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

如何找到A包含至少一个向量元素的向量B？

例:

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

途径

随着arrayfun和ismember-

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

或与- arrayfun和bsxfun-

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

或与- arrayfun和setdiff-

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

或与- arrayfun和intersect-

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

也可以cellfun在这里使用,这样四个cellfun基于对应的解决方案就像这样 -

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))

采取不同的路线[使用bsxfun掩蔽能力]

而不是进入那些arrayfun或cellfun它们实际是多圈的方法为基础的方法,可以使溶液很多通过将矢量A为2D数字数组.因此,这里的想法是有一个2D数组,其中行数是最大元素A 数和列数作为单元格数A.该数组的每一列都将保存每个单元格中的元素,A并NaNs填充空白空间.

使用这种方法的解决方案代码看起来像这样 -

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));

验证

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

标杆

对于有兴趣了解运行时性能的人来说,这是一个快速的基准测试,具有足够大的数据量 -

%// 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

在我的系统上获得的结果是 -

************************  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.

可以看出,cellfun基于解决方案的速度比基础解决方案快一点arrayfun!此外,基于掩码的bsxfun方法看起来很有趣,但请记住它的内存饥饿性.