How would I extract the first and last nonzero numbers from this array?

So I am trying to grab the first and last numbers in the array multiple times. For example if I wanted to do it on this array.

[0 0 0 0 0 0 0 1 2 2 2 3 0 0 0 0 0 3 2 2 2 2 7 0 0 0 0 0 0 0]

Lets say I wanted the 1 and 3 in the first set of numbers and the 3 and 7 in the next set of numbers. How would I do this?

Answer 1

[Stephen23]

Simpler:

>> V = [0,0,0,0,0,0,0,1,2,2,2,3,0,0,0,0,0,3,2,2,2,2,7,0,0,0,0,0,0,0];
>> D = diff([0,V,0]==0);
>> [V(D<0);V(D(2:end)>0)] % each column is one pair of values
ans =
     1     3
     3     7
>>

[Brandon Fox]

Thank you this works beautifully!

[Stephen23]

@Brandon Fox: my pleasure, I hope it helps. You can accept the answer that helped you most resolve your original question: this is the easiest way for you to show your appreciation to the volunteers who helped you.

[Brandon Steinlein]

Is it possible to pull the associated position (column/row) in the old matrix of the numbers in the new matrix?

Answer 2

[Robert U]

Hi Brandon Fox:

There are several cases that should be covered in case you want to create a robust function:

1. vector starts with a number not zero
2. vector ends with a number not zero
3. single numbers in line not zero

A starting approach is to capture changes in the vector might be

 function [ numbers ] = extract_nonzero( varargin )
    if nargin < 1 
        input = [0 0 0 0 0 0 0 1 2 2 2 3 0 0 0 0 0 3 2 2 2 2 7 0 0 0 0 0 0 0];
    elseif nargin > 1
        error('Number of inputs is not supported.')
    else
        input = varargin{1};
    end

      % Append zeros to capture first and last number
      input = [0 input 0];

      % create a shifted vector
      input_1 = [input(2:end) 0];

      % find changes in input vector and create output
      numbers(:,1) = input_1( diff( abs( input~=0 ) )>0 );
      numbers(:,2) = input( diff( abs( input~=0 ) )<0 );
 end

Here, single values are counted twice: first as start of elements of interest and as end.

Kind regards,

Robert

[Stephen23]

Note that:

• varargin is not required: a simple variable should be used. Code hinting and code-completion are not useful with varargin. Compare varargin:

vs. a named input argument:

• Why bother to check the number of inputs (a bit pointless, see above), but not check what the inputs are: Type, size, range, ...?
• Shadowing the inbuilt input is not a good idea.
• Logical vectors only contain equivalent of zero and one, so abs is totally superfluous.
• Unnecessary duplication of data: input and input1 are almost the same. This is not good data design.

See my answer for much simpler code.

[Robert U]

True, that was sloppy.

Answer 3

[Andrei Bobrov]

a = [0 0 0 0 0 0 0 1 2 2 2 3 0 0 0 0 0 10 2 2 2 2 7 0 0 0 0 0 0 0];
b = a > 0;
out = [a(strfind(b,[0 1])+1);a(strfind(b,[1 0]))];

[Andrei Bobrov]

fixed