find 1s in the Matrix

Hi All,

I have a matrix 31996x66 and i need to write a if/else and for loop to do following.

the for loop must got through column 48 to 65 and anywhere the number is 1 it goes 36 cells back ()to the same row and save the number and if zero doesnt do anything.

for example:

if in row 500 coulumn 50 number is 1 then (50-36=14) it should go to row 500 column 14 and take the number and save in new table.

At the end the new Matrix should be 31996x(number of values for 1 in each row.)

[Matt J]

and i need to write a if/else and for loop to do following.

Surely this is an assignment if you are so constrained in how the solution is to be implemented. And if itis assignment, surely you do not want us to do it for you.

[Matin jaberi]

Surly im electrical engineer and this is part of data I received from client and im trying save some time by automating them. i finished studing while ago kido.

[Walter Roberson]

If your client is imposing a requirement that you use a particular iteration structure, then we would need to know what other restrictions they are imposing in order to advise you fully.

[Walter Roberson]

Provided that there are no nan or infinities, there is a fairly simple vectorized expression for this calculation. If there are nan or infinities then a fix-up statement would need to be added.

Is the requirment truly that it must be done with a loop, or is the requirement just that it be done?

[Matin jaberi]

It has got nothing to do with my clients requirements. its me wanting to automate the excel sheet so the matlab run and find the numbers for me rather than I go through excel sheet myself. And there is no othere structure.

[Matin jaberi]

there might be many different ways to find the numbers I need but thats the simplest script i came up with to get what I need.

Answer 1

[Walter Roberson]

firstcol = 48;
lastcol = 65;
offsetcol = 14;
idx = firstcol:lastcol;
offsetidx = idx-offsetcol;
result = YourMatrix(:,offsetidx) .* (YourMatrix(:,idx) == 1);

at the end the new Matrix should be 31996x(number of values for 1 in each row.)

You cannot do that; there could be a different number of 1's in each row and numeric matrices cannot have a different number of columns in each row.

The above code outputs an array the size of the subset, with 0 for the entries where the condition was not met.

I notice, by the way, that 62-14 = 48, so columns 48, 49, 50, 51 both act as numeric sources (values to be fetched) and as control information about which values are to be fetched. Is that overlap desired?

[Matin jaberi]

Thanks for your answer, the offset call is 36 not 14. As i mentioned before if column number is 50 then take 36 out and the desired col would be col 14.

[Walter Roberson]

firstcol = 48;
lastcol = 65;
offsetcol = 36;
idx = firstcol:lastcol;
offsetidx = idx-offsetcol;
result = YourMatrix(:,offsetidx) .* (YourMatrix(:,idx) == 1);

Answer 2

[Matin jaberi]

so the box highlighted yellow is col 48 where number ==1 and the green one is column 14 is the same row. this should apply only when number is ==1