how to compute blocks of ones in a row

Dear All,

I have this matrix.

P=[1   0   1   1   0   0   1   1   1
   0   0   1   0   1   1   0   0   0];

I need to accumulate the ones numbers but as blocks of ones with the length of the block, here I have in the first row 3 blocks 1 , 1 1 , 1 1 1 , so the sum is 1+1+1 for the blocks and 2 for the length 1 1 and 3 for 1 1 1 so

sum=1+1+1+2+3

in the second row

sum=1+1+2

I wrote this

ind =strfind([0, P(1,:)],[0 1])
or
ind =find (diff([0 P(1,:)])==1)

it gives me the number of blocks but not the length of every block. I need to write the formula but is rather confuse. could anybody help me please

regards,

Imola

[Sean de Wolski]

I don't understand how you got the 2,3. Wouldn't it be 1+1+1, and 1+2+3? Or do you not count blocks with only one 1?

[imola]

Dear Sean,

I regards every block of ones as 1, then account the length of the block, see my code, it gives me the index of the blocks if I account them, I got 3 blocks for first row and 2 for second row, but how to find the length for every block.

thanks for answering me.

regards,

[Guillaume]

You've not really answered Sean question.

In your example, for the first row you've got three blocks (so first part of sum is 3*1). Three block would seem to imply three lengths: [1 2 3]. Hence we would expect the complete sum to be 3*1 + 1 + 2 + 3. Why are you not accounting for the blocks of length 1?

[Guillaume]

And by the way, don't call your variable sum

Answer 1

[Guillaume]

A simple way to get the lengths and number of blocks:

P = [1   0   1   1   0   0   1   1   1];

transitions = diff([0 P 0]); 
%transitions is 1 when going from 0 to 1 and -1 when going from 1 to 0.
%because 0 is added before and after, the first transition is always 0 to 1 and last always 1 to 0
%so there's always the same numbers of 1 and -1 in the diff
numblocks = sum(transitions == 1)  %or sum(transitions == -1)
blocklengths = find(transitions == -1) - find(transitions == 1)

To operate on a whole matrix at once is a bit trickier:

P = [1   0   1   1   0   0   1   1   1
     0   0   1   0   1   1   0   0   0];

transitions = diff([zeros(size(P, 1), 1) P zeros(size(P, 1), 1)], [], 2);
[starts, rows] = find(transitions' == 1);
[ends, ~] = find(transitions' == -1);
blocklengths = ends - starts;
sumlengthsbyrow = accumarray(rows, blocklengths, [size(P, 1) 1])
numblocksbyrow = accumarray(rows, 1, [size(P, 1) 1])
totalsumbyrow = sumlengthsbyrow + numblocksbyrow

If you don't want to include the blocks of length 1 in sumlengthsbyrow:

sumlengthsbyrow = accumarray(rows(blocklengths > 1), blocklengths(blocklengths > 1), [size(P, 1) 1])

[imola]

Dear Guillaume,

Thank you for the useful answer. that's great. But if I don't want to account the single 1 length, and I need to set it to zero such that the row [1 0 0 1 1] will equal 4 not 5 and [1 0 1 1 0 0 1 1 1] will equal 8

I mean the result should be

sumlengthsbyrow =[5 ; 2]
numblocksbyrow =[3;2]
totalsumbyrow =[8;4]

thanks

[Guillaume]

As I wrote at the end of my answer, if you don't want to include the lengths of one, it's:

sumlengthsbyrow = accumarray(rows(blocklengths > 1), blocklengths(blocklengths > 1), [size(P, 1) 1])
numblocksbyrow = accumarray(rows, 1, [size(P, 1) 1])
totalsumbyrow = sumlengthsbyrow + numblocksbyrow

[imola]

Dear Guillaume,

Thanks a lot. I really appreciate your help.

regards