如何找到小于阈值的矩阵每一行中的元素数

Question

The question is just as the title. I can do it by using loops. However, I wonder is there a better way to achieve this without a loop. Thanks. In case I did not make my question clear, the following code shows what I want using loops. By the way, the data in each row is sorted from the smallest to the largest, from the left to the right.

Mat = rand(20);
Mat = sort(Mat,2);
Mat(:,1) = 0; % in case these is no element smaller than the threshold
result = zeros(20,1);
threshold = 0.2;
for i = 1:20
    result(i) = length(find(Mat(i,:) < threshold));
end

Answer 1

您可以使用条件构造一个logical矩阵，然后sum每一行sum以找到符合条件的项目数

result = sum(Mat < threshold, 2)