How can I take the last value and sum the values of a row of ones in a matrix?

Hi!

I have the following (large) dataset

figure 1

and I would like to produce the following output

figure 2

In practice, when there are more than 1 zeros in column 1 (see rows 17:19, figure 1) I would like to replace in row 17 (see figure 2)

• the value of 0 in row 19 and column 1 of figure 1
• the value of 250 in row 20 and column 2 of figure 1
• the sum of the values in form row 17 to row 19 and column 3 of figure 1
• the value of -1 in row 19 and column 4 of figure 1

and I want to do this every time when there are more than 1 consecutive zeros in column 1.

To clarify, I would replace row 17 of figure 1 a row composed by the following values: [0 (row 19, col 1); 250 (row 20, col 2); the sum of the values from row 17 to row 19 of col 3; -1 (row 19, col 4)] and repeat this every time there are more than one zeros in column 1.

Thanks in advance for the help!

[Dyuman Joshi]

"the value of 0 in row 20 and column 1 of figure 1"

There is no 0 in row 20, col 1 of figure 1.

"and repeat for each row of the matrix."

Just to clarify, you want to do this every time when there are more than 1 consecutive zeros in column 1?

And the data in the 2nd column should be shifted up by 1 place every time? Then what should be the values in the spaces left at the bottom in 2nd column?

[Image Analyst]

What are you doing with row 20??? It lookos like you just want to replace row 17 with things and then delete rows 18 and 19. Is that correct?

If you have the Image Processing Toolbox you can easily find the rows with 2 or more consecutive zeros in column 1 using bwlabel and regionprops.

[Francesco]

Hi, sorry for the typos. Now it is "the value of 0 in row 20 and column 1 of figure 1" and "the value of -1 in row 20 and column 4 of figure 1".

"Just to clarify, you want to do this every time when there are more than 1 consecutive zeros in column 1?"

Yes, exactly.

"And the data in the 2nd column should be shifted up by 1 place every time? Then what should be the values in the spaces left at the bottom in 2nd column?"

I would replace row 17 of figure 1 a row composed by the following values: [0 (row 19, col 1); 250 (row 20, col 2); the sum of the values from row 17 to row 19 of col 3; -1 (row 19, col 4)] and repeat this every time there are more than one zeros in column 1.

Thank you.

Answer 1

[Sufiyan]

data = rand(11, 3);
data(:, 1) = [1; 0; 0; 0; 2; 0; 0; 0; 0; 3; 0]; 
sampleTable = table(data(:, 1), data(:, 2), data(:, 3), 'VariableNames', {'Column1', 'Column2', 'Column3'})

sampleTable = 11×3 table
    Column1    Column2     Column3
    _______    ________    _______

       1        0.15411    0.46297
       0        0.98663    0.92216
       0         0.6106    0.25468
       0       0.016536    0.94741
       2        0.50032     0.8343
       0       0.067772    0.21512
       0         0.3922    0.12722
       0        0.86009    0.58253
       0        0.73474    0.69901
       3         0.9907     0.6209
       0        0.42106    0.49568

consecutiveZeros = diff([0; sampleTable.Column1 == 0; 0]);
startIndices = find(consecutiveZeros == 1)

startIndices = 3×1

2 6 11

endIndices = find(consecutiveZeros == -1) - 1

endIndices = 3×1

4 9 11

Now you know the start and end indices, you can do any operations (like sum of col 3 from starting index to ending index) and modify the values and at the end delete the useless rows.

[Francesco]

Thank you very much!