for循环在while循环内

时间:2019-06-29 07:48:53

标签: matlab for-loop while-loop

让我描述一个任务:我有3个矩阵(M1,M2,M3),它们每个都有lenght(Mi)行和2列。给我们一个函数g(x,s),其中s是一个二维参数,并且给出了xeta。我想检查第一个矩阵M1,如果存在诸如g(x,M1(i,:)>eta之类的,我想结束算法并设置s_new=M1(i,:)。如果M1中的这样的s不存在,我想转到矩阵M2并在其中搜索。下一个矩阵M3。如果所有矩阵中都不存在这样的s_new,我想破坏一下。 我的第一次尝试:

function[s_new]= checking(M1,M2,M3,x)
bool1=0;
eta = 10^-8;
g = @(x,s) x-s(1)-s(2);
while bool1==0
            for i=1:length(M1)
                if g(x,M1(i,:))>eta
                    s_new=M1(i,:);
                    bool1=1;
                end
            end
            for i=1:length(M2)
                 if g(x,M2(i,:))>eta
                    s_new=M2(i,:);
                    bool1=1;
                 end
            end
            for i=1:length(M3)
                 if g(x,M3(i,:))>eta
                    s_new=M3(i,:);
                    bool1=1;
                 end
            end
            bool1=1;
        end

我的第二次尝试涉及一些break选项,但是也没有用。问题是:当alghoritm在M1中找到s(例如我们的条件成立)时,它不会停止,它转到M2,如果找到了s,它将更改s_new。另外,为了节省时间,如果M1中存在这样的算法,我不希望算法通过矩阵M2。

示例为何效果不佳:

M1=[0,-1;0,-1], M2=[0,-2;0,-2], M3=[0,0;0,0], x=0 

它应该返回向量[0,-1]并返回[0,-2]。任何帮助表示赞赏。 编辑:for循环中的bool1 = 1用红色下划线表示状态bool1可能未使用,就好像bool1 = 0时从启动条件无法识别它一样

2 个答案:

答案 0 :(得分:2)

我想我找到了问题

您打算在bool1=1;情况下中断while循环

您可以在每个部分之后添加if bool1, break;end

%function[s_new]= checking(M1,M2,M3,x)
M1=[0,-1;0,-1]; 
M2=[0,-2;0,-2];
M3=[0,0;0,0];
x=0;

bool1=0;
eta = 10^-8;
g = @(x,s) x-s(1)-s(2);
while bool1==0
    for i=1:length(M1)
        if g(x,M1(i,:))>eta
            s_new=M1(i,:);
            bool1=1;
        end
    end
    if bool1, break;end

    for i=1:length(M2)
         if g(x,M2(i,:))>eta
            s_new=M2(i,:);
            bool1=1;
         end
    end
    if bool1, break;end

    for i=1:length(M3)
         if g(x,M3(i,:))>eta
            s_new=M3(i,:);
            bool1=1;
         end
    end
    bool1=1;
end

display(s_new)

没有while循环会更优雅:

%function[s_new]= checking(M1,M2,M3,x)
M1=[0,-1;0,-1]; 
M2=[0,-2;0,-2];
M3=[0,0;0,0];
x=0;

bool1=0;
eta = 10^-8;
g = @(x,s) x-s(1)-s(2);

for i=1:length(M1)
    if g(x,M1(i,:))>eta
        s_new=M1(i,:);
        bool1=1;
    end
end

if ~bool1
    for i=1:length(M2)
         if g(x,M2(i,:))>eta
            s_new=M2(i,:);
            bool1=1;
         end
    end
end

if ~bool1
    for i=1:length(M3)
         if g(x,M3(i,:))>eta
            s_new=M3(i,:);
            bool1=1;
         end
    end
end

display(s_new)

答案 1 :(得分:2)

添加到@Rotem的解决方案中,您可以完全解决问题的for循环。您可以使用find函数及其功能来报告满足您条件的第一个索引,而不是遍历所有索引。同样,您也可以完全避免使用变量bool1。我提出以下代码:

%function [s_new]= checking(M1,M2,M3,x)
M1 = [0,-1; 0,-1]; 
M2 = [0,-2; 0,-2];
M3 = [0,0; 0,0];
x = 0;
eta = 10^-8;
% Here I have vectorized your function g so that it can work over all the rows of the s matrix. This is the step which actually removes the for-loop. sum(s, 2) always sums over the 2 columns in your matrix.
g = @(x, s) x - sum(s, 2);
% Now we assign an empty matrix to s_new. This step gets rid of the bool1 because now we can check whether s_new is still empty or not. It also returns empty matrix if no value of s_new is found in any of your matrices M1, M2 or M3
s_new = [];
% Now we find if an s_new exists in M1. We first calculate the g function over the entire M1 matrix. Then we check if your condition is satisfied. Then the find function returns the first row of M1 that satisfies your condition. If it doesnot find any row that satisifies this condition, find will return an empty matrix which we can check to assign the value to s_new
s_new_index = find(g(x, M1) > eta, 1, 'first');
if ~isempty(s_new_index) % the isempty function checks for empty matrices. ~ stands for NOT
    s_new = M1(s_new_index, :);
end
% Now we check if s_new was assigned earlier. If not then we repeat the same thing with M2 and then M3
if isempty(s_new)
    s_new_index = find(g(x, M2) > eta, 1, 'first');
    if ~isempty(s_new_index)
        s_new = M2(s_new_index, :);
    end
end
if isempty(s_new)
    s_new_index = find(g(x, M3) > eta, 1, 'first');
    if ~isempty(s_new_index)
        s_new = M3(s_new_index, :);
    end
end