find a same element inside a cell

I have a cell=
{[1,2,5],[2,3,4],[1,2,4,5],[4,7,9],[1,2,4,5,6,9,10,11],[6,12,13],[1,2,4,5,6,9,13,14]}
I want to find array in this cell that has a other array inside. For my example [1,2,5] is repeated in [1,2,4,5] and [1,2,4,5,6,9,10,11] and [1,2,4,5,6,9,10,11]. but [2,3,4] is not in others. My result
{[1,2,4,5],[1,2,4,5,6,9,10,11],[1,2,4,5,6,9,13,14]}

1 comentario

Stephen23
Stephen23 el 17 de Oct. de 2018
@Naime Ahmadi: is the order significant? For example, does [1,2,5] match both of these?:
[1,2,4,5] % same order
[5,2,4,1] % different order

Iniciar sesión para comentar.

 Respuesta aceptada

Jan
Jan el 17 de Oct. de 2018
Editada: Jan el 17 de Oct. de 2018
C = {[1,2,5],[2,3,4],[1,2,4,5],[4,7,9], ...
[1,2,4,5,6,9,10,11],[6,12,13],[1,2,4,5,6,9,13,14]};
keep = false(1, numel(C));
for i1 = 1:numel(C)
for i2 = 1:numel(C)
if i1 ~= i2 && ~keep(i2)
keep(i2) = all(ismember(C{i1}, C{i2}));
end
end
end
Result = C(keep);
What should happen for {[1,2,5], [1,2,5]} ?

1 comentario

Stephen23
Stephen23 el 18 de Oct. de 2018
Editada: Stephen23 el 18 de Oct. de 2018
"I do not have c"
C is just your cell array.
You told us that you have a cell array, but you did not tell us its name. So Jan just used C for its name.

Iniciar sesión para comentar.

Más respuestas (3)

Stephen23
Stephen23 el 17 de Oct. de 2018
Editada: Stephen23 el 18 de Oct. de 2018
You could do this in two lines, but here I show it on four lines:
>> A = {[1,2,5],[2,3,4],[1,2,4,5],[4,7,9],[1,2,4,5,6,9,10,11],[6,12,13],[1,2,4,5,6,9,13,14]};
>> F = @(r,c)r~=c&all(ismember(A{r},A{c}));
>> [R,C] = ndgrid(1:numel(A));
>> X = any(arrayfun(F,R,C),1);
>> B = A(X)
B =
1 2 4 5
1 2 4 5 6 9 10 11
1 2 4 5 6 9 13 14
KSSV
KSSV el 17 de Oct. de 2018
C = {[1,2,5],[2,3,4],[1,2,4,5],[4,7,9],[1,2,4,5,6,9,10,11],[6,12,13],[1,2,4,5,6,9,13,14]};
D = {[1 2 5]} ;
iwant = cell([],1) ;
count = 0 ;
for i = 1:numel(C)
idx = ismember(C{i},D{1}) ;
if nnz(idx)==numel(D{1})
count = count+1 ;
iwant{count} = C{i} ;
end
end

4 comentarios

NA
NA el 18 de Oct. de 2018
sorry I do not have D. it means that I want a code that by looking at c, can recognize [1,2,4,5], [1,2,4,5,6,9,10,11],[1,2,4,5,6,9,13,14] has [1,2,5] repeated.
KSSV
KSSV el 18 de Oct. de 2018
D=C{1}; or any number you want.
Stephen23
Stephen23 el 18 de Oct. de 2018
Editada: Stephen23 el 18 de Oct. de 2018
"is there any possibility that by checking all array we recognize [1,2,5] is repeated"
My answer does that.
Stephen23
Stephen23 el 18 de Oct. de 2018
"result:"
[2,4,6,10,12,16,17],[2,4,6,10,12,15,18,19,20],[2,4,6,10,12,15,22,23,25]
My answer does that too. Did you actually try any of the answers, apart from this one?

Iniciar sesión para comentar.

Bruno Luong
Bruno Luong el 18 de Oct. de 2018
Editada: Bruno Luong el 18 de Oct. de 2018
C = {[1,2,5],[2,3,4],[1,2,4,5],[4,7,9],...
[1,2,4,5,6,9,10,11],[6,12,13],[1,2,4,5,6,9,13,14]};
% uncomment this line to try with random data
% C = arrayfun(@(i) randi(10,1,3+randi(7)), 1:1000, 'unif', 0);
x = unique(cat(2,C{:}));
% runing ID
c = cumsum([1 cellfun('length', C)]);
I = cumsum(accumarray(c(:),1));
% Map values of C{} to position in x
J = cellfun(@(a) ismembc2(a,x), C, 'unif', 0);
J = cat(2,J{:});
% "Matrix" of set belonging
m = length(C);
B = zeros([m 1 length(x)]);
B(I(1:end-1) + (J(:)-1)*m) = 1;
% check for inclusion
B = all(B>=permute(B,[2 1 3]),3);
% discard self-inclusion
B(1:m+1:end) = false;
% Keep sets that does comprise at least one other set
Result = C(any(B,2));

2 comentarios

Stephen23
Stephen23 el 18 de Oct. de 2018
Note that ismembc2 is an undocumented MATLAB function: its behavior and/or presence can change without warning between MATLAB versions.
Bruno Luong
Bruno Luong el 18 de Oct. de 2018
Editada: Bruno Luong el 18 de Oct. de 2018
True, but never see it changes in probably 10+ years. ISMEMBER though documented has it behavior changes (order of location).
If fear replace the statement with
[~,J] = cellfun(@(a) ismember(a,x), C, 'unif', 0);

Iniciar sesión para comentar.

Preguntada:

NA
el 17 de Oct. de 2018

Editada:

el 18 de Oct. de 2018

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by