[matlab] 2.数据可视化

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选择匿名的用户   2021-5-22 18:38   81   0
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   <pre class="blockcode">t&#61;(0:0.01:2)*<span style="color:#000000;">pi;
x</span>&#61;<span style="color:#000000;">sin(t);
y</span>&#61;<span style="color:#000000;">cos(t);
z</span>&#61;cos(2*<span style="color:#000000;">t);
plot3(x,y,z,</span><span style="color:#800000;">&#39;</span><span style="color:#800000;">r-</span><span style="color:#800000;">&#39;</span>,<span style="color:#800000;">&#39;</span><span style="color:#800000;">linewidth</span><span style="color:#800000;">&#39;</span>,1.5<span style="color:#000000;">);
box on; </span>%<span style="color:#000000;">打开坐标系边框
legend(</span><span style="color:#800000;">&#39;</span><span style="color:#800000;">三维曲线作图</span><span style="color:#800000;">&#39;</span>);</pre>
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  <span class="cnblogs_code_collapse">三维曲线绘图</span>
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<p><img alt="" src="https://beijingoptbbs.oss-cn-beijing.aliyuncs.com/cs/5606289-da8f3f845d8068b8bb3170ce7a17ff21.png"></p>
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   <pre class="blockcode">x&#61;0:0.2:1<span style="color:#000000;">;
y</span>&#61;<span style="color:#000000;">x;
[x y]</span>&#61;<span style="color:#000000;">meshgrid(x,y);
plot(x,y,</span><span style="color:#800000;">&#39;</span><span style="color:#800000;">b&#43;</span><span style="color:#800000;">&#39;</span><span style="color:#000000;">)
</span>%这就是所有(x,y)的取值 ,然后计算所有的z&#61;f(x,y)就可以了</pre>
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  <span class="cnblogs_code_collapse">利用meshgrid生成网格节点</span>
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<p> <img alt="" src="https://beijingoptbbs.oss-cn-beijing.aliyuncs.com/cs/5606289-72156db56d54093c26aadcb48faf34bb.png"></p>
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   <pre class="blockcode">dd&#61;0.001;n&#61;1&#43;1/<span style="color:#000000;">dd;
[x,y]</span>&#61;meshgrid(0:dd:1,0:dd:1<span style="color:#000000;">);
z</span>&#61;<span style="color:#000000;">0;
</span><span style="color:#0000ff;">for</span> j&#61;1<span style="color:#000000;">:n
    </span><span style="color:#0000ff;">for</span> i&#61;1<span style="color:#000000;">:n
        </span><span style="color:#0000ff;">if</span> x(i,j)&lt;sqrt(1-y(i,j).^2<span style="color:#000000;">)
            z(i,j)</span>&#61;x(i,j).^2&#43;y(i,j).^2<span style="color:#000000;">;
        </span><span style="color:#0000ff;">else</span><span style="color:#000000;">
            z(i,j)</span>&#61;<span style="color:#000000;">0;
        end
    end
end
surf(x,y,z);
shading flat;</span></pre>
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  <span class="cnblogs_code_collapse">z&#61;x^2&#43;y^2被圆柱面x^2&#43;y^2&#61;1以及三坐标平面所截得的在第一卦限的图像</span>
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<p><img alt="" src="https://beijingoptbbs.oss-cn-beijing.aliyuncs.com/cs/5606289-e55c0ab63b08ea3c4a999140581a4380.png"></p>
<p> </p>
<p><a href="https://pan.baidu.com/s/1HUhgTi-cstetsKYTDhEgVw" title="附件">附件链接</a>( 密码ju5z ) </p>
<p>下载附件 a.csv 给出了高程数据,共有 2913 列,2775 行</p>
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  <div class="cnblogs_code_hide" id="cnblogs_code_open_4d8cf2a1-3896-4358-9754-9d1d6cdf1bc9">
   <pre class="blockcode">filename &#61; <span style="color:#800000;">&#39;</span><span style="color:#800000;">a.csv</span><span style="color:#800000;">&#39;</span><span style="color:#000000;">;
z </span>&#61;<span style="color:#000000;"> csvread(filename);
x0</span>&#61;1:1:2775<span style="color:#000000;">;
y0</span>&#61;1:1:2913<span style="color:#000000;">;
[xx0,yy0]</span>&#61;<span style="color:#000000;">meshgrid(x0,y0);
xx</span>&#61;xx0<span style="color:#800000;">&
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