Question

我有这段代码工作正常，但我不喜欢我实现这3个单独for循环的方式。任何人都可以向我建议我如何将它们合并在一起，使其更加高效和紧凑？谢谢

clear variables;
close all;
clc;

ilambda=5;
mu=2;

n=2;
jmax=n+1;
P= sym('P',[jmax,jmax]);

for j1 = 1:jmax
        for j2 = 2:jmax-1
            [c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2);    
            if j1<jmax
                E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c3 * P(j1, j2+1) - c4 * P(j1, j2-1);
            else
                E(j1, j2) = c1 * P(j1, j2) - c3 * P(j1, j2+1) - c4 * P(j1, j2-1);
            end

        end
end

    j2=1;
    for j1=1:jmax;
        [c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2);    
        if (j1<jmax)
            E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c3 * P(j1, j2+1);
        else
            E(j1, j2) = c1*P(j1, j2) - c3 * P(j1, j2+1);
        end
    end


    j2=jmax;
    for j1=1:jmax;
        [c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2); 
        if (j1==1)
            E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c4 * P(j1, j2-1);
        elseif (j1==jmax)
            c1=((j1-1)*mu)+((j2-1)*mu);
            E(j1, j2) = c1 * P(j1, j2) - c4 * P(j1, j2-1) - c5 * P(j1-1, j2);
        else
            E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2)- c4 * P(j1, j2-1)- c5 * P(j1-1, j2);
        end
    end

Answer 1

由于j1始终从1转到jmax，因此不应该那么难。
你现在基本上做的是处理矩阵的内部部分，然后是第一列，然后是最后一列，包括所有行（即按列构建矩阵）。
您可以按行（使用索引j1）构建矩阵，然后根据j2单独构建列，如下所示：

for j1 = 1:jmax

    % set j2 at 1 and work on the first column
    j2=1;
    [c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2);
    if (j1<jmax)
        E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c3 * P(j1, j2+1);
    else
        E(j1, j2) = c1*P(j1, j2) - c3 * P(j1, j2+1);
    end

    % set j2 as variable in for loop
    for j2 = 2:jmax-1
        [c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2);
        if j1<jmax
            E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c3 * P(j1, j2+1) - c4 * P(j1, j2-1);
        else
            E(j1, j2) = c1 * P(j1, j2) - c3 * P(j1, j2+1) - c4 * P(j1, j2-1);
        end
    end

    % finally set j2 as jmax and work on the last column
    j2=jmax;
    if (j1==1)
        E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c4 * P(j1, j2-1);
    elseif (j1==jmax)
        c1=((j1-1)*mu)+((j2-1)*mu);
        E(j1, j2) = c1 * P(j1, j2) - c4 * P(j1, j2-1) - c5 * P(j1-1, j2);
    else
        E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2)- c4 * P(j1, j2-1)- c5 * P(j1-1, j2);
    end

% basically now you have completed the j1-th row
% the following loop will work in the very same way, that is:
% working on the j1-st row and then working separately on the 1st
% column, then on the inner columns and then on the last column.
end

另一种（在我看来非常优雅）方法是使用switch/case。在这种情况下，伪代码将类似于

for j1=1:jmax
    for j2=1:jmax
        switch j2
            case 1 
            % enter here what to do if j2==1
            case jmax
            % enter here what to do if j2==jmax
            otherwise
            % enter here what to do if j2 is between 2 and jmax-1
        end
    end
end

注意：我尝试使用系数c1 ... c5的任意值，我希望这些（这些）解决方案与您的coefficients()函数一起工作。

在Matlab中将3个for循环组合成1个

1 个答案: