Count the number of trend changes in a vector.

2012 7 12
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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.
Tips please!
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.

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Sean de Wolski
Sean de Wolski 2012-7-12
If zero doesn't count, then I only see three chanfges
[0 into 1]
[0 into -2]
[-2 into 1]
please clarify.
Jan
Jan 2012-7-12
I still do not see, why this diagram includes "4 height changes". Please define "height change" exactly.
Jan
Jan 2012-7-16
Please, Sebastian, add informations, which are necessary to define the question, by editing the question, not as comment.
I see your point there. I edited accordingly.

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Andrei Bobrov
Andrei Bobrov 2012-7-16
编辑：Andrei Bobrov 2012-7-16

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EDIT
a = diff(vec(:));
out = nnz(unique(cumsum([true;diff(a)~=0]).*(a~=0)));

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I just realized that is what Doug mentioned earlier. This (nor mine above) solution is no different/no better than his and struggles the same on a vector e.g
vec = [ 1 2 3 3 4 5 6]
Wow. That's really condensed! Would you mind breaking down the logic for me? I can't seem to wrap my head round it.
Andrei Bobrov
Andrei Bobrov 2012-7-17
编辑：Andrei Bobrov 2012-7-17
Hi Sebastian! Thank you for "Accepted..".
Let
vec = [-3 -2 3 3 2 0 0 1 2 5]';
Our goal is to determine the number of groups of identical nonzero elements located near in a:
a = diff(vec);
In our case: out = 6.
t - indexs of all groups (with zeros's groups).
t = cumsum([true;diff(a)~=0]);
non-zero elements in a:
t1 = t(a~=0); % or t1 = cumsum([true;diff(a)~=0]).*(a~=0);
index of groups with none-zero element ( t21 ):
t21 = unique(t1); % or t22 = unique(cumsum([true;diff(a)~=0]).*(a~=0));
and lost:
out1 = numel(t21); % or out2 = nnz(unique(cumsum([true;diff(a)~=0]).*(a~=0)));

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更多回答（2 个）

Doug Hull
Doug Hull 2012-7-12

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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.

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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.

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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!

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