Question

我有一个函数ranker，它接受一个向量并按升序为其分配数字排名。例如，
ranker([5 1 3 600]) = [3 1 2 4]或
ranker([42 300 42 42 1 42] = [3.5 6 3.5 3.5 1 3.5]。

我正在使用矩阵variable_data，我想将排名函数应用于variable data中所有行的每一行。这是我目前的解决方案，但我觉得有一种方法可以对其进行矢量化并使其同样快速：p

variable_ranks = nan(size(variable_data));
for i=1:1:numel(nmac_ids)
    variable_ranks(i,:) = ranker(abs(variable_data(i,:)));
end

Answer 1

如果将矩阵行放入单元格数组中，则可以将函数应用于每个单元格。

考虑将SORT函数应用于每一行的简单示例

a = rand(10,3);
b = cell2mat( cellfun(@sort, num2cell(a,2), 'UniformOutput',false) );
%# same as: b = sort(a,2);

你甚至可以这样做：

b = cell2mat( arrayfun(@(i) sort(a(i,:)), 1:size(a,1), 'UniformOutput',false)' );

同样，使用for循环的版本可能更快..

Answer 2

与Amro和Jonas合作

variable_ranks = tiedrank(variable_data')';

Ranker已经被Stat工具箱中的Matlab函数所取代（对那些没有它的人来说很抱歉），

[R，TIEADJ] = tiedrank（X）计算向量X中的值的等级。如果任何X值被绑定，则绑定计算他们的平均排名。该返回值TIEADJ是一个调整对于非参数所需的联系测试signrank和rankum，并为斯皮尔曼等级的计算的相关性。

TIEDRANK将沿着Matlab 7.9.0（R2009b）中的列进行计算，但它没有记录。因此，通过转置输入矩阵，行将变为列并对它们进行排名。然后使用第二个转置以与输入相同的方式组织数据。实质上是一个非常优雅的黑客：p

Answer 3

一种方法是重写ranker以获取数组输入

sizeData = size(variable_data);

[sortedData,almostRanks] = sort(abs(variable_data),2);
[rowIdx,colIdx] = ndgrid(1:sizeData(1),1:sizeData(2));
linIdx = sub2ind(sizeData,rowIdx,almostRanks);
variable_ranks = variable_data;
variable_ranks(linIdx) = colIdx;

%# break ties by finding subsequent equal entries in sorted data
[rr,cc] = find(diff(sortedData,1,2) == 0);
ii = sub2ind(sizeData,rr,cc);
ii2 = sub2ind(sizeData,rr,cc+1);
ii = sub2ind(sizeData,rr,almostRanks(ii));
ii2 = sub2ind(sizeData,rr,almostRanks(ii2));
variable_ranks(ii) = variable_ranks(ii2);

修改

相反，你可以使用TMW的TIEDRANK（谢谢，@ Amro）：

variable_rank = tiedrank(variable_data')';

Answer 4

我编写了一个执行此操作的函数，它位于FileExchange tiedrank_(X,dim)上。它看起来像这样......

%[Step 0a]: force dim to be 1, and compress everything else into a single 
%dimension. We will reverse this process at the end.
if dim > 1 
    otherDims = 1:length(size(X));
    otherDims(dim) = [];
    perm = [dim otherDims];
    X = permute(X,perm);
end
originalSiz = size(X);
X = reshape(X,originalSiz(1),[]);
siz = size(X);

%[Step 1]: sort and get sorting indicies
[X,Ind] = sort(X,1);

%[Step 2]: create matrix [D], which has +1 at the start of consecutive runs
% and -1 at the end, with zeros elsewhere.
D = zeros(siz,'int8');
D(2:end-1,:) = diff(X(1:end-1,:) == X(2:end,:));
D(1,:) = X(1,:) == X(2,:);
D(end,:) = -( X(end,:) == X(end-1,:) );

clear X

%[Step 3]: calculate the averaged rank for each consecutive run
[a,~] = find(D);
a = reshape(a,2,[]);
h = sum(a,1)/2;

%[Step 4]: insert the troublseome ranks in the relevant places
L = zeros(siz);
L(D==1) = h;
L(D==-1) = -h;
L = cumsum(L);
L(D==-1) = h; %cumsum set these ranks to zero, but we wanted them to be h

clear D h

%[Step 5]: insert the simple ranks (i.e. the ones that didn't clash)
[L(~L),~] = find(~L);

%[Step 6]: assign the ranks to the relevant position in the matrix
Ind = bsxfun(@plus,Ind,(0:siz(2)-1)*siz(1)); %equivalent to using sub2ind + repmat
r(Ind) = L;

%[Step 0b]: As promissed, we reinstate the correct dimensional shape and order
r = reshape(r,originalSiz);
if dim > 1
    r = ipermute(r,perm);
end

我希望能帮助别人。

将函数应用于所有行

4 个答案: