Thread Subject: working with matrix

a=1 2 3 4
    3 4 5 7
    2 4 5 4
b=[4 4 4 8 8 6 6 9]

c=[1 3 5 6 7 2 8 4]
 here in b the same elements are grouped.. and c denotes the indices of the repeating elements in the original matrix
d=[4 6 4 9 4 8 8 6]
 is there any method to take the elements of c as indices and substitute the values of a in that place ie e= 1 3 1 4 1 2 2 3
                                                              2 5 2 7 2 4 4 5
                                                              3 5 3 4 3 4 4 5

"anil chandy" <anilvchandy@gmail.com> wrote in message <gnbbh6$946$1@fred.mathworks.com>...
> a=1 2 3 4
> 3 4 5 7
> 2 4 5 4
> b=[4 4 4 8 8 6 6 9]
>
> c=[1 3 5 6 7 2 8 4]
> here in b the same elements are grouped.. and c denotes the indices of the repeating elements in the original matrix
> d=[4 6 4 9 4 8 8 6]
> is there any method to take the elements of c as indices and substitute the values of a in that place ie e= 1 3 1 4 1 2 2 3
> 2 5 2 7 2 4 4 5
> 3 5 3 4 3 4 4 5

"anil chandy"
> a=1 2 3 4
> 3 4 5 7
> 2 4 5 4
> b=[4 4 4 8 8 6 6 9]
> c=[1 3 5 6 7 2 8 4]
> here in b the same elements are grouped.. and c denotes the indices of the repeating elements in the original matrix
> d=[4 6 4 9 4 8 8 6]
> is there any method to take the elements of c as indices and substitute the values of a in that place ie e= 1 3 1 4 1 2 2 3
> 2 5 2 7 2 4 4 5
> 3 5 3 4 3 4 4 5

your example is not clear...

in general, thought, creating an array based on a
- template of values -> your a
- a list of indices -> your c
can be achieved by

% the list of indices into...
     lst=[
           3,2,1,16
          16,16,16,16
     ];
% ...a template of values
     v=magic(4)
%{
% v =
    16 2 3 13
      5 11 10 8
      9 7 6 12
      4 14 15 1
%}
     r=v(lst)
%{
% r =
     9 5 16 1 % <- note: same size as LST
     1 1 1 1
%}

us

"us"
> in general, thought, creating an array based on a

although i gave some thought to this problem, this should have read
..., though,...

just a thought...
us

"anil chandy" <anilvchandy@gmail.com> wrote in message <gnbbh6$946$1@fred.mathworks.com>...
> a=1 2 3 4
> 3 4 5 7
> 2 4 5 4
> b=[4 4 4 8 8 6 6 9]
>
> c=[1 3 5 6 7 2 8 4]
> here in b the same elements are grouped.. and c denotes the indices of the repeating elements in the original matrix
> d=[4 6 4 9 4 8 8 6]
> is there any method to take the elements of c as indices and substitute the values of a in that place ie e= 1 3 1 4 1 2 2 3
> 2 5 2 7 2 4 4 5
> 3 5 3 4 3 4 4 5

  It is difficult to understand what you are saying, Anil. You explanation is very unclear. However, I found a way of producing arrays d and e from a, b, and c, provided one correction is made to e. Is this the kind of thing you wanted to do? I see no way of deducing either b or c from a. The following depends on c being a valid permutation of 1:8.

 d = zeros(size(c));
 d(c) = b;
 p = cumsum([1,diff(b)~=0]);
 p(c) = p;
 e = a(:,p);

The d and e produced are:

 d = [4 6 4 9 4 8 8 6]

 e = [1 3 1 4 1 2 2 3
      3 5 3 7 3 4 4 5
      2 5 2 4 2 4 4 5]

You seem to have the 2's and 3's in the second two rows of e interchanged. Is that a mistake?

Roger Stafford

its a mistake.. wat i actually meant was i ve a matrix A=[ 4 6 4 9 4 8 8 6 ]
and another matrix B=[4 4 4 6 6 8 8 9] which is formed by grouping the similar terms in D .. and another matrix c=[1 3 5 2 8 6 7 4] which denotes the indices of 4 6 8 and 9 in the matrix D..
i got one more matrix [ 1 2 3 4
                                  3 4 5 7
                                  2 4 5 4
 now i want to put the first coulmn ie 1 in the positions 1 3 5 in a new matrix
                                                     3 then 2nd coulmn in the position 2
                                                     2 and 8 3rd coulmn at 6 7 ,final one at 4
 desired output
       1 2 1 4 1 3 3 2
       3 4 3 7 3 5 5 4
       2 4 2 4 2 5 5 4

there is some problem when what i typed gets displayed in the forum...

"its a mistake.. wat i actually meant was i ve a matrix A=[ 4 6 4 9 4 8 8 6 ]
and another matrix B=[4 4 4 6 6 8 8 9] which is formed by grouping the similar terms in D .. and another matrix c=[1 3 5 2 8 6 7 4] which denotes the indices of 4 6 8 and 9 in the matrix D..
i got one more matrix 1 2 3 4
                                   3 4 5 7
                                   2 4 5 4
 now i want to put the first coulmn ie 1 in the positions 1 3 5 in a new matrix
                                                     3 then 2nd coulmn in the position 2
                                                     2 and 8 3rd coulmn at 6 7 ,final one at 4
 
desired output
       1 2 1 4 1 3 3 2
       3 4 3 7 3 5 5 4
       2 4 2 4 2 5 5 4

A=[ 4 6 4 9 4 8 8 6 ]
[B, C] = sort(A);
idx = cumsum(logical([1 diff(B)]));
T = [ 1 2 3 4;3 4 5 7;2 4 5 4];

output = T(:,accumarray(C',idx',[],@(x) x))



'nu lbb@aoh.ammpo eY avoo n khgaa'eau oyptt iMEca:eeochtil

Another approach calls sort instead of accumarray.

[B, C] = sort(A);
idx = cumsum(logical([1 diff(B)]));
[idx2,idx2] = sort(C)
T(:,idx(idx2))




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idx = cumsum(logical([1 diff(B)]));
i didnt undestand this line..

and why ur using the variable x?

"anil chandy" <anilvchandy@gmail.com> wrote in message <gnetra$327
> and why ur using the variable x?

The @(x) x is an anonymous function that simply returns what it is fed. That is the way accumarray works for this type of problem.
To understand the other line, take it apart and look at it's components (a technique which will serve you well with Matlab).



idx = cumsum(logical([1 diff(B)]));

can be expanded to:

tmp1 = diff(B) % Look at the help for diff if needed.
tmp2 = [1 tmp1]
tmp3 = logical(tmp2) % Look at the help for logical if needed.
idx = cumsum(tmp3) % Look at the help for cumsum if needed.