Arrange a set of elements in an array

Hello,

So i have an array that goes like this:

 A=
 [1
  4
  3
  2
  1
  4
  3
  2
  1
  4
  3
  2
  1
  4
  3
  2
  .
  .
  .

    ]

I want to arrange it as

A=
[4
3
2
1
4
3
2
1
4
3
2
1 
4
3
2
1 
4
3
2
1
]

How can I do this?

[Adam]

So you just want to move the 1 from the start to the end?!

[Soumyatha Gavvala]

No, I want to move all the 1's to the 4th position. They all correspond to another set of data and I use the dataset array for which this is one of the column.

Answer 1

[Walter Roberson]

t = reshape(A, 4, []);
t = t([2 3 4 1], :);
A = t(:);

[Soumyatha Gavvala]

My array length is not a multiple of 4.

[Walter Roberson]

Your question is not well defined yet.

If you just want to move the first point to end, as Adam questioned, then

A = [A(2:end); A(1)];

[Walter Roberson]

The desired outcome is not defined for the case where the array is not a multiple of 4 long. Consider the very last group and suppose it is only 3 long instead of 4, then it would be [1 4 3] for the input, and the 1 needs to move 3 places further down, so it would have to become [4 3 X 1] to match the pattern, but what goes where the X is? Should it become [4 3 2] to match the length? If so then where does the 2 come from? Should it become [4 3 1], moving the 1 to the end of the local group? Maybe, but you don't say what should happen. Perhaps it needs to be left as [1 4 3] if there is not the full set of 4 -- we don't know.

Answer 2

[John BG]

Soumyatha

have you tried a circshift?

A =       1.00
          4.00
          3.00
          2.00
          1.00
          4.00
          3.00
          2.00
          1.00
          4.00
          3.00
          2.00
          1.00
          4.00
          3.00
          2.00
circshift(A,-1)
 =
          4.00
          3.00
          2.00
          1.00
          4.00
          3.00
          2.00
          1.00
          4.00
          3.00
          2.00
          1.00
          4.00
          3.00
          2.00
          1.00

or perhaps you mean that there may be any amount of each number:

L=uint64(randi([1 4],randi([10 27],1,1),1))  % this is just a test matrix

h1=histogram(L)

Val=h1.Values;
L0={};
for k=1:1:length(Val) L1=k.*ones(1,Val(k)); L0=[L0;L1]; end

% size matters: measure max chain length
dist1=zeros(1,length(Val)); for k=1:1:length(Val) dist1(k)=numel(L0{k});   end

L2=zeros(length(Val),max(dist1))

for k=1:1:length(Val)  L2(k,[1:numel(L0{k})])=L0{k}; end

L3=flip(L2);
L4=uint64(nonzeros(L3(:)));

If you find this answer of any help solving your question,

please mark my answer as accepted

thanks in advance

John

[e-mail]