matlab - 如何找到至少包含矢量 B 的一个元素的单元格 A 的矢量？

Question

如何找到至少包含向量 B 一个元素的 A 向量？

例子:

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

方法

与 arrayfun和 ismember -

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

或用arrayfun和 bsxfun -

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

或用arrayfun和 setdiff -

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

或用arrayfun和 intersect -

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

也可以使用 cellfun在这里，这样四个对应项cellfun基于解决方案最终像这样 -

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))

走不同的路线[使用bsxfun的屏蔽能力]

而不是进入那些arrayfun或 cellfun基于本质上是循环方法的方法，可以通过转换 A 使解决方案很多矢量化成一个二维数值数组。所以，这里的想法是有一个 2D行数为 A 中最大元素数的数组和列数作为 A 中的单元格数.该数组的每一列将包含 A 的每个单元格中的元素。和 NaNs会填满空白。

采用这种方法的解决方案代码如下所示 -

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

---

基准测试

对于有兴趣查看运行时性能的人，这里有一个具有足够大数据量的快速基准测试 -

%// 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

这样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他们的快一点 arrayfun基础同行!此外，基于掩码的 bsxfun这种方法看起来很有趣，但请记住它是需要大量内存的。