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copied!<h2>Updated Solution:</h2> <p>From the comments below, the new solution is:</p> <pre><code>idx = argsort(A[:, 0:2], axis=0)[:,1] kidx = where(sum(A[idx,:][:-1,0:2]!=A[idx,:][1:,0:2], axis=1)==0)[0] kidx = unique(concatenate((kidx,kidx+1))) for n in arange(0,A.shape[0],1): if n not in kidx: print A[idx,:][n] > [1 3 4 5 6 7] [3 3 3 3 3 3] [3 4 6 7 7 6] [456 6 5 343 435 5] </code></pre> <p><code>kidx</code> is a index list of the elements you <em>don't</em> want. This preserves rows where the first two inner elements do not match any other inner element. Since everything is done with indexing, it should be fast(ish), though it requires a sort on the first two elements. Note that original row order is not preserved, though I don't think this is a problem.</p> <h2>Old Solution:</h2> <p>If I understand it correctly, you simply want to filter out the results of a list of lists where the first element of each inner list is equal to the second element.</p> <p>With your input from your update <code>A=[[1,1,3,5,6,6],[1,1,4,4,5,6],[1,3,4,5,6,7],[3,4,6,7,7,6],[1,1,4,6,88,7],[3,3,3,3,3,3],[456,6,5,343,435,5]]</code>, the following line removes <code>A[0]</code>,<code>A[1]</code> and <code>A[4]</code>. <code>A[5]</code> is also removed since that seems to match your criteria. </p> <pre><code>[x for x in A if x[0]!=x[1]] </code></pre> <p>If you can use numpy, there is a really slick way of doing the above. Assume that <code>A</code> is an array, then</p> <pre><code>A[A[0,:] == A[1,:]] </code></pre> <p>Will pull out the same values. This is probably faster than the solution listed above if you want to loop over it.</p>

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