Matrix manipulation / if else loops

Hi,

I would like to scan through each row of my data matrix (made of 0s and 1s only) and create a new matrix following the rules below:

• If all columns are 0 - assign 0s to new matrix
• If all columns are 1s - assign 1 to column 2 and 0s to column 1 and 3 of new matrix
• If two columns = 1, then:
• scan up to rows above until you reach a 0 in one of the two columns.
• assign a 1 to the column in the new matrix that was on 0 most recently.
• If both were on 0 equally recently, then assign 1 to both columns.
• (Can assume that first row will be 0 0 0)

This is the code I have so far, but I am struggling on the last section (if two columns both = 1)... any help would be most appreciated. Thank you!

matrix = [0 0 0; 1 0 0; 1 1 0; 1 0 0; 1 0 0; 1 1 1; 0 1 1; 0 1 1; 1 1 1; 1 1 0; 1 1 0; 1 0 0; 1 1 0; 1 0 0]; % data matrix

A = zeros(size(matrix)); % new matrix of 0s to ammend, same size as data matrix

for i = 1:length(matrix(:,1)) % iterate through every row in the data matrix

if sum(matrix(i,:))==0; % if all the columns = 0, assign all columns in new matrix to 0

A(i,:)=0;

elseif sum(matrix(i,:))==3; % if all the columns = 1, assign atrium column in new matrix to 1 and the others to 0

A(i,2)=1;

elseif sum(matrix(i,:))==1; % if only 1 column = 1, assign this column in new matrix to 1 and the others to 0

idx = find(matrix(i,:)); % (find which column it is that =1)

A(i,idx) = 1; % (assign corresponding column in new matrix to 1)

else sum(matrix(i,:))==2; % if two columns =1

idx = find(matrix(i,:)); % find the index of the nonzero elements in this row

idx_1 = (matrix(i-1,idx(1))); % find the values of the row above elements in these columns

idx_2 = (matrix(i-1,idx(2)));

if idx_1 ==0 % if the element in the row above = 0, then assign a 1 to the corresponding column in row i

A(i,idx(1))=1;

if idx_2 ==0 % if the element in the row above = 0, then assign a 1 to the corresponding column in row i

A(i,idx(2))=1;

end

end

end

end

Answer 1

[David Hill]

Misunderstood you previously.

m = [0 0 0; 1 0 0; 1 1 0; 1 0 0; 1 0 0; 1 1 1; 0 1 1; 0 1 1; 1 1 1; 1 1 0; 1 1 0; 1 0 0; 1 1 0; 1 0 0];
s=sum(m,2);
M=zeros(size(m));
M(s==3,2)=1;
M(s==1,:)=m(s==1,:);
f=find(s==2);
for k=1:length(f)
   F=find(m(f(k),:));
   t=M(1:f(k)-1,F);
   f1=find(~t(:,1),1,'last');
   f2=find(~t(:,2),1,'last');
   if f2>f1
       M(f(k),F(2))=1;
   elseif f2==f1
       M(f(k),F)=1;
   else
       M(f(k),F(1))=1;
   end
end
M

M = 14×3

0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 0 1 0 0

[Laura Steel]

Hi David,

Thanks for your answer, however, unfortunatey it doesn't work as intended, as it outputs the following matrix:

[0 0 0

1 0 0

1 0 0

1 0 0

1 0 0

0 1 0

0 1 0

0 1 0

0 1 0

0 1 0

0 1 0

1 0 0

1 0 0

1 0 0]

My current code works fine for the first two points of my original aim, but it is just the following section I am struggling to write...

• If two columns both = 1, then:
• scan up to rows above until you reach a 0 in one of the two columns.
• assign a 1 to the column in the new matrix that was on 0 most recently and a 0 to the other column.
• If both were on 0 equally recently, then assign 1 to both columns.

i.e. [0 0 1; 0 1 1] should go to [0 0 1; 0 1 0]

or, i.e. [0 0 1; 0 1 1; 0 1 1] should go to [0 0 1; 0 1 0; 0 1 0]

If you had any ideas as to how to code this specific section that would be very useful. Thanks :)

[Voss]

Replace

t=M(1:f(k)-1,F);
f1=find(t(:,1),1,'last');
f2=find(t(:,2),1,'last');

with

t=m(1:f(k)-1,F);
f1=find(t(:,1)==0,1,'last');
f2=find(t(:,2)==0,1,'last');

[Laura Steel]

Brilliant, thank you both very much!