Vector problem

Hi,

I've got a problem, and I hope that someone here can help me. Consider the following vector 'v'.

0 0

1 0

1 0

1 1

1 0

I have to create a third column to this vector, and the value of that column has to depend on the combination of column 1 and 2. If the combination is (1,1), like in the fourth row, the value of the third column has to be 1. If the combination is (0,0), like in the first row, the value third column has to be 0.

The problem arises when the combination is (1,0). In this situation, we have to look at the row before a consecutive series of (1,0). If before the (1,0)series, the combination is (0,0), the value of the third column of the (1,0) combination has to be 0. If before the (1,0)series the combination is (1,1), the value has to be 1.

The outcome should be the following:

0 0 0

1 0 0

1 0 0

1 1 1

1 0 1

Does anyone know how I can program this in Matlab? Thanks,

Pieter

[Matt Fig]

BTW, and for future reference, a N-by-2 variable is not a vector unless N is 1. You have got yourself a matrix, or more generally, an array. A vector has all of its dimension sizes equal to 1 except one dimension.

Answer 1

[Jan]

Very similar to Paulo's implementation, but faster on my computer:

a = [0 0; 1 0; 1 0; 1 1; 1 0];

function c = AddColumn(a)
n    = size(a, 1);
b    = zeros(n, 1);
e    = a(:, 1) == a(:, 2);  % Equal elements
b(e) = a(e, 1);

q = 0;  % What is wanted if first row is [1, 0]?
for i = 1:n
   if e(i)
      q = b(i);
   else
      b(i) = q;
   end
end

c = [a, b];

And a vectorized approach:

a = [0 0; 1 0 ; 1 0 ; 1 1 ; 1 0];

function c = AddColumn(a)
n     = size(a, 1);
aEq   = a(:, 1) == a(:, 2);  % Equal elements
aFull = a(aEq, 1);
c     = [a, aFull(cumsum(aEq)];

[Paulo Silva]

That vectorized approach is awesome, I had to spend a couple of minutes to learn how it worked so fine :)

[Paulo Silva]

+1 vote

Answer 2

[Paulo Silva]

a=[ 0 0; 1 0 ; 1 0 ; 1 1 ; 1 0] %inputs
c=zeros(size(a,1),3); %pre-allocate table
c(:,1:2)=a; %put the inputs on the table
c(1,3)=a(1,1) & a(1,2); %first value doesnt depend on the last one
%the other values depend on the last output
for b=2:size(a,1) %for every line starting on the second
c(b,3)=(a(b,1) & a(b,2)) | c(b-1,3); %compute the rest of the values
end
c; %your truth table

Just for my own record here's my version that works

%a=[ 1 1; 1 0 ; 1 0 ; 1 0 ; 1 0] %inputs
a=[ 0 0; 1 0 ; 1 0 ; 1 1 ; 1 0] %inputs
%a=randi([0 1],2870,2); %user situation
%tic
c=zeros(size(a,1),3); %pre-allocate table
c(:,1:2)=a; %put the inputs on the table
b=1:size(a,1);
c(b,3)=a(b,1) & a(b,2); %find all 1 1 situations and give output 1
%the other values depend on the last output
%find those special 1 0 situations
for b=2:size(a,1) 
d=0;
%find if special situation occurs and find the result before consecutive
%series
while ((a(b-d,1)==1) & (a(b-d,2)==0) & ((b-d)>1))
d=d+1;
end
c(b,3)=c(b-d,3); %compute the rest of the values
end
c %your truth table
%toc %shouldnt take long

[Pieter]

Thanks for the answer, but I'm afraid my problem isn't solved yet.

The example I've given is just a short sample of my total dataset, I'm sorry I didn't mention that before. My dataset contains 2870 rows and 2 colums, so I need a code which calculates everything automatically.

Pieter

[Paulo Silva]

and the problem now is what? you have the dataset, use it instead of the variable a I gave in the code! the rest of the code can remain the same.

[Sean de Wolski]

Paulo's example is generalized and should work for any array you have of size nx2 called 'a'.

[Paulo Silva]

I checked the code, it doesn't work as required, my mistake, Jan's answer is working just fine.

Answer 3

[Wolfgang Schellenberger]

Why not c=abs(a(:,1)-a(:,2)) ?

[Paulo Silva]

it gives the wrong answer for the first two inputs 1 0 and doesn't take into account the last output value.

[Sean de Wolski]

-1 Vote!

|1-1| = 0; which yields the wrong result. It also doesn't account for previous rows.

Answer 4

[Pieter]

It works! Thank you very much for the help and the patience. I just started working in Matlab this week, so I'm still a beginner.