's where I get stuck.

[Image Analyst]

Yeah, what do you do in a situation like this? The 0 could be a member of the consecutive decreasing sequence (4,3,2,1,0) but also a member of the increasing sequence (0,1,2,3,4,5). So do you want to count the 0 twice -- once in each group? Or just in the first group? So would the result be

4 for the 1,2,3,4

1 for going from the first 4 to the second 4 which doesn't increase or decrease by 1

1 for going from second 4 to the third 4 which doesn't increase or decrease by 1

5 for the decreasing sequence [4,3,2,1,0] (which starts with the third 4)

6 for the increasing sequence [0,1,2,3,4,5] or 5 if you didn't want to include the 0 again.

So that would give a result of [4,1,1,5,6].

I guess, Peter, we'd need a better explanation of your rules. Also, can we assume that the numbers area strictly integers, and that this is not your homework?

I don't understand why you said

A = input('Enter a vector: ');
diff(find(diff([nan A nan])~=1));
% By entering the vector [1 2 3 4 4 4 3 2 1], the output would be 
% ans =
%    4     1     1     1     1     1

I can see the first 4, and maybe two of the 1's, but since the tail end of the vector was [4,3,2,1] (4 consecutive numbers) and your code was clearly trying to allow for differences of -1 as well as +1, why is there not a 4 as the last element of your answer?

[Bruno Luong]

Agree with Image Analyst. The question is unclear.