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copied!<p>[Edited to put the simpler approach first.]</p> <p>In practice I'd probably just do:</p> <pre><code>In [166]: d = dict(m_out[:,::-1]) In [167]: [(k, d[k]) for k in eqn_out if k in d] Out[167]: [(7, 6), (9, 8), (11, 10), (13, 12), (15, 14), (17, 16)] </code></pre> <p>But for fun, sticking in numpy, how about something like:</p> <p>[Updated: better numpy method]:</p> <pre><code>In [15]: m_out[np.in1d(v, eqn_out)][:, ::-1] Out[15]: array([[ 7, 6], [ 9, 8], [11, 10], [13, 12], [15, 14], [17, 16]]) </code></pre> <p>Or my original numpy approach:</p> <pre><code>In [150]: import numpy as np In [151]: m_out = np.arange(50).reshape(25,2) In [152]: v = m_out[:,1] In [153]: eqn_out = np.arange(7, 18) In [154]: eqn_out Out[154]: array([ 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]) </code></pre> <p>Keep only the values we know about:</p> <pre><code>In [155]: np.intersect1d(eqn_out, v) Out[155]: array([ 7, 9, 11, 13, 15, 17]) </code></pre> <p>Find where they're located at (assumes the data is sorted!!): </p> <pre><code>In [156]: v.searchsorted(np.intersect1d(eqn_out, v)) Out[156]: array([3, 4, 5, 6, 7, 8]) </code></pre> <p>Use these indices for selection purposes:</p> <pre><code>In [157]: m_out[v.searchsorted(np.intersect1d(eqn_out, v))] Out[157]: array([[ 6, 7], [ 8, 9], [10, 11], [12, 13], [14, 15], [16, 17]]) </code></pre> <p>Flip:</p> <pre><code>In [158]: m_out[v.searchsorted(np.intersect1d(eqn_out, v))][:,::-1] Out[158]: array([[ 7, 6], [ 9, 8], [11, 10], [13, 12], [15, 14], [17, 16]]) </code></pre>

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