please i need help with Matrix manipulation

hello everyone, i need help on writing a script that does the following:

data=[1 1 2;
      2 1 3;
      3 3 4;
      3 3 5;
      4 3 6]

the script need to check if a number in the second column is repeated ,means its a layer , for example 1 2 and 1 3 will be layer 1 containing 2 and 3 etc... then the final result should be something like this

result=[1 0 0 0 0;
        2 3 0 0 0;
        0 0 4 5 6]

any suggestions?

thank you

[Walter Roberson]

The 3 4 is not the same length as the other rows so your data array cannot be constructed.

Your second column contains only 1s and 3s, so why do you have more than 2 rows of output? Why is the first row being output? Why does the last row start with 0 0 unlike the others that are filled from the left?

[sami elahj]

hey Robert thanx for the quick answer, 3 2 is the same length ive just edited it, the first doesn't really matter, and the fact that the last row starts with zeros its just a way to fill the empty values with zeros. i have attached a sketch to explain what i am really trying to get at

the real exercise its matrix extracted from a tree form data points it has like 40 points but its repeated once i figure how to do the first and the second layers the rest will be easy.

Thanx again

[KSSV]

Hi

result matrix is not bit clear. You want layers to be filled depending on its values? I mean there is layer 1 and 3. Do you want layer 2 to be completely zero? You need to elaborate result matrix.

[sami elahj]

hi Dr. Siva For each repeated value in the second column, I'd like to all values in the third column that share each repeated value in the second column as entries in the correct column. (from the pic)

1--->2
1--->3

so layer 1 will contain 2 and 3

3--->4
3--->5
3--->6

layer 2 will contain the 4 , 5 and 6

resulting in

[2 3 0 0 0; 
0 0 4 5 6]

or transposed

once layer is filled with the corresponding values the rest of the values should be zeros

Answer 1

[sami elahj]

it seems like this works

mx=max(data(:,3)); %// maximal node index
ly = zeros(mx,1); 
for ii=2:mx, 
    ly(ii) = ly( data( data(:,3)==ii, 2 ) )+1; %// get the layer of the parent
end
for ci=1:max(ly), 
    c{ci} = find(ly==ci); %// put all nodes of layer ci in a cell
end; 
mxc = max(cellfun(@numel,c));
res=zeros(mxc,max(ly));
for ci=1:numel(c), 
    res(1:numel(c{ci}),ci)=c{ci}(:);
end;

Answer 2

[the cyclist]

I think I kinda understand what you mean, but there is still too much guesswork here.

If your result were

result=[2 3 0 0 0;
        0 0 4 5 6]

then I would understand how you got it. But where does your first row

result=[1 0 0 0 0 ...

come from? How do we get that?

And why is the result not

result=[1 0 0 0 0 0;
        0 2 3 0 0 0;
        0 0 0 4 5 6]

with the 2nd row offset in the same way that the first one is?

Your one example does not adequately explain the rule/algorithm for going from data to result.

[sami elahj]

Thank you the cyclist, i apologize for not explaining accurately, in the example i ve attached comes a tree form data where from each ramification (root=the repeated number in the data matrix) comes out a number of lines and points. the first column [1 0 0 0 0 0] its a mistake of mine so it can be ignored, and the wanted result can rather the way i implemented it (as columns) or the way you suggested(as rows), en both way columns/rows to me they ll be meaning a layer. one more thing the resulting matrix dimensions are not important once the data is classified as asked. hoping i explained it better this time :)

Answer 3

[KSSV]

Hi sami elahj

You can follow the below code to find the repetitions, which gives you layers. Here I have generated random numbers to create data. You can replace data with your matrix.

N = 10 ;
a = 1 ; b = 20 ;
row1 = linspace(1,N,N)' ;
row2 = round(a + (b-a).*rand(N,1)) ;
row3 = round(a + (b-a).*rand(N,1)) ;
data = [row1 row2 row3] ;
% check for repetition 
eps = 1.e-5 ;
repeat = cell(N,1) ;
repeat_val = cell(N,1) ;
for i = 1:N
   diff_row2 = repmat(data(i,2),[N,1])-data(:,2) ;
   % Arrange the distances in ascending order
   [val, pos] = sort(abs(diff_row2)) ;
   % Pick the points which are repeated 
   repeat{i} = pos(val<=eps) ;
   repeat_val{i} = data(repeat{i},2) ;
end

repeat, repeat_val gives you the indices and values respectively. Coming about the result matrix. You have to clear few questions.

Regards

Answer 4

[KSSV]

Hope the attached file gives what you want. Replace matrix data as your data matrix. PS: In the present file I have taken random numbers in data matrix. Some times it may throw error, run the file, it will work.

[sami elahj]

hey Dr Siva thanx for the script, i have try it and this is what i got:

ans =

     1     2     3     0     0     0
     3     0     0     4     5     6

the problem that occurs is when trying to expand the data matrix (as putting different possibilities) for example

data = [1 1 2;
        2 1 3;
        3 2 4;
        4 3 5;
        5 3 6;
        6 4 7;
        7 5 8]

will be equivalent on the sketch as shown, so basically the problem when there is no repeated value in the second column,say the 2--->4 example, the script will need to check if number 2 belongs to the previous layer if so whatever numbers comes from it will belong to the next layer. i don't know if that make sense to you? thanx

Ps: this is the original system i am trying to work out, in school they recommended to use a simple loop to solve this but i don't seem to figure out the best logic as i was getting errors all the time