Counting runs in a vector

Hi all, I have the following vector for which I would like to count "runs", i.e. successive increments in each direction.

 v = [0 0 1 2 3 2 2 1 2 2 1 2 3 4 3];

The desired output would look like

 out = [3 2 1 1 3 1]

.. because from the first to the fifth element, v moves 3, and from the fifth to the eighth, switches direction and retracts 2, and so on.

[Jan]

The description is neither clear nor unique. How is "run" exactly defines?

[Hamad Alsayed]

On reflection, this would amount to measuring the absolute value of the vertical distance between each peak and trough (0 first element .... 3 fifth element ... 1 eighth element ...). So a "run" is defined as each one of those individual arcs. I hope this clears it up for you. I see from other Answers that the function findpeaks may be useful here, but I have no access to the Signal Processing toolbox. If you require further clarification please do ask.

[Jan]

Sorry, I do not understand the explanation. What is the "vertical distance", what are the "peaks" and what is "trough"?

Are you looking for local maxima and minima?

[Hamad Alsayed]

The vector is oscillating up and down - plotting it will illustrate this clearly. So the question boils down to... "it moved up this much"... then "moved down this much"... etc

Question has now been answered. Thank you for your input, Jan.

Answer 1

[Stephen23]

>> V = [0,0,1,2,3,2,2,1,2,2,1,2,3,4,3];
>> W = V([true,0~=diff(V)]); % remove repeats
>> diff(W([true,0~=diff(sign(diff(W))),true]))
ans =
   3  -2   1  -1   3  -1

Take the absolute value if you need to.

[Hamad Alsayed]

Thank you very much, Stephen. Works perfectly for me.

Answer 2

[Damo Nair]

Hi,

I hate to trouble you, but is there any way to retrieve the index of the last value? I mean, in the above example 3 would correspond to an index of 5 & -2 to 8.

Thanks Damo.

[Stephen23]

Of course, just keep track of the indices:

>> V = [0,0,1,2,3,2,2,1,2,2,1,2,3,4,3];
>> X = [true,0~=diff(V)];
>> W = V(X);
>> Y = [true,0~=diff(sign(diff(W))),true];
>> diff(W(Y)) % original answer
ans =

     3  -2   1  -1   3  -1

Now you can identify the indices:

>> Z = false(size(X));
>> Z(X) = Y;
>> find(Z)
ans =
    1    5    8    9   11   14   15

Note that 1 is included as it defines the start value.

[Damo Nair]

Outstanding! I couldn't work out the relation between Y & V. Now that you make it so simple I feel a bit silly.

Thanks very much. Goodday Damo.