Matlab 我需要帮助在同一绘图上以不同颜色绘制if/else命令的不同排列

基本上我有一个代码，它生成了成本和可靠性之间所有可能的排列图。总共有864个数据点被分成8行。其中五行有2个选项，其中三行有3个选项

这是我的代码副本。我试图让“其他相机”和“深度与结构测试”的排列与其他六种可能性的颜色不同。我尝试使用“gscatter”命令，但运气不太好

我相信我需要在if/else语句中使用scatter命令，尽管我不太确定在“scatter”命令的“X”和“Y”中绘制什么。目前，我的代码设置为以一种颜色打印所有数据。我用“gscatter”删除了我的代码，因为我遇到了很多错误，当我试图修复它们时，情节最终没有按计划进行

% Pareto_Eval
baseline_cost = 45;
nrows = 8;
%Initialize Variables
for aa = 1:nrows
   cost_delta(aa) = 0;
   reliability(aa) = 1;
end
icount = 1;

   %Propulsion
for row1 = 1:2  
    if row1 == 1
        cost_delta(1)= -7;
        reliability(1) = 0.995;
    elseif row1==2
        cost_delta(1)=0;
        reliability(1)=.99;
    end


    %Entry Mode
for row2 = 1:2
    if row2 == 1
        cost_delta(2) = -3;
        reliability(2) = .99;
    else
        cost_delta(2) = 0;
        reliability(2) = .98;
    end


    %Landing Method
for row3 = 1:3
    if row3 == 1                %if needs declaration
        cost_delta(3)= 0;
        reliability(3) = .99;
    elseif row3 == 2            %elseif needs declaration
        cost_delta(3) = 4;
        reliability(3) = .995;
    else                        %else does not need declaration
        cost_delta(3) = -2;
        reliability(3) = .95;
    end


    %Lander Type
for row4 = 1:3    
    if row4 == 1
        cost_delta(4)= 10;
        reliability(4) = .99;
    elseif row4 == 2
        cost_delta(4) = 0;
        reliability(4) = .99;
    else
        cost_delta(4) = 15;
        reliability(4) = .95;
    end


    %Rover Type
 for row5 = 1:2
    if row5 == 1
        cost_delta(5)= -2;
        reliability(5) = .98;
    else
        cost_delta(5) = 0;
        reliability(5) = .975;
    end


    %Power Source
for row6 = 1:2
    if row6 == 1
        cost_delta(6) = -3;
        reliability(6) = .95;
    else
        cost_delta(6) = 0;
        reliability(6) = .995;
    end   

    %Depth & Structure Testing
for row7 = 1:2
    if row7 == 1
        cost_delta(7) = 0;
        reliability(7) = .99;
    else 
        cost_delta(7) = 2;
        reliability(7) = .85;
    end      

      %Other Cameras
for row8 = 1:3    
    if row8 == 1
        cost_delta(8)= -1;
        reliability(8) = .99;
    elseif row8 == 2
        cost_delta(8) = -1;
        reliability(8) = .99;
    else
        cost_delta(8) = 0;
        reliability(8) = .9801;
    end

    cost_delta_total = 0;
    reliability_product = 1;

    for bb=1:nrows
        cost_delta_total = cost_delta_total + cost_delta(bb);
        reliability_product = reliability_product*reliability(bb);
    end

    total_cost(icount) = baseline_cost + cost_delta_total;
    total_reliability(icount) = reliability_product;
    icount = icount + 1;

end; end; end;      %Rows 1,2,3
end; end; end;      %Rows 4,5,6 
end; end;           %Rows 7,8


%Plot the Pareto Evaluation    
fignum=1;
figure(fignum)
sz = 5;
scatter(total_reliability, total_cost, sz, 'blue')
xlabel('Reliability')
ylabel('Cost')
title('Pareto Plot')   

感谢您的帮助。我对Matlab没有太多的经验，我也试着四处寻找帮助，但没有任何效果

下面是一个示例代码，可以使我创建的问题更简单：

% Pareto_Eval
baseline_cost = 55;
nrows = 3;


%Initialize Variables
for aa = 1:nrows
   cost_delta(aa) = 0;
   reliability(aa) = 1;
end
icount = 1;

%Group 1
for row1 = 1:2
    if row1 == 1
        cost_delta(1)= 5;
        reliability(1) = 0.999;  
    elseif row1==2
        cost_delta(1) = 0;      
        reliability(1) = .995;  
    end

    %Group 2
    for row2 = 1:2         
      if row2 == 1
        cost_delta(2) = 0;    
        reliability(2) = .98;
      else              
        cost_delta(2) = -2;
        reliability(2) = .95;
      end

      %Group 3
      for row3 = 1:2
        if row3 == 1
          cost_delta(3) = 3;   
          reliability(3) = .997;
         else                  
          cost_delta(3) = 0;
          reliability(3) = .96;
        end

       %initializing each row      
       cost_delta_total = 0;
       reliability_product = 1;

        for bb = 1:nrows   
          cost_delta_total = cost_delta_total + cost_delta(bb);  
          reliability_product = reliability_product*reliability(bb); 
        end


       total_cost(icount) = baseline_cost + cost_delta_total;
       total_reliability(icount) = reliability_product;
       icount = icount + 1;
      end
    end
end

fignum=1;
figure(fignum)
sz = 25;
scatter(total_reliability, total_cost, sz)
xlabel('Reliability')
ylabel('Cost')
title('Pareto Plot')

---

基本上，我需要在每个if循环中绘制一个图，但我不确定如何做到这一点，并将它们全部放在同一个图上。

听起来像是一个有趣的项目！我不确定我是否正确理解了你想要的图，但希望下面的代码能让你更接近你想要的

我从嵌套for循环的混乱开始（正如您所做的），但通过构建置换矩阵使其更加简洁

counter = 0;
for propulsion_options = 1:2
    for entry_mode = 1:2
        for landing_method = 1:3
            for lander_type = 1:3
                for rover_type = 1:2
                    for power_source = 1:2
                        for depth_testing = 1:2
                            for other_cameras = 1:3
                                counter = counter +1
                                permutations(counter,:) = [...
                                    propulsion_options,...
                                    entry_mode,...
                                    landing_method,...
                                    lander_type,...
                                    rover_type,...
                                    power_source,...
                                    depth_testing,...
                                    other_cameras];
                            end
                        end
                    end
                end
            end
        end
    end
end

通过这种方式，我将实际得分排除在循环之外，并且可能更容易调整值。我将成本和可靠性数组初始化为与排列数组相同的大小：

cost_delta = zeros(size(permutations));
reliability = zeros(size(permutations));

然后，对于每个度量，我在排列数组中搜索每个可能值的所有发生率，并分配适当的分数：

%propulsion
propertyNo = 1;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -7;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.995;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.99;

%entry_mode (2)
propertyNo = 2;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -3;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.98;

%landing_method (3) 
propertyNo = 3;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = 0;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 4;
cost_delta(find(permutations(:,propertyNo)==3),propertyNo) = -2;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.995;
reliability(find(permutations(:,propertyNo)==3),propertyNo) = 0.95;

%lander_type (3)
propertyNo = 4;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = 10;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
cost_delta(find(permutations(:,propertyNo)==3),propertyNo) = 15;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==3),propertyNo) = 0.95;

%rover_type (2)
propertyNo = 5;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -2;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.98;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.975;

%power_source (2)
propertyNo = 6;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -3;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.95;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.995;

%depth_testing (2)
propertyNo = 7;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = 0;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 2;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.85;

%other_cameras (3)
propertyNo = 8;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -1;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = -1;
cost_delta(find(permutations(:,propertyNo)==3),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==3),propertyNo) = 0.9801;

然后，通过沿着第二维度对产品进行求和和，每个排列可以获得总成本/可靠性得分：

cost_delta_total = sum(cost_delta,2);
reliability_product = prod(reliability,2);

最后，您可以绘制所有点（根据原始点）：

或者，您可以通过搜索特定的属性值来创建排列索引，并绘制这些不同的颜色（实际上，此位回答了您最具体的问题，即如何在同一轴上绘制两个对象-您只需使用

按住；

命令）：

---

祝任务顺利！发射日是什么时候

听起来是个有趣的项目！我不确定我是否正确理解了你想要的图，但希望下面的代码能让你更接近你想要的

我从嵌套for循环的混乱开始（正如您所做的），但通过构建置换矩阵使其更加简洁

counter = 0;
for propulsion_options = 1:2
    for entry_mode = 1:2
        for landing_method = 1:3
            for lander_type = 1:3
                for rover_type = 1:2
                    for power_source = 1:2
                        for depth_testing = 1:2
                            for other_cameras = 1:3
                                counter = counter +1
                                permutations(counter,:) = [...
                                    propulsion_options,...
                                    entry_mode,...
                                    landing_method,...
                                    lander_type,...
                                    rover_type,...
                                    power_source,...
                                    depth_testing,...
                                    other_cameras];
                            end
                        end
                    end
                end
            end
        end
    end
end

通过这种方式，我将实际得分排除在循环之外，并且可能更容易调整值。我将成本和可靠性数组初始化为与排列数组相同的大小：

cost_delta = zeros(size(permutations));
reliability = zeros(size(permutations));

然后，对于每个度量，我在排列数组中搜索每个可能值的所有发生率，并分配适当的分数：

%propulsion
propertyNo = 1;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -7;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.995;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.99;

%entry_mode (2)
propertyNo = 2;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -3;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.98;

%landing_method (3) 
propertyNo = 3;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = 0;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 4;
cost_delta(find(permutations(:,propertyNo)==3),propertyNo) = -2;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.995;
reliability(find(permutations(:,propertyNo)==3),propertyNo) = 0.95;

%lander_type (3)
propertyNo = 4;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = 10;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
cost_delta(find(permutations(:,propertyNo)==3),propertyNo) = 15;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==3),propertyNo) = 0.95;

%rover_type (2)
propertyNo = 5;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -2;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.98;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.975;

%power_source (2)
propertyNo = 6;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -3;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.95;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.995;

%depth_testing (2)
propertyNo = 7;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = 0;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = 2;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.85;

%other_cameras (3)
propertyNo = 8;
cost_delta(find(permutations(:,propertyNo)==1),propertyNo) = -1;
cost_delta(find(permutations(:,propertyNo)==2),propertyNo) = -1;
cost_delta(find(permutations(:,propertyNo)==3),propertyNo) = 0;
reliability(find(permutations(:,propertyNo)==1),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==2),propertyNo) = 0.99;
reliability(find(permutations(:,propertyNo)==3),propertyNo) = 0.9801;

然后，通过沿着第二维度对产品进行求和和，每个排列可以获得总成本/可靠性得分：

cost_delta_total = sum(cost_delta,2);
reliability_product = prod(reliability,2);

最后，您可以绘制所有点（根据原始点）：

或者，您可以通过搜索特定的属性值来创建排列索引，并绘制这些不同的颜色（实际上，此位回答了您最具体的问题，即如何在同一轴上绘制两个对象-您只需使用

按住；

命令）：

---

祝任务顺利！发射日是什么时候

你的问题不清楚，你能问一个简单的问题，写一个简单的代码来解释吗？是的，对不起。下面是一个与之类似的示例代码，参数较少：构建数据的方式是错误的。当行号由for循环递增时，对行号调用

if/elseif/else

。循环期间将触发所有

if

条件。在所有这些循环中，您只分配最后一个

else

条件的最后一个值。。。您可以在一条指令中建立您的

总可靠性

，因为您的简化代码非常有用。请把它放在原代码的地方，不要显示原代码，这会让人困惑和分心。让你的问题简单明了，切中要害！谢谢你的问题不清楚，你能问一个简单的问题，写一个简单的代码来解释吗？是的，对不起。下面是一个与之类似的示例代码，参数较少：构建数据的方式是错误的。当行号由for循环递增时，对行号调用

if/elseif/else

。循环期间将触发所有

if

条件。在所有这些循环中，您只分配最后一个

else

条件的最后一个值。。。您可以在一条指令中建立您的

总可靠性

，因为您的简化代码非常有用。请把它放在原代码的地方，不要显示原代码，这会让人困惑和分心。让你的问题简单明了，切中要害！谢谢您可以使用

ndgrid

更轻松地构建置换矩阵。非常感谢您的输入，很抱歉反应太晚。不过这帮了大忙。不幸的是，没有实际的发射，更多的是一个理论项目/问题来模拟真实世界的任务以及如何选择顶级参数。再次感谢您，您可以使用

ndgrid

更轻松地构建置换矩阵。感谢您的输入，很抱歉反应太晚。不过这帮了大忙。不幸的是，没有实际的发射，更多的是一个理论项目/问题来模拟真实世界的任务以及如何选择顶级参数。再次感谢