如何找到包含至少一个向量B的元素的单元格A的向量？

Question

How to find the vectors of A that contain at least one element of the vector B ?

example:

A = {[2 5],[8 9 2],[33 77 4],[102 6],[10 66 17 7 8 11],[110 99],[1 4 3],[15 41 88]}

B = [5 77 41 66 7]

Result = {[2 5],[33 77 4],[10 66 17 7 8 11],[15 41 88]}

Answer 1

Approaches

With arrayfun and ismember -

Result = A(arrayfun(@(n) any(ismember(B,A{n})),1:numel(A)))

Or with arrayfun and bsxfun -

Result = A(arrayfun(@(n) any(any(bsxfun(@eq,B(:),A{n}),2)),1:numel(A)))

Or with arrayfun and setdiff -

Result = A(arrayfun(@(n) numel(setdiff(B,A{n})) < numel(B),1:numel(A)))

Or with arrayfun and intersect -

Result = A(arrayfun(@(n) ~isempty(intersect(B,A{n})),1:numel(A)))

One could also use cellfun here, such that the four counterpart cellfun based solutions end up like these -

Result = A(cellfun(@(x) any(ismember(B,x)), A))

Result = A(cellfun(@(x) any(any(bsxfun(@eq,B(:),x),2)),A))

Result = A(cellfun(@(x) numel(setdiff(B,x)) < numel(B),A))

Result = A(cellfun(@(x) ~isempty(intersect(B,x)),A))

Taking different route [Using bsxfun 's masking capability]

Rather than going into those arrayfun or cellfun based approaches that are essentially loopy approaches, one can make the solution alot vectorized by converting A into a 2D numeric array. So, the idea here is to have a 2D array in which number of rows is the maximum number of elements in A and number of columns as number of cells in A . Each column of this array would hold elements from each cell of A and NaNs would fill up the empty spaces .

The solution code with such an approach would look like this -

lens = cellfun('length',A); %// number of elements in each cell of A
mask = bsxfun(@ge,lens,(1:max(lens))'); %//'# mask of valid places in the 2D array
A_arr = NaN(size(mask)); %//initialize 2D array in which A elements are to be put
A_arr(mask) = [A{:}]; %// put the elements from A

%// Find if any element from B is in any element along the row or dim-3       
%// locations in A_arr. Then logically index into A with it for the final
%// cell array output
Result = A(any(any(bsxfun(@eq,A_arr,permute(B,[1 3 2])),1),3));

---

Verification

>> celldisp(Result)
Result{1} =
     2     5
Result{2} =
    33    77     4
Result{3} =
    10    66    17     7     8    11
Result{4} =
    15    41    88

---

Benchmarking

For people interested in seeing the runtime performances, here's a quick benchmarking test with a sufficiently huge datasize -

%// Create inputs
N = 10000; %// datasize
max_num_ele = 100; %// max elements in any cell of A
num_ele = randi(max_num_ele,N,1); %// number of elements in each cell of A
A = arrayfun(@(n) randperm(N,num_ele(n)), 1:N, 'uni', 0); 
B = randperm(N,num_ele(1));

%// Warm up tic/toc.
for k = 1:100000
    tic(); elapsed = toc();
end

%// Start timing all approaches
disp('************************  With arrayfun **************************')
disp('------------------------  With arrayfun + ismember')
tic
Result = A(arrayfun(@(n) any(ismember(B,A{n})),1:numel(A)));
toc, clear Result

disp('------------------------  With arrayfun + bsxfun')
tic
Result = A(arrayfun(@(n) any(any(bsxfun(@eq,B(:),A{n}),2)),1:numel(A)));
toc, clear Result

disp('------------------------  With arrayfun + setdiff')
tic
Result = A(arrayfun(@(n) numel(setdiff(B,A{n})) < numel(B),1:numel(A)));
toc, clear Result

disp('------------------------  With arrayfun + intersect')
tic
Result = A(arrayfun(@(n) ~isempty(intersect(B,A{n})),1:numel(A)));
toc, clear Result

disp('************************  With cellfun **************************')
disp('------------------------  With cellfun + ismember')
tic
Result = A(cellfun(@(x)any(ismember(B,x)), A));
toc, clear Result

disp('------------------------  With cellfun + bsxfun')
tic
Result = A(cellfun(@(x) any(any(bsxfun(@eq,B(:),x),2)),A));
toc, clear Result

disp('------------------------  With cellfun + setdiff')
tic
Result = A(cellfun(@(x) numel(setdiff(B,x)) < numel(B),A));
toc, clear Result

disp('------------------------  With cellfun + setdiff')
tic
Result = A(cellfun(@(x) ~isempty(intersect(B,x)),A));

disp('************************  With masking bsxfun **************************')
tic
lens = cellfun('length',A); %// number of elements in each cell of A
mask = bsxfun(@ge,lens,(1:max(lens))'); %//'
A_numarr = NaN(size(mask));
A_numarr(mask) = [A{:}];
Result = A(any(any(bsxfun(@eq,A_numarr,permute(B,[1 3 2])),1),3));
toc

The results thus obtained on my system were -

************************  With arrayfun **************************
------------------------  With arrayfun + ismember
Elapsed time is 0.409810 seconds.
------------------------  With arrayfun + bsxfun
Elapsed time is 0.157327 seconds.
------------------------  With arrayfun + setdiff
Elapsed time is 1.154602 seconds.
------------------------  With arrayfun + intersect
Elapsed time is 1.081729 seconds.
************************  With cellfun **************************
------------------------  With cellfun + ismember
Elapsed time is 0.392375 seconds.
------------------------  With cellfun + bsxfun
Elapsed time is 0.143341 seconds.
------------------------  With cellfun + setdiff
Elapsed time is 1.101331 seconds.
------------------------  With cellfun + setdiff
************************  With masking bsxfun ********************
Elapsed time is 0.067224 seconds.

As one can see, cellfun based solutions are wee bit faster than their arrayfun based counterparts! Also, mask-based bsxfun approach looks like an interesting one, but do keep in mind it's memory-hungry nature .