我需要帮助在同一图上以不同颜色绘制 if/else 命令的不同排列

Question

基本上我有一个代码，它可以生成成本和可靠性之间所有可能排列的图表。共有 864 个数据点分为 8 行。其中五行有 2 个选项，其中三行有 3 个选项。

这里是我的代码的副本。我试图让 'Other Cameras' 和 'Depth & Structure Testing' 的排列与其他六种可能性具有不同的颜色。我尝试使用 'gscatter' 命令，但运气不佳。

我相信我需要在 if/else 语句本身中使用分散命令，尽管我不太确定在 'X' 和 'Y' 中为 'scatter' 命令。目前我的代码设置为以一种颜色绘制所有数据。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')

感谢任何帮助。我没有太多使用 Matlab 的经验，我尝试四处寻找帮助，但没有任何效果。

这是我创建的一个示例代码，可以简化问题：

% 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，但我不确定该怎么做并将它们都放在同一个图上

Answer 1

听起来是个有趣的项目！不确定我是否正确理解了您的预期情节，但希望下面的代码能让您更接近您正在寻找的内容。

我开始时使用了相当深的嵌套 for 循环（就像您所做的那样），但通过构建排列矩阵使其更加简洁。

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));

然后对于每个指标，我在排列数组中搜索每个可能值的所有出现并分配适当的分数：

%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;

然后每个排列都可以通过沿第二个维度对产品求和并获得总成本/可靠性分数：

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

最后，您可以绘制所有点（按照您的原件）：

%Plot the Pareto Evaluation    
fignum=1;
figure(fignum)
sz = 5;
scatter(reliability_product, cost_delta_total, sz, 'b')
xlabel('Reliability')
ylabel('Cost')
title('Pareto Plot')

或者您可以通过搜索特定的属性值并绘制这些不同的颜色来创建排列的索引（实际上，这个位回答了您最具体的问题，即如何在同一轴上绘制两个东西 - 您只需要 hold on; 命令）：

propertyNo = 7;
indexDepth1 = find(permutations(:,propertyNo)==1);
indexDepth2 = find(permutations(:,propertyNo)==2);
fignum=2;
figure(fignum)
sz = 5;
scatter(reliability_product(indexDepth1), cost_delta_total(indexDepth1), sz, 'k');
hold on;
scatter(reliability_product(indexDepth2), cost_delta_total(indexDepth2), sz, 'b');
xlabel('Reliability')
ylabel('Cost')
title('Pareto Plot')   
legend('Depth & Structure Test 1','Depth & Structure Test 2')

propertyNo = 8;
indexCam1 = find(permutations(:,propertyNo)==1);
indexCam2 = find(permutations(:,propertyNo)==2);
indexCam3 = find(permutations(:,propertyNo)==3);
fignum=3;
figure(fignum)
sz = 5;
scatter(reliability_product(indexCam1), cost_delta_total(indexCam1), sz, 'k');
hold on;
scatter(reliability_product(indexCam2), cost_delta_total(indexCam2), sz, 'b');
scatter(reliability_product(indexCam3), cost_delta_total(indexCam3), sz, 'g');
xlabel('Reliability')
ylabel('Cost')
title('Pareto Plot')   
legend('Other Camera 1','Other Camera 2','Other Camera 3')

祝你任务顺利！发射日是什么时候？

我需要帮助在同一图上以不同颜色绘制 if/else 命令的不同排列

I need help plotting different permutations of an if/else command in different colors on the same plot

matlab

if-statement

scatter-plot

pareto-chart