问题
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]}
回答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.
来源：https://stackoverflow.com/questions/28152718/how-to-find-the-vectors-of-the-cell-a-that-contain-at-least-one-element-of-the-v