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Counting runs in a vector

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Hamad Alsayed
Hamad Alsayed 2017-12-4
编辑： Stephen23 2017-12-7
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.
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Jan
Jan 2017-12-4
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
Hamad Alsayed 2017-12-4
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.

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采纳的回答

Stephen23
Stephen23 2017-12-4
编辑：Stephen23 2017-12-4
>> 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.

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Damo Nair
Damo Nair 2017-12-6
编辑：Stephen23 2017-12-6
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.
  2 个评论
Stephen23
Stephen23 2017-12-6
编辑：Stephen23 2017-12-7
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
Damo Nair 2017-12-6
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.

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