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    <p>If you want to keep with scipy, you have to convex hull (you did so)</p> <pre><code>&gt;&gt;&gt; from scipy.spatial import ConvexHull &gt;&gt;&gt; points = np.random.rand(30, 2) # 30 random points in 2-D &gt;&gt;&gt; hull = ConvexHull(points) </code></pre> <p>then build the list of points on the hull. Here is the code from doc to plot the hull</p> <pre><code>&gt;&gt;&gt; import matplotlib.pyplot as plt &gt;&gt;&gt; plt.plot(points[:,0], points[:,1], 'o') &gt;&gt;&gt; for simplex in hull.simplices: &gt;&gt;&gt; plt.plot(points[simplex,0], points[simplex,1], 'k-') </code></pre> <p>So starting from that, I would propose for computing list of points on the hull</p> <pre><code>pts_hull = [(points[simplex,0], points[simplex,1]) for simplex in hull.simplices] </code></pre> <p>(although i did not try)</p> <p>And you can also come with your own code for computing the hull, returning the x,y points. </p> <p><strong>If you want to know if a point from your original dataset is on the hull</strong>, then you are done.</p> <p><strong>I what you want is to know if a any point is inside the hull or outside</strong>, you must do a bit of work more. What you will have to do could be</p> <ul> <li><p>for all edges joining two simplices of your hull: decide whether your point is above or under</p></li> <li><p>if point is below all lines, or above all lines, it is outside the hull</p></li> </ul> <p>As a speed up, as soon as a point has been above one line and below one another, it is inside the hull.</p>
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    1. POWhat's an efficient way to find if a point lies in the convex hull of a point cloud?
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    1. COI want to find out, if an arbitrary point is in the convex hull of the point-cloud or outside of it. :)
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    2. COso are you satisfied with answer ?
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    3. COYour answer for inside or outside the hull isn't correct in that above and below is not a sufficient test. For example, if a point is just outside the hull but, say, midway along a 45deg diagonal, then your test will fail. Instead, sum the angles between the test point and all the points of the convex hull: if it's inside the angles will sum to 2pi, and if it's outside they'll sum to 0 (or I might have some detail of this wrong, but that's the basic idea).
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