如何绘制这个？ MATLAB [英] How do I plot this? MATLAB

问题描述

我有一个矩阵，X，其中我想使用kmeans函数绘制它。我想要的：如果行在第4列中的值为1，我想它是正方形如果该行在第4列中的值为2我想它+形状BUT如果该行的值为0，列5，它必须是蓝色的，如果该行在第5列中的值为1，它必须是黄色的

I have a matrix, X, in which I want to plot it using the kmeans function. What I would like: If row has a value of 1 in column 4 I would like it to be square shaped If the row has a value of 2 in column 4 I would like it + shaped BUT If the row has a value of 0 in column 5 it must be blue and if the row has a vale of 1 in column 5 it must be yellow

（您不需要使用这些确切的颜色和形状，我只想区分这些。）我试过这个，它没有工作：

(You don't need to use these exact colors and shapes, I just want to distinguish these.) I tried this and it did not work:

plot(X(idx==2,1),X(idx==2,2),X(:,4)==1,'k.');

谢谢！

推荐答案

根据 kmeans 文档页面我建议这个嵌套逻辑：

Based on the example on the kmeans documentation page I propose this "nested" logic:

X = [randn(100,2)+ones(100,2);...
     randn(100,2)-ones(100,2)];
opts = statset('Display','final');

% This gives a random distribution of 0s and 1s in column 5: 
X(:,5) = round(rand(size(X,1),1));

[idx,ctrs] = kmeans(X,2,...
                    'Distance','city',...
                    'Replicates',5,...
                    'Options',opts);

hold on
plot(X(idx==1,1),X(idx==1,2),'rs','MarkerSize',12)
plot(X(idx==2,1),X(idx==2,2),'r+','MarkerSize',12)

% after plotting the results of kmeans, 
% plot new symbols with a different logic on top:

plot(X(X(idx==1,5)==0,1),X(X(idx==1,5)==0,2),'bs','MarkerSize',12)
plot(X(X(idx==1,5)==1,1),X(X(idx==1,5)==1,2),'gs','MarkerSize',12)
plot(X(X(idx==2,5)==0,1),X(X(idx==2,5)==0,2),'b+','MarkerSize',12)
plot(X(X(idx==2,5)==1,1),X(X(idx==2,5)==1,2),'g+','MarkerSize',12)

$ b b

上面的代码是一个最小的工作示例，因为统计工具箱是可用的。

关键功能是绘图的嵌套逻辑。例如：

The above code is a minimal working example, given that the statistics toolbox is available.
The key feature is the nested logic for the plotting. For example:

X(X(idx==1,5)==0,1)

内部 X（idx == 1,5） X（：，5）为 idx == 1 。从那些，只有 0 的值被考虑： X（X（...）== 0,1）。基于问题中的逻辑，这应该是一个蓝色方块： bs 。

您有四种情况，因此有四条曲线。

The inner X(idx==1,5) selects those values of X(:,5) for which idx==1. From those, only values which are 0 are considered: X(X(...)==0,1). Based on the logic in the question, this should be a blue square: bs.
You have four cases, hence there are four additional plot lines.

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