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    <p>Provided the list of points remains sufficiently small, a linear search should do the trick:</p> <pre><code>def dist_sq(a, b): # distance squared (don't need the square root) return (a[0] - b[0])**2 + (a[1] - b[1])**2 def find(l, coord): return min(l, key=lambda p:dist_sq(coord, p)) l = [(35.9879845760485, -4.74093235801354), (35.9888687992442, -4.72708076713794), (35.9889733432982, -4.72758983150694), (35.9915751019521, -4.72772881198689), (35.9935223025608, -4.72814213543564), (35.9941433944962, -4.72867416528065), (35.9946670576458, -4.72915181755908), (35.995946587966, -4.73005565674077), (35.9961479762973, -4.7306870912609), (35.9963563641681, -4.7313535758683), (35.9968685892892, -4.73182757975504), (35.9976738530666, -4.73194429867996) ] coord = (35.9945570576458, -4.73110757975504) print find(l, coord) </code></pre> <p>The same solution using <code>numpy</code>:</p> <pre><code>import numpy as np l = np.array([(35.9879845760485, -4.74093235801354), (35.9888687992442, -4.72708076713794), (35.9889733432982, -4.72758983150694), (35.9915751019521, -4.72772881198689), (35.9935223025608, -4.72814213543564), (35.9941433944962, -4.72867416528065), (35.9946670576458, -4.72915181755908), (35.995946587966, -4.73005565674077), (35.9961479762973, -4.7306870912609), (35.9963563641681, -4.7313535758683), (35.9968685892892, -4.73182757975504), (35.9976738530666, -4.73194429867996) ]) coord = np.array((35.9945570576458, -4.73110757975504)) print l[np.argmin(np.apply_along_axis(np.linalg.norm, 1, l - coord))] </code></pre> <p>If that's not feasible, I suggest you look into <a href="http://en.wikipedia.org/wiki/Nearest_neighbor_search#Methods" rel="nofollow">better algorithmic approaches</a>.</p>
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    1. COIf you don't do this very often, or if the list changes frequently, it's likely that the overhead cost of a "better algorithmic approach" will outweigh the benefit, even with a very long list. Every point in the list will have to be processed at least once, whatever approach you use; the question is whether information about the relative position of points in the list is stored in a way that enables quick look-up.
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    2. COthanks! I like the numpy solution! Will evaluate it in terms of performance with @samplebias solution
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