Remove repeated numbers in series

Hi everyone,

I have a sequence of number like

x = [3 3 3 3 3 2 2 2 2 2 2 2 1 1 1 1 3 3 3 3 3 3];

And I would like to reduce it to be:

x = [3 2 1 3];

Does anyone know how to do that?

Thanks a lot,

Eric

Answer 1

[Star Strider]

Try this:

x = [3 3 3 3 3 2 2 2 2 2 2 2 1 1 1 1 3 3 3 3 3 3];
dx = diff([0 x]);
fx = find(dx ~= 0);
Result = x(fx)

producing:

Result =
     3     2     1     3

Note: This works here, and may not scale up to other problems.

[Stephen23]

Note that this code will not work if the first value happens to be zero. This bug is easy to fix by prepending a logical value after diff (rather than prepending a fake data value before diff). It is also easy get rid of the superfluous find as well:

>> x = [0 0 0 3 3 3 3 3 2 2 2 2 2 2 2 1 1 1 1 3 3 3 3 3 3];
>> y = x([true,diff(x)~=0])
y =
   0   3   2   1   3

[Star Strider]

@Stephen —

Noted. Thank you.

Answer 2

[madhan ravi]

Try this:

ii = [find(x(1:end-1) ~= x(2:end)) numel(x)]; 
l = diff([0 ii]); % just as a bonus but can be neglected if your not interested how many times the number appears consecutively
u = x(ii)  

[Stephen23]

What is l used for?

[madhan ravi]

Ah , it’s not needed but it gives the number of consecutive times the number appears in a sequence.