擅长:python、mysql、java
<blockquote>
<p>I then need to check to see if the points fall inside a circle of
radius one by using squareroot(x^2 + y^2) < 1.</p>
</blockquote>
<p>您可以使用<a href="https://stackoverflow.com/a/48737579/8767209">array filtering</a></p>
<pre><code>pt_norm = (squareDataOne + squareDataTwo)
r_inside_circle = np.sqrt(pt_norm[pt_norm < 1])
</code></pre>
<p>这将给出<code>r_inside_circle</code>中圆内所有点的半径。当您增加值<code>plots</code>时,您将看到<code>(4.0*len(r_inside_circle))/len(dataOne)</code>将接近PI。在</p>