average distance between column values in a matrix

I have a 16 row, 20,000 column matrix of 0's and 1's. 1's represent activity in each row, 0's represent a lack there of. I want to find the average distance (# of columns) between points of activity in the rows and then have an overall average of all 16 rows. So if, for example, a row had 01110001110, the distance would be 6, (columnar distance between centers of groups of random consecutive ones (or single ones). Any ideas?

Thank you

[Brandon]

anybody?

[Jan]

Perhaps you could start to explain, what you have tried so far.

[Brandon]

I don't know where to begin on even these simple issues. I'm brand new to coding and learning as I go along.

[Brandon]

I just get the pros input and then go read up on whatever functions you suggest (regexp, etc.)

Answer 1

[per isakson]

See the answer, How to convert 0's and 1's to intergers, and then try:

    cac = regexp( '000001111111111000000000011111', '(0+)|(1+)', 'match' );
    num = cellfun( @numel, cac );

This is more than half way to a solution.

--- CONT. ---

Given that Y is a character array, .<16x20000 char>, try first with one row

    cac = regexp( Y(1,:), '(0+)|(1+)', 'match');

if that works I guess the simplest (:KISS:) is to use a loop and operate on one row at a time.

--- CONT. ---

I put these lines in the editor and select the cell (yellow background) and click Evaluate cell ( or Cntrl+Enter)

  cac = regexp( '000001111111111000000000011111', '(0+)|(1+)', 'match' );
  num = cellfun( @numel, cac );
  disp( num )

that gives me the following line in the command window

     5    10    10     5

I have R2012a but that shouldn't matter. Why not try

    Y = repmat( '11100011111111111110000000000011111111110000000', 6, 1 );
    num = cell( 6, 1 );
    for rr = 1 : 6
        cac = regexp( Y(rr,:), '(0+)|(1+)', 'match');
        num{rr} = cellfun( @numel, cac );
    end
    num{:}

that produces in the command window

ans =
     3     3    13    11    10     7
ans =
     3     3    13    11    10     7
ans =
     3     3    13    11    10     7
ans =
     3     3    13    11    10     7
ans =
     3     3    13    11    10     7
ans =
     3     3    13    11    10     7

[Brandon]

Thank you

[Brandon]

So for my matrix (Y), i put cac=regexp('Y', '(0+)|(1+)', 'match');

This is returning a 0x0 cell, is that suggesting that within Y there are no 0's which turn into 1's in subsequent columns? ...Sorry for being so terrible at this coding stuff, I'm new!

[per isakson]

The major problem is the blips around Y. You must distinguish between the name of the variable and the value of the variable.

[Brandon]

I tried Y(1,:) already, it tells me "??? Undefined function or method 'regexp' for input arguments of type

'double'."

[Brandon]

I tried copy and pasting the contents of the matrix into the regexp line, it gave me a 1x80,000 matrix. It didn't separate it out like with the '000001111111111000000000011111' example for some reason..

[per isakson]

If it returns anything you must be aware that Matlab works column wise.

Answer 2

[Andrei Bobrov]

use function regionprops from Image Processing Toolbox

x =[...
1     0     1     1     1     1     0     0     0     1     1     1     0     0     0     1     1];

s = regionprops(x > 0,'Centroid' );
out0 = cat(1,s.Centroid);
out = diff(out0(:,1));

OR:

ii = find([true;diff(x(:))~=0]);
i2 = [ii,[ii(2:end)-1;numel(x)]];
out = diff(mean(i2(x(ii)>0,:),2));