sort element in cell

A={[1,2,3,4,5,8,9,39],[2,3,17,18,25,26,27],[3,4,14,15,16,17,18],[4,5,6,11,12,13,14],[5,6,7,8],...
    [10,11,12,13],[16,21,22,23,24],[26,28,29],[2,30],[6,31],[10,32],[19,33],[20,34],[22,35],[23,36],[25,37],[29,38]};
ref=31;
base = cellfun(@(m)any(ismember(m,ref)),A,'uni',0);
base_mes=A{find([base{:}]==1)};
include_base = cellfun(@(m)any(ismember(m,base_mes)),A,'uni',0);
result=cell(1,numel(include_base));
index_other=find([include_base{:}]==1);
for i=1:size(index_other,2)
        result{i}=A{index_other(i)};
end

base_mes=[6,31], I want to find 6 and 31 in A, after this sort A according to 31 and 6.

result={[6,31],[4,5,6,11,12,13,14],[5,6,7,8],[1,2,3,4,5,8,9,39],[2,3,17,18,25,26,27],[3,4,14,15,16,17,18],[10,11,12,13],[16,21,22,23,24],[26,28,29],[2,30],[10,32],[19,33],[20,34],[22,35],[23,36],[25,37],[29,38]}

[Jos (10584)]

What does "sort A according to 31 and 6" mean, exactly?

[NA]

ref=31, I need to find an array that includes 31. so [6,31] is found, after this, as [6,31] is included 6, so need to find arrays that is included 6. in this case it becomes [4,5,6,11,12,13,14], [5,6,7,8].

finally the order of array in cell A changes.[6,31] become first, [4,5,6,11,12,13,14] second and [5,6,7,8] third and put other left array in A

[Jos (10584)]

Apparently you question has been answered, but what if

• there are more cells that contain 31
• what if the cell(s) containing 31, contain more than 2 numbers

I am just curious ...

[NA]

good point. How should I fix it?

[Stephen23]

• "there are more cells that contain 31" -> "How should I fix it?" -> only you can decide how to "fix" that, or if it needs "fixing" at all. You can tell us what you want to happen, but we cannot tell you what you want to happen in that situation.
• "what if the cell(s) containing 31, contain more than 2 numbers" -> my answer does not assume anything about how many elements the vectors have.

Answer 1

[Stephen23]

You can easily use logical indexing for this:

A = {[1,2,3,4,5,8,9,39],[2,3,17,18,25,26,27],[3,4,14,15,16,17,18],[4,5,6,11,12,13,14],[5,6,7,8],[10,11,12,13],[16,21,22,23,24],[26,28,29],[2,30],[6,31],[10,32],[19,33],[20,34],[22,35],[23,36],[25,37],[29,38]};
ref = 31
idr = cellfun(@(v)any(ismember(v,ref)),A);
vec = A{idr};
idv = cellfun(@(v)any(ismember(v,vec)),A);
Z = [A(idr),A(idv&~idr),A(~idv)];

Giving:

>> Z{:}
ans =
    6   31
ans =
    4    5    6   11   12   13   14
ans =
    5    6    7    8
ans =
    1    2    3    4    5    8    9   39
ans =
    2    3   17   18   25   26   27
ans =
    3    4   14   15   16   17   18
ans =
   10   11   12   13
ans =
   16   21   22   23   24
ans =
   26   28   29
ans =
    2   30
ans =
   10   32
ans =
   19   33
ans =
   20   34
ans =
   22   35
ans =
   23   36
ans =
   25   37
ans =
   29   38

[Stephen23]

PS: following on from Jos' comment, if more than one ref match is possible, then you can take them all into account using this line:

vec = [A{idr}];

[NA]

Thank you.

How can I change order inside each cell

Z{:}
ans =
    6   31
ans =
    4    5    6   11   12   13   14
ans =
    5    6    7    8

I want to change above to this

    Z{:}
ans =
    31   6
ans =
    6   4   5   11   12   13   14
ans =
    6    5    7    8

as 31 is ref, I want to be first element in each array. Also 6 should be first element.

[Stephen23]

>> fun = @(v) sort(0-ismember(v,vec)-(v==ref));
>> [~,ids] = cellfun(fun,Z,'uni',0);
>> Z1 = cellfun(@(v,x)v(x),Z,ids,'uni',0);
>> Z1{:}
ans =
   31    6
ans =
    6    4    5   11   12   13   14
ans =
    6    5    7    8
ans =
    1    2    3    4    5    8    9   39
ans =
    2    3   17   18   25   26   27
ans =
    3    4   14   15   16   17   18
ans =
   10   11   12   13
ans =
   16   21   22   23   24
ans =
   26   28   29
ans =
    2   30
ans =
   10   32
ans =
   19   33
ans =
   20   34
ans =
   22   35
ans =
   23   36
ans =
   25   37
ans =
   29   38

[Stephen23]

@Naime Ahmadi: you can probably do that using a loop or two. Try it!