# need general coding/ command to separate matrix data

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Commented: dpb on 7 Sep 2022
I have a for loop y= 0:2:8 and whenever the for loop value chnaged it generates two values which was stored in a matrix. Now I have a complete matrix in which column one have y=0 then several values in a rows then y changed and several vlaues in a row as shown in fig.
final matrix is:
0 5 9
0 7 3.3
0 5.5 9
1 4.0 3.5
1 2 3.3
...................... so on.
Question: I need a general coding that store number of rows in a seperate matrix/ vector when y value is fixed. I mean when y=0 then all rows with y=0 stored in a seperate vector/ matrix. Then again y changed and results in several rows. These rows stored in a seperate vector/matrix.
Your support will be highly appreciated.
Regards
Now I send you my coding but formula is shorter my formula to simpler. So, when v exceute it give us matrix of three column. Important is value of first column is fixed upto seperate rows. I need all those values in a seperate matrix. reason is that for a fixed value of beta there are certian values in column 2 and 3. I need to draw graph using these values. Then again beta change and i have certian set of values and i need again draw graph using "hold on" command.
clear all
c1= 0.4;
v=[];
betavalues= 0:2:4;
a= zeros(length(betavalues),1);
for i = 1:length(betavalues)
a(i)= betavalues(i);
beta=a(i)
lamdavalues= 1:1:10;
b= zeros(length(lamdavalues),1);
for j=1:length(lamdavalues)
b(j)= lamdavalues(j);
lamda=b(j);
x1= beta*lamdavalues*c1;
v(end+1,:)=[beta lamda x1];
end
end

Bruno Luong on 7 Sep 2022
Edited: Bruno Luong on 7 Sep 2022
Try this:
% Fake data
finalmatrix=sortrows(randi(10,30,3),1)
finalmatrix = 30×3
1 7 1 2 6 10 2 7 9 3 10 9 3 7 1 3 3 5 3 2 4 4 4 5 4 8 8 4 6 2
y = finalmatrix(:,1);
n=diff(find([true; diff(y)~=0; true]))
n = 10×1
1 2 4 3 1 5 3 5 4 2
splitmatrix=mat2cell(finalmatrix,n,size(finalmatrix,2));
splitmatrix{:}
ans = 1×3
1 7 1
ans = 2×3
2 6 10 2 7 9
ans = 4×3
3 10 9 3 7 1 3 3 5 3 2 4
ans = 3×3
4 4 5 4 8 8 4 6 2
ans = 1×3
5 2 1
ans = 5×3
6 2 2 6 10 1 6 1 6 6 6 2 6 5 5
ans = 3×3
7 3 9 7 3 7 7 4 8
ans = 5×3
8 9 7 8 8 2 8 3 3 8 2 5 8 9 8
ans = 4×3
9 6 6 9 9 7 9 4 7 9 1 6
ans = 2×3
10 8 7 10 6 3
dpb on 7 Sep 2022
I got totally lost in the explanation above, but if @Bruno Luong's solution for splitmatrix is the right data, then
figure
hold on
cellfun(@(c)plot(c(:,2),c(:,3)),splitmatrix)
should be close to what you're looking for.
As always, "salt to suit"...