find array inside a cell

I have a A={[1,2,4],[2,3,7],[2,5],[4,5,6]} and B=[1;2;3;6;8]

I want to write a code that find B inside A.

the result should be c={[1,2],[2,3],[2],[6]}.

I try this code

c=cell(1, size(A,2));
for i=1:size(A,2)
    current=A{i};
    for j=1:size(B,1)
        index_X = find(A{i} == B(j));       
        if isempty(index_X)==0 
           c{i}(end+1)=B(index_X);
        end
    end
end

but it does not work

Answer 1

[Stephen23]

Simpler in one line:

>> A = {[1,2,4],[2,3,7],[2,5],[4,5,6]};
>> B = [1;2;3;6;8];
>> C = cellfun(@(m)intersect(m,B),A,'uni',0);
>> celldisp(C)
C{1} =
     1
     2
C{2} =
     2
     3
C{3} = 
     2
C{4} =
     6

[NA]

if B is matrix

how can I find row of B that one of the element (whether column 1 or 2 of B) is include A

A = {[1,2,4],[2,3,7],[2,5],[4,5,6]};

B=[1,2;1,5;2,3;2,4;2,5;3,4;4,7]

result should be C_out={[2,3,5,6,7],[1,3,4,5,7],[1,2,3,4],[2,4,5,6,7]}

[Stephen23]

@Naime Ahmadi: I do not follow the logic of your example. When I check for the "row of B that one of the element (whether column 1 or 2 of B) is include A" for the first cell of A, I get this:

[
   1  % B(1,1)==1==A{1}(1)
   2  % B(2,1)==1==A{1}(1)
   3  % B(3,1)==2==A{1}(2)
   4  % B(4,1)==2==A{1}(2)
   5  % B(5,1)==2==A{1}(2)
   6  % B(6,2)==4==A{1}(3)
   7  % B(7,1)==4==A{1}(3)
]

Please explain how you derived your examples.

[NA]

A={[1,2,4]}

B=[1,2; this row is not acceptable as both 1 and 2 are include in A

1,5; this row is acceptable as one element (1) include in A

2,3; this row is acceptable as one element (2) include in A

2,4] this row is not acceptable as both 2 and 4 are include in A

answer should be C_out={[1,4]}

if A={[1,2,4],[2,3],[1,2,5]}

answer should be C_out={[2,3],[1,2,4],[3,4]}

[Stephen23]

>> A = {[1,2,4],[2,3,7],[2,5],[4,5,6]};
>> B = [1,2;1,5;2,3;2,4;2,5;3,4;4,7];
>> F = @(v)find(sum(any(v==permute(B,[1,3,2]),2),3)==1);
>> C = cellfun(F,A,'uni',0);
>> celldisp(C)
C{1} =
   2
   3
   5
   6
   7
C{2} =
   1
   4
   5
   6
   7
C{3} =
   1
   2
   3
   4
C{4} =
   2
   4
   5
   6
   7

Your C_out{2} example seems to be incorrect.

[NA]

Thank you.

But I have a problem with permute(B,[1,3,2]).

I understand permute, rearrange the dimensions but I do not understand how?

I try permute(B,[1,2,3]) but I do not know why the ansewer is

1 2

1 5

2 3

2 4

2 5

3 4

4 7

[Stephen23]

"I try permute(B,[1,2,3]) but I do not know why the ansewer is"

That vector won't do anything. That vector specifies that the dimensions should be arranged 1->1, 2->2, and 3->3, so nothing changes. In contrast, the vector I used [1,3,2] swaps the dimensions like this: 1->1, 2->3, and 2->3.

[NA]

I have aquestion about this

A = {[1,2,4],[2,3,7],[2,5],[4,5,6]};

B = [1;2;3;6;8];

C = cellfun(@(m)intersect(m,B),A,'uni',0);

insted of the element I want to find a rows. so I try this one but it is not give me following answer

C = cellfun(@(m)find(m==B),A,'uni',0);

C{1} =

1

2

C{2} =

2

3

C{3} =

2

C{4} =

4

[Stephen23]

Just use the outputs of intersect:

>> [C,X,Y] = cellfun(@(m)intersect(m,B),A,'uni',0);
>> Y{:}
ans =
   1
   2
ans =
   2
   3
ans =
   2
ans =
   4

[NA]

Thank you. What does X give?

[Stephen23]

"What does X give?"

Open the intersect documentation, then find the description of the second output argument: that is what it gives. Reading the documentation is how MATLAB users know what functions do.

[NA]

A = {[1,2,4],[2,3,7],[2,5],[4,5,6]};

B = [1,2;1,5;2,3;2,4;2,5;3,4;4,7];

F = @(v)find(sum(any(v==permute(B,[1,3,2]),2),3)==1);

C = cellfun(F,A,'uni',0);

How can I change this code to only look at first column, element should be only in the first column

A={[1,2,4]}

B=[3,4 this row is not acceptable as first element is not in the A

answer should be={[1,2,5,7],[4,5,6],[3,4],[]}

[Stephen23]

@Naime Ahmadi: you really are going to have to explain the logic of that. Given these inputs:

A = {[1,2,4],[2,3,7],[2,5],[4,5,6]};
B = [1,2;1,5;2,3;2,4;2,5;3,4;4,7];

"answer should be ={[1,2,5,7],[4,5,6],[3,4],[]} "

Lets compare the values in the first column of B against the first cell of A:

1 -> matches A{1}(1)
1 -> matches A{1}(1)
2 -> matches A{1}(2)
2 -> matches A{1}(2)
2 -> matches A{1}(2)
3 -> no match
4 -> matches A{1}(3)

So this would give matches for rows [1,2,3,4,5,7]. Where do you get [1,2,5,7] from?

And for the second cell of A:

1 -> no match
1 -> no match
2 -> matches A{2}(1)
2 -> matches A{2}(1)
2 -> matches A{2}(1)
3 -> matches A{2}(2)
4 -> no match

So this would give matches for rows [3,4,5,6]. Where do you get [4,5,6] from?

Please explain carefully. I cannot read your mind, only what you write.

[NA]

first column of B against the first cell of A:

A=[1,2,4]

B=[1,2;1,5;2,3;2,4;2,5;3,4;4,7]

1,2 -> Not matches as both (1,2) is in A

1,5 -> matches A{1}(1) (first column is in A but second column is not in A)

2,3 -> matches A{1}(2) (first column is in A but second column is not in A)

2,4-> Not matches A{1}(2) as both (2,4) is in A

2,5-> matches A{1}(2) (first column is in A but second column is not in A)

3,4-> no match (4 is in A but as it is in second column of B it is not match)

4,7 -> matches A{1}(3) (first column is in A but second column is not in A)

[Stephen23]

@Naime Ahmadi: in your earlier comment you wrote "answer should be={[1,2,5,7],[4,5,6],[3,4],[]} "

Now you seem to indicate that it should be [2,3,5,7] for the first cell. Which is correct?

Perhaps this does what you want:

>> A = {[1,2,4],[2,3,7],[2,5],[4,5,6]};
>> B = [1,2;1,5;2,3;2,4;2,5;3,4;4,7];
>> F = @(v)find(any(v==B(:,1),2)&~any(v==B(:,2),2));
>> C = cellfun(F,A,'uni',0)
>> C{:}
ans =
   2
   3
   5
   7
ans =
   4
   5
   6
ans =
   3
   4
ans =
   7

Answer 2

[madhan ravi]

A={[1,2,4],[2,3,7],[2,5],[4,5,6]} ;
B=[1;2;3;6;8];
Result=cell(1,numel(A)); % preallocate
for i = 1:numel(A)
    idx=ismember(A{i},B);
    Result{i}=A{i}(idx);
end
celldisp(Result)

Gives:

Result{1} =
 
     1     2
 
Result{2} =
 
     2     3
 
Result{3} =
 
     2

Result{4} =
 
     6