i want to quantize the matrix values in terms of 0s and 1s...plz help me

i've matrix of 3x3

a = [ 12 21 13; 32 45 67; 43 22 91 ]

my condition is

 if:

   a(i,j)<=a(i,j+1)             a(i,j)=0

 elseif  

   a(i,j)>a(i,j+1)          a(i,j)=1

i've written the code as:

 for i=1:length(a)
    for j=1:length(a)-1
        if(a(i,j)<=a(i,j+1))
            a(i,j)=1;
        else
            a(i,j)=0;
        end
    end
end
a

the output is:

 a =  1     0    13
      1     1    67
      0     1    91

now i want to retain only those columns which are in 0s and 1s i.e only column 1 and 2 i.e

a = 1 0 1 1 0 1

i know that a(:,1) gives me the only first column and a(:,2) gives me the second column.

but how to get both the columns like i said above

can anybody tell me how to do so programatically

plz help me

Answer 1

[Iain]

General answer:

 b = a(:,vector_of_columns_i_want_to_keep)

Specfic answers:

 b = a(:,[1 2])
 b = a(:,1:2)
 b = a(:,1:end-1)

[Raman]

thank u

Answer 2

[dpb]

>> a = [ 12 21 13; 32 45 67; 43 22 91 ]
a =
    12    21    13
    32    45    67
    43    22    91
>> for i=1:size(a,2)-1,a(:,i)=a(:,i)<a(:,i+1);end
>> a
a =
     1     0    13
     1     1    67
     0     1    91
>> bsxfun(@lt,a(:,2:3),a(:,1:2))
ans =
     1     0
     0     0
     0     0
>>

[dpb]

Sorry, forgot to reset a for second and also reversed the columns...

>> a = [ 12 21 13; 32 45 67; 43 22 91 ];
>> bsxfun(@lt,a(:,1:2),a(:,2:3))
ans =
     1     0
     1     1
     0     1
>>

NB for generality, instead of using hardcoded column indices, could write

bsxfun(@lt,a(:,1:end-1),a(:,2:end))

etc.

[Raman]

when i write this

bsxfun(@lt,a(:,1:end-1),a(:,2:end))

the output it produces is not correct. it gives me

     0     1
     0     1
     1     1

instead of giving:

    1      0
    1      1
    0      1

can u tell me how this can be done in order to correct it...

[dpb]

Sorry for the screwup first time--I had both reassigned a and inadvertently reversed the order of the rows in the test. Then to compound the problem just did a cut'n paste w/o paying attention --

So, clean start--

          >> a = [ 12 21 13; 32 45 67; 43 22 91 ]
          a =
              12    21    13
              32    45    67
              43    22    91
          >> bsxfun(@lt,a(:,1:end-1),a(:,2:end))
          ans =
               1     0
               1     1
               0     1

    This is the same as explicitly writing

      >> [a(:,1)<a(:,2) a(:,2)<a(:,3)]
      ans =
           1     0
           1     1
           0     1
      >>

hth...

--dpb