How can I separate the elements of a vector ?

Hello,

I have a row vector i of size say 1x300 and that its indices can be any number of the range [1 5] in a descending order. For example:

i = [5 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 0 0 0 0 0]

As you can notice number 4 is not always in i. And I'm trying to separate the similar numbers into 4 groups in this example.

for jj = max(i):-1:0
    t = i==jj;
    idx = find(t~=0);
    .
    .
    .
end

In this particular example, jj should take only values 5,3,2,1,0. But it is not since the step is -1. And the resulting idx is empty.

When i is like the following, my code works fine.

i = [6 6 5 4 4 4 3 3 3 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 0 0 0 0 0]

I would appreciate your help.

Answer 1

[Stephen23]

for jj = i(diff(i)~=0)

or

for jj = unique(i,'stable')

[Rayan Glus]

Thank you so much, Stephen!

Sometimes many of i's elements have a value of -1. I'm not interested in -1 values. Can I do the following before the for loop?

exclude = find(i == -1);
i(exclude) = []; 

Is there another way to do this using the same the for loop?

Thanks again

[Stephen23]

"Sometimes many of i's elements have a value of -1. I'm not interested in -1 values."

Aaah, extra information that was not explained in the question... but still insufficient information:

• Are the values of i always monotonic? Counter example: [2,2,2,5,5,1,1,1,3,3,3]
• Is -1 always last? Or can it occur spread throughout i ? How is it distributed?
• Can values occur in multiple groups? Or is does each value define exactly one group?
• What happens to one valkue with -1 in the middle, example: [2,2,2,-1,-1,2,2,2], is that one group or two?

[Rayan Glus]

• Yes it is always monotonic. example: [5 5 3 3 3 2 1 1 1 1 -1 -1 -1 -1] descending order
• Yes. -1 is always the last.
• No they cannot occur in multiple groups.
• the 2's in your example cannot be interrupted by any other value. [2 2 2 2 2 0 0 0 0 -1 -1 -1 -1]

[Stephen23]

Then before the loop use some logical indexing (simpler and more efficient than FIND):

i(i<0) = [];

or without modifying i at all:

for jj = unique(i(i>=0),'stable')

[Rayan Glus]

I'm not sure why when I use

for jj = unique(i(i>0),'stable')

i will get updated when jj reaches 1, hence jj will have another value.

i = [6 6 5 4 4 4 3 3 3 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 0 0 0 0 0]

Once jj reaches 0, i will get updated and the for loop will start all over again withh jj starting from value 5

[Rayan Glus]

This will exclude the 0's and -1's correct? if I want to keep the 0's I can just add an = sign, am I correct?

for jj = unique(i(i>=0),'stable')

[Stephen23]

i = [6,6,5,4,4 4 3 3 3 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 0 0 0 0 0,-1,-1,-1];
%         vv             I forgot that you want zeros.
unique(i(i>=0),'stable')

ans = 1×7

6 5 4 3 2 1 0

[Rayan Glus]

Thanks, Stephen. Do you know by any chance why when jj reaches 0, i get updated and the for loop starts all over again

[Stephen23]

"do you know by any chance why when jj reaches 0, i get updated and the for loop starts all over again"

I don't see that (nor do I see your code):

i = [6,6,5,4,4,4,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0,0,0,0,0,-1,-1,-1];
for jj = unique(i(i>=0),'stable')
    display(jj)
end

jj = 6

jj = 5

jj = 4

jj = 3

jj = 2

jj = 1

jj = 0

[Rayan Glus]

My bad.

Thank you so much. I really appreciate it

Answer 2

[Walter Roberson]

File Exchange, run length encoding. If I recall correctly, contribution is from Jan.

[Rayan Glus]

Thanks Walter. I will check it out.