Giving a Different Value to Elements of Array

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Gözde Üstün 2020-7-8

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编辑： Stephen23 2020-7-14

Hello,

I have a value for A(:,:,3,1) and another value for A(:,:,3,2). Similarly, I have different values for A(:,:,4,1) and A(:,:,4,2) But in this for loop I cant give values to elements.

How can I do that?

for i=3:basis_number
    for k =1:2
        A(:,:,i,k) = repmat(blkdiag(temp,temp2),d/4,d/4);
        A(:,:,i,k) = repmat(blkdiag(temp2,temp),d/4,d/4);
        A(:,:,i,k) = repmat(blkdiag(temp3,temp4),d/4,d/4);
        A(:,:,i,k) = repmat(blkdiag(temp4,temp3),d/4,d/4);
    end
end

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Mahesh Taparia 2020-7-13

Hi

Can you elaborate more, what you want to do and what error you are getting?

Gözde Üstün 2020-7-13

Here I want to do that:

A(:,:,3,1) = repmat(blkdiag(temp,temp2),d/4,d/4);

A(:,:,3,2) = repmat(blkdiag(temp2,temp),d/4,d/4);

A(:,:,4,1) = repmat(blkdiag(temp3,temp4),d/4,d/4);

A(:,:,4,2) = repmat(blkdiag(temp4,temp3),d/4,d/4);

But instead of writing by one by, I wanted to use for loop

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Answer 1

Stephen23 2020-7-13

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编辑：Stephen23 2020-7-14

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Using numbered variables is a sign that you are doing something wrong.

In this case, using indexing would allow you to use a loop, i.e. instead of this

A(:,:,3,1) = repmat(blkdiag(temp1,temp2),d/4,d/4);
A(:,:,3,2) = repmat(blkdiag(temp2,temp1),d/4,d/4);
A(:,:,4,1) = repmat(blkdiag(temp3,temp4),d/4,d/4);
A(:,:,4,2) = repmat(blkdiag(temp4,temp3),d/4,d/4);

you can do something like this (of course you should actually just create the cell array C right from the start using basic indexing and NOT create variables with numbers in the their names):

V = 3:basis_num;
C = {[],[];[],[];temp1,temp2;temp3,temp4}; % size = basis_num * 2
A = preallocate to correct size
for ii = 3:basis_num
    x = C{ii,1};
    y = C{ii,2};
    A(:,:,ii,1) = repmat(blkdiag(x,y),d/4,d/4);
    A(:,:,ii,2) = repmat(blkdiag(y,x),d/4,d/4);
end

EDIT: based on the comments follwing this question:

n = d/2;
r = pow2(n);
C = arrayfun(@(~)rand(2,2),1:r/2,'uni',0);
% generate indices:
X = [1+eye(n);4-eye(n)];
X = fliplr([X;2+X]);
% generate output array:
%A = zeros(d,d,r,2); % preallocate!
k = 0;
for ii = 3:basis_number
    for jj = 1:2
        k = k+1;
        A(:,:,ii,jj) = blkdiag(C{X(k,:)});
    end
end

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Hello, Thank you very much for your answer it worked :) Can I ask some question?

1)Why did not you use V anywehere?

2)A = preallocate to correct size

what is this sentence I didnt understand anything from this sentence I just close it

3)I wrote code like that:

function [A,B] = trying(d,basis_number)
%A=zeros(d,d,basis_number,d);    
%A=zeros(d,d,2,2);
A=zeros(d,d,basis_number,2);    
B=A;
...
...    Here there are something for basis number smaller than 3
if basis_number>2 && d==4
    V = 3:basis_number;
    C = {[],[];[],[];temp,temp2;temp3,temp4}; % size = basis_num * 2
    C2= {[],[];[],[];temp5,temp6;temp7,temp8};
%    A = preallocate to correct size
    for ii = 3:basis_number
        x = C{ii,1};
        y = C{ii,2};
        x2 = C2{ii,1};
        y2 = C2{ii,2};
        A(:,:,ii,1) = repmat(blkdiag(x,y),d/4,d/4);
        A(:,:,ii,2) = repmat(blkdiag(y,x),d/4,d/4);
        B(:,:,ii,1) = repmat(blkdiag(x2,y2),d/4,d/4);
        B(:,:,ii,2) = repmat(blkdiag(y2,x2),d/4,d/4);
    end
   elseif basis_number>2 && d==6
       D = {[],[];[],[];temp,temp2;temp3,temp4}; % size = basis_num * 2
       D2= {[],[];[],[];temp5,temp6;temp7,temp8};
       for ii = 3:basis_number
        x3 = D{ii,1};
        y3 = D{ii,2};
        x4 = D2{ii,1};
        y4 = D2{ii,2};
        A(:,:,ii,1) = repmat(blkdiag(x3,y3,x3),d/6,d/6);
        A(:,:,ii,2) = repmat(blkdiag(x3,x3,y3),d/6,d/6);
        B(:,:,ii,1) = repmat(blkdiag(x4,y4,y4),d/6,d/6);
        B(:,:,ii,2) = repmat(blkdiag(y4,x4,x4),d/6,d/6);
       end
   end    

But as you can see it seems too much. And If I have 6 d then the permutations should go till the 8 basis count and this time I should have:

A(:,:,3,1) = repmat(blkdiag(x3,x3,y3),d/6,d/6);
A(:,:,3,2) = repmat(blkdiag(x3,y3,x3),d/6,d/6);
A(:,:,4,1) = repmat(blkdiag(y3,x3,x3),d/6,d/6);
A(:,:,4,2) = repmat(blkdiag(y3,y3,x3),d/6,d/6);
A(:,:,5,1) = repmat(blkdiag(y3,x3,y3),d/6,d/6);
A(:,:,5,2) = repmat(blkdiag(x3,y3,y3),d/6,d/6);
A(:,:,6,1) = repmat(blkdiag(x4,y4,x4),d/6,d/6);
.
.
.

And I cannot go more than 4 basis even if I change the size of C

Best regards

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1) I already closed the sentence V thanks

2)ok I understood

3) Sorry but no I didnt understand. My aim is to not to use dimension in if statement but now it is impossible

If I dont use this if

elseif basis_number>2 && d==6

the program cannot understand which dimension I am studying so I added this block

     elseif basis_number>2 && d==6
       D = {[],[];[],[];temp,temp2;temp3,temp4}; % size = basis_num * 2
       D2= {[],[];[],[];temp5,temp6;temp7,temp8};
       for ii = 3:basis_number
        x3 = D{ii,1};
        y3 = D{ii,2};
        x4 = D2{ii,1};
        y4 = D2{ii,2};
        A(:,:,ii,1) = repmat(blkdiag(x3,y3,x3),d/6,d/6);
        A(:,:,ii,2) = repmat(blkdiag(x3,x3,y3),d/6,d/6);
        B(:,:,ii,1) = repmat(blkdiag(x4,y4,y4),d/6,d/6);
        B(:,:,ii,2) = repmat(blkdiag(y4,x4,x4),d/6,d/6);
       end
   end    

Then the code includes too many sentences and repetition. And each time I have to redefine the size of C For instance for 4basis C is eqaul {[],[];[],[];temp,temp2;temp3,temp4} but for 6 basis C is equal {[],[];[],[],[];[],[][];temp,temp2;temp3,temp4}

And for the fifth basis I cant write these sentence A(:,:,ii,1) or A(:,:,ii,2)

Stephen23 2020-7-13

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"And each time I have to redefine the size of C For instance for 4basis "

I don't see any reason why you need to redefine C for different d, just generate the appropriate indices on each loop iteration (I have no idea what your required permutations are). Your approach of creating lots of variables and using if restricts you from writing generalized code which expands with d or basis_number.

The generalized solution would be something like this:

C = {all required values/arrays in here};
for ii = 1:whatever
    X = permutation indices based on d and whatever else
    Y = ditto
    A(:,:,ii,1) = blkdiag(C{X});
    A(:,:,ii,2) = blkdiag(C{Y});
end

Presumably you know the rules for those permutations that you want to generate.

"And If I have 6 d then the permutations should go till the 8 basis count"

I have no idea what that means.

"My aim is to not to use dimension in if statement but now it is impossible"

I don't see why: if you know the logic then you can define the indices that access the required arrays from C, use a comma-separated list to provide them to blkdiag. There seems to be nothing forcing you to use if.

"And for the fifth basis I cant write these sentence..."

Why not?

I do not have your data, I cannot run your example code,and your explanations are not very clear, so I cannot provide you with a complete, working, tested solution. I have shown you a general approach that you can probably use.

Avoid creating numbered variables, lots of C, D, etc. and using if. They will not help you. Using arrays and indexing will help you.

Gözde Üstün 2020-7-13

Hello, thanks for your answer.

For 4d, code is ok. I have this values:(2^2)

A(:,:,3,1) = repmat(blkdiag(temp1,temp2),d/4,d/4);

A(:,:,3,2) = repmat(blkdiag(temp2,temp1),d/4,d/4);

A(:,:,4,1) = repmat(blkdiag(temp3,temp4),d/4,d/4);

A(:,:,4,2) = repmat(blkdiag(temp4,temp3),d/4,d/4);

But for 6d this time I should these values: (the all triple permutations of temp1 and temp2 and temp3 and temp4 ) Here we have 2^3 basis

A(:,:,3,1) = repmat(blkdiag(temp1,temp2,temp1),d/6,d/6);

A(:,:,3,2) = repmat(blkdiag(temp2,temp1,temp1),d/6,d/6);

A(:,:,4,1) = repmat(blkdiag(temp1,temp1,temp2),d/6,d/6);

A(:,:,4,2) = repmat(blkdiag(temp2,temp2,temp1),d/6,d/6);

A(:,:,5,1) = repmat(blkdiag(temp1,temp2,temp2),d/6,d/6);

A(:,:,5,2) = repmat(blkdiag(temp2,temp1,temp2),d/6,d/6);

A(:,:,6,1) = repmat(blkdiag(temp3,temp4,temp3),d/6,d/6);

A(:,:,6,2) = repmat(blkdiag(temp4,temp3,temp3),d/6,d/6);

A(:,:,7,1) = repmat(blkdiag(temp3,temp3,temp4),d/6,d/6);

A(:,:,7,2) = repmat(blkdiag(temp4,temp4,temp3),d/6,d/6);

A(:,:,8,1) = repmat(blkdiag(temp4,temp3,temp4),d/6,d/6);

A(:,:,8,2) = repmat(blkdiag(temp3,temp4,temp4),d/6,d/6);

And for 8d I should have all fourt permutations of temp with temp2 and temp3 with temp4. Here we have 2^4 basis

I really dont know how it could be possible with your ideas. It will not a smarter code It will be massive And I am still thinking for your code how it could be possible Anyway Thank you very much

Stephen23 2020-7-13

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Your examples keep changing. In your earlier comment you gave this table as the expected output for d=6:

% 1st example:
A(:,:,3,1) = repmat(blkdiag(x3,x3,y3),d/6,d/6);
A(:,:,3,2) = repmat(blkdiag(x3,y3,x3),d/6,d/6);
A(:,:,4,1) = repmat(blkdiag(y3,x3,x3),d/6,d/6);
A(:,:,4,2) = repmat(blkdiag(y3,y3,x3),d/6,d/6);
A(:,:,5,1) = repmat(blkdiag(y3,x3,y3),d/6,d/6);
A(:,:,5,2) = repmat(blkdiag(x3,y3,y3),d/6,d/6);
A(:,:,6,1) = repmat(blkdiag(x4,y4,x4),d/6,d/6);
...

Now you gave in your most recent comment this table for d=6:

% 2nd example:
A(:,:,3,1) = repmat(blkdiag(temp1,temp2,temp1),d/6,d/6);
A(:,:,3,2) = repmat(blkdiag(temp2,temp1,temp1),d/6,d/6);
A(:,:,4,1) = repmat(blkdiag(temp1,temp1,temp2),d/6,d/6);
A(:,:,4,2) = repmat(blkdiag(temp2,temp2,temp1),d/6,d/6);
A(:,:,5,1) = repmat(blkdiag(temp1,temp2,temp2),d/6,d/6);
A(:,:,5,2) = repmat(blkdiag(temp2,temp1,temp2),d/6,d/6);
A(:,:,6,1) = repmat(blkdiag(temp3,temp4,temp3),d/6,d/6);
...

The order has changed: if we assume that x3==temp1 and y3==temp2 then here are the first lines of those two examples::

A(:,:,3,1) = repmat(blkdiag(temp1,temp1,temp2),d/6,d/6); % 1st example
A(:,:,3,1) = repmat(blkdiag(temp1,temp2,temp1),d/6,d/6); % 2nd example

They are clearly not the same order. Most/all of the other lines have changed order too.

Why do the examples keep changing order? Is the order random? Are any of them actually correct?

Stephen23 2020-7-13

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"I really dont know how it could be possible with your ideas"

I really don't know how it would be possible with lots of numbered variables, IFs, etc.

"It will not a smarter code It will be massive..."

I doubt that. Here is a simple working example (unlike your examples, none of which can be run):

% random fake data:
d = 6; % must be even
n = d/2;
r = pow2(n);
C = arrayfun(@(~)rand(2,2),1:r/2,'uni',0);
% generate indices:
X = [1+eye(n);2-eye(n)];
X = fliplr([X;2+X]);
% generate output array:
A = zeros(d,d,r,2); % preallocate!
k = 0;
for ii = 3:r
    for jj = 1:2
        k = k+1;
        A(:,:,ii,jj) = blkdiag(C{X(k,:)});
    end
end

Considering that this code works for any d I would not consider it particularly "massive". It is certainly much more versatile than your hard-coded approach using lots of variables and IFs. Here is your 1st example (I removed the superfluous repmat, mapped x3->C{1}, y3->C{2}, etc., corrected the order of A(:,:,6,1) to match the first row, and followed the same pattern for the remaining rows):

B(:,:,3,1) = blkdiag(C{1},C{1},C{2});
B(:,:,3,2) = blkdiag(C{1},C{2},C{1});
B(:,:,4,1) = blkdiag(C{2},C{1},C{1});
B(:,:,4,2) = blkdiag(C{2},C{2},C{1});
B(:,:,5,1) = blkdiag(C{2},C{1},C{2});
B(:,:,5,2) = blkdiag(C{1},C{2},C{2});
B(:,:,6,1) = blkdiag(C{3},C{3},C{4});
B(:,:,6,2) = blkdiag(C{3},C{4},C{3});
B(:,:,7,1) = blkdiag(C{4},C{3},C{3});
B(:,:,7,2) = blkdiag(C{4},C{4},C{3});
B(:,:,8,1) = blkdiag(C{4},C{3},C{4});
B(:,:,8,2) = blkdiag(C{3},C{4},C{4});

Lets compare the outputs:

>> isequal(A,B)
ans = 1

So far everything seems to be working as expected, no "massive" code required as far as I can tell.

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Hello

Regarding my example the orders mean that the permutation each of them has a different block. They are correct For A(:,:,3,1) I want to see that: repmat(blkdiag(temp1,temp1,temp2),d/6,d/6);and for the A(:,:,3,2) I want to see that: repmat(blkdiag(temp2,temp1,temp1),d/6,d/6); This means permutation. Lets do it clear

In two dimension I have that:

alice(:,:,1,1) = %this is the forst block
    0.5000   -0.5000
   -0.5000    0.5000
alice(:,:,2,1) = %third block
     0     0
     0     1
alice(:,:,1,2) = %this is the second block
    0.5000    0.5000
    0.5000    0.5000
    
alice(:,:,2,2) = %5. block
     1     0
     0     0

Now In 4d and 2 basis (A(4,4,2,2)) I should see that:

     alice(:,:,1,1) =
    0.5000   -0.5000         0         0
   -0.5000    0.5000         0         0
         0         0    0.5000   -0.5000
         0         0   -0.5000    0.5000
alice(:,:,2,1) =
     0     0     0     0
     0     1     0     0
     0     0     0     0
     0     0     0     1
alice(:,:,1,2) =
    0.5000    0.5000         0         0
    0.5000    0.5000         0         0
         0         0    0.5000    0.5000
         0         0    0.5000    0.5000
alice(:,:,2,2) =
     1     0     0     0
     0     0     0     0
     0     0     1     0
     0     0     0     0

Up to now my cod eis perfectly fine

And now for 4d and 3 basis I should see that: (A(4,4,3,2))

          alice(:,:,1,1) =
    0.5000   -0.5000         0         0
   -0.5000    0.5000         0         0
         0         0    0.5000   -0.5000
         0         0   -0.5000    0.5000
alice(:,:,2,1) =
     0     0     0     0
     0     1     0     0
     0     0     0     0
     0     0     0     1
alice(:,:,3,1) = %combination of the first block and second block
    0.5000   -0.5000         0         0
   -0.5000    0.5000         0         0
         0         0    0.5000    0.5000
         0         0    0.5000    0.5000
alice(:,:,1,2) =
    0.5000    0.5000         0         0
    0.5000    0.5000         0         0
         0         0    0.5000    0.5000
         0         0    0.5000    0.5000
alice(:,:,2,2) =
     1     0     0     0
     0     0     0     0
     0     0     1     0
     0     0     0     0
alice(:,:3,2) %another permutation of fisrt block and second block 
    0.5000    0.5000         0         0
    0.5000    0.5000         0         0
         0         0    0.5000    -0.5000
         0         0    -0.5000    0.5000

And if I have A(4,4,4,2) In addition the above result we should see that :

alice(:,:,4,1) = %permutation of the thord and fourth block
   0     0     0
   1     0     0
   0     1     0
   0     0     0

and

     alice(:,:,4,2) = %
another permutation of the third and fourth block
     1     0     0     0
     0     0     0     0
     0     0     0     0
     0     0     0     1

For 6 d this time we have 3 block and their permutations.

I saved this block as temp.

for instance temp =

 0.5000   -0.5000
-0.5000    0.5000

temp2 =

 0.5000    0.5000
 0.5000    0.5000

temp3 =

 0     0
 0     1

temp4 =

 1     0
 0     0

By the way thank you very much for your code

And that is why I dont know what should I write inside C?

Should C be

 C={temp,temp2,temp3} ????
 Fun array is a very different structure How can I use it in our case

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Finally I have that:

n = d/2;
        r = pow2(n);
        C= {temp,temp2,temp3,temp4}
        C2={temp5,temp6,temp7,temp8}
    %    C = arrayfun(@(~)rand(2,2),1:r/2,'uni',0);
        % generate indices:
        X = [1+eye(n);4-eye(n)];
        X = fliplr([X;2+X]);
        % generate output array:
        %A = zeros(d,d,r,2); % preallocate!
        k = 0;
        for ii = 3:basis_number
            for jj = 1:2
                k = k+1;
                A(:,:,ii,jj) = blkdiag(C{X(k,:)});
                B(:,:,ii,jj) = blkdiag(C2{X(k,:)});
            end
        end