Question

我有一个实数的列向量（V1），如：

如果我添加，让我们说，这个向量中每行的数字1，我会得到（V2）：

我需要找出V2中创建的一些错误窗口中有多少来自V2的元素。例如，error-window = 0.1（但在我的情况下，V1中的每个元素都有它自己的错误窗口）：

123.1100 123.3100
125.0290 125.2290
...
954.1190 954.3190

我可以创建一些像这样的代码：

% x - my vector
% ppm - a variable responsible for error-window
window = [(1-(ppm/1000000))*x, (1+(ppm/1000000))*x]; % - error-window
mdiff = 1:0.001:20; % the numbers I will iteratively add to x 
                    % (like the number 1 in the example)
cdiff = zeros(length(mdiff),1); % a vector that will contain counts of elements
                                % corresponding to different mdiff temp = 0;

for i = 1:length(mdiff)
    for j = 1:size(window,1)
        xx = x + mdiff(i);
        indxx = find( xx => window(j,1) & xx <= window(j,2) );
        if any(indxx)
            temp = temp + length(indxx); %edited
        end
    end
    cdiff(i) = temp;
    temp = 0;
end

所以，最后cdiff将包含与mdiff对应的所有计数。唯一的，我想让代码更快。或者有没有办法避免使用第二个循环（使用j）？我的意思是直接使用多维条件。

修改

我决定简化这样的代码（感谢我在这里得到的反馈）：

% x - my vector
% ppm - a variable responsible for error-window
window = [(1-(ppm/1000000))*x, (1+(ppm/1000000))*x]; % - error-window
mdiff = 1:0.001:20; % the numbers I will iteratively add to x 
                    % (like the number 1 in the example)
cdiff = zeros(length(mdiff),1); % a vector that will contain counts of elements
                                % corresponding to different mdiff temp = 0;

for i = 1:length(mdiff)
    xx = x + mdiff(i);
    cdiff(i) = sum(sum(bsxfun(@and,bsxfun(@ge,xx,window(:,1)'),bsxfun(@le,xx,window(:,2)'))));
end

在这种情况下，代码运行得更快，看起来正确

Answer 1

认为这可能适合你 -

%%// Input data
V1 = 52+rand(4,1)

V2 = V1+1;
t= 0.1;

low_bd = any(abs(bsxfun(@minus,V2,[V1-t]'))<t,2); %%//'
up_bd = any(abs(bsxfun(@minus,V2,[V1+t]'))<t,2); %%//'
count = nnz( low_bd | up_bd )

也可以将其写成 -

diff_map = abs(bsxfun(@minus,[V1-t V1+t],permute(V2,[3 2 1])));
count = nnz(any(any(diff_map<t,2),1))

修改1：

low_bd = any(abs(bsxfun(@minus,V2,window(:,1)'))<t,2); %%//'
up_bd = any(abs(bsxfun(@minus,V2,window(:,2)'))<t,2); %%//'
count = nnz( low_bd | up_bd )

编辑2：已编辑代码的矢量化表单

t1 = bsxfun(@plus,x,mdiff);
d1 = bsxfun(@ge,t1,permute(window(:,1),[3 2 1]));
d2 = bsxfun(@le,t1,permute(window(:,2),[3 2 1]));
t2 = d1.*d2;
cdiff_vect = max(sum(t2,3),[],1)';

Answer 2

add = 1; %// how much to add
error = .1; %// maximum allowed error

V2 = V1 + add; %// build V2   
ind = sum(abs(bsxfun(@minus, V1(:).', V2(:)))<error)>1; %'// index of elements
%// of V1 satisfying the maximum error condition. ">1" is used to because each 
%// element is at least equal to itself
count = nnz(ind);

MATLAB：搜索匹配多维条件的数组中的元素

2 个答案: