# Count the number of trend changes in a vector.

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Sebastian Holmqvist on 12 Jul 2012
I need to count the number of times a vector increases/decreases. E.g
vec = [0 0 1 2 4 4 2 0 1 2 3]
diff(vec)
ans =
[0 1 1 2 0 -2 -2 1 1 1]
would evidently yield 4 trend changes. I don't wish to count the zero as a trend.
Update:
If you were to plot(vec) it's easier to see what I mean. I wish to simply count the number of row changes in the plot. A flat line (derivative of 0) isn't considered a row change in my application.
Take this as an example:
vec = [3 3 3 5 5 4 3 2 4 5 5 5];
plot(vec)
As you can see from the plot, it's 4 row changes (even though it's 8 in pure diff).
Update:
Ok, I'll try to do better.
Imagine a matrix that describes position over time, where row is position and column is time. When going in a straight line (along a single row) I wish to do nothing. But when changing course over to another row-line I wish to count each new "course" as an occurrence.
E.g
• Travelling at row 5 and then moving up to row 6 and then going straight, that's one course change.
• [5 5 5 6 6 6] => 1 "course change"
• Travelling at row 5 and then moving up to row 7 is similarly one change.
• [5 5 5 7 7 7] => 1 "course change"
• Travelling at row 5, moving up to 7 and then directly back to row 6 is then two changes.
• [5 5 5 7 6 6 6] => 2 "course changes"
• Travelling at row 5, moving up to 7 and then to 12 and then going straight is two changes as the row increment is 2 and then 5 (not a straight course).
• [5 5 5 7 12 12 12] => 2 "course changes"
So basically I wish to count each time you change position (or row) in a straight course.
Sebastian Holmqvist on 30 Jul 2012
I see your point there. I edited accordingly.

Andrei Bobrov on 16 Jul 2012
Edited: Andrei Bobrov on 16 Jul 2012
EDIT
a = diff(vec(:));
out = nnz(unique(cumsum([true;diff(a)~=0]).*(a~=0)));
Sebastian Holmqvist on 17 Jul 2012
Great explanation! Thanks!

Doug Hull on 12 Jul 2012
You would need to remove the zeros from the first diff vector and diff it again, then count number of non-zeros. The only complication is the case where the first diff vector looks like:
[1 1 0 1 1]
Once the zero is removed, it looks constant. That can be dealt with, but would be messy.
Sebastian Holmqvist on 12 Jul 2012
Good idea. But it doesn't work with this example:
>> vec = [3 3 3 5 5 4 3 2 4 5 5 5];
>> d_vec = diff(vec)
d_vec =
0 0 2 0 -1 -1 -1 2 1 0 0
>> d_vec(d_vec==0) = []
d_vec =
2 -1 -1 -1 2 1
>> diff(d_vec)
ans =
-3 0 0 3 -1
>> sum(diff(d_vec)~=0)
ans =
3
Which would be 3 changes with your argumentation. But with my (somewhat fuzzy) specification there should be 4. plot(vec) should make it more clear.

Sebastian Holmqvist on 16 Jul 2012
I think I've cracked it. My logic follows from;
1. I wish to count the number of first derivative changes.
2. I wish to exclude derivatives == 0.
3. Derivative element trends are to be treated as single elements.
First off, I wrote a small function to remove all local dupes.
function vec = rmDupes(vec)
% Remove all duplicate subsequential elements in a vector
% Iterate backwards to avoid changing indexing
for i = length(vec):-1:2
if vec(i) == vec(i-1)
vec(i) = [];
end
end
By not removing the zero elements before removing the dupes I don't get the case where [1 2 3 3 4 5] counts as one.
vec = [1 1 2 2 3 4 1 1]
d_vec = diff(vec)
d_vec = rmDupes(d_vec)
sum(d_vec ~= 0)
ans =
3
If anyone can detect any flaws in my solution, or has a better suggestion, I would be more than grateful if you would give me some feedback!