Matlab

Question

我有一段考试代码，他们要我做的是为了取得更好的成绩而无需第二个 "for" 语句即可实现同样的事情。

密码是：

piv = 1:n;                  %// piv: position vector
for k = 1:n-1               %// for each column :
   if ((max(abs(A(piv(k:n),k)))) > eps(normA))    %// if pivot is non zero
       [~, I] = max(A(piv(k:n),k));       %// find the max index
       I = I + (k-1);                 
       piv([k,I]) = piv([I,k]);       %// swap pivot elements
       A(piv(k+1:n),k) = A(piv(k+1:n),k)/A(piv(k),k);    %// calculate the multipliers and save them in the column 
       for j = k+1:n
            A(piv(j),k+1:n) = A(piv(j),k+1:n) - (A(piv(k),k+1:n)*A(piv(j),k));   %// multiply for multipliers and subtract them by the row 
       end 
    end
end

这是高斯分解方法，但没关系，问题是我需要得到相同的结果，但 e 和 j 变量没有第二个。

Answer 1

你当然可以杀死最里面的循环bsxfun. I am leaving it to you to explain to your prof on how it does what it does. Going through the bsxfun docs would be a good idea and in this process you might learn some vectorization techniques。这是实现 -

parte2 = bsxfun(@times,A(piv(k),k+1:n),A(piv(k+1:n),k))
A(piv(k+1:n),k+1:n) = A(piv(k+1:n),k+1:n) - parte2

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