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    <p>You can try something like this:</p> <pre><code>list1=[[['B', 10], 1], [['C', 15], 1], [['F', 30], 1]] list2=[[['G', 20], 2], [['D', 25], 1]] list2_iter = iter(list2) for item1 in list1: in_list2 = False for item2 in list2_iter: if item1[1] == item2[1]: print "Match" if item1 == item2 else "Modified", item1, item2 in_list2 = True break else: print "Inserted", item2 if not in_list2: print "Deleted", item1 </code></pre> <p>Note that the inner loop for <code>list2</code> is using an iterator, so it does not start from the beginning of <code>list2</code> each time, but from the last element it stopped in the previous iteration of the outer loop. Everything else is rather straightforward.</p> <p>Output:</p> <pre><code>Inserted [['G', 20], 2] Modified [['B', 10], 1] [['D', 25], 1] Deleted [['C', 15], 1] Deleted [['F', 30], 1] </code></pre> <p>Note that this will not always find the best match, as it will only do one pass through both, <code>list1</code> and <code>list2</code>. For a more complete algorithm, take a look at <a href="http://en.wikipedia.org/wiki/Diff#Algorithm" rel="nofollow">Diff</a> and <a href="http://en.wikipedia.org/wiki/Longest_common_subsequence_problem" rel="nofollow">Longest Common Subsequence</a>.</p> <hr> <p><em>Update:</em> I just realized how this problem is in fact virtually the same as finding the <a href="https://en.wikipedia.org/wiki/Levenshtein_distance" rel="nofollow">minimum edit distance</a> of two strings, for which I just happened to have some code lying around. After some generalization:</p> <pre><code>import operator def diff(s1, s2, match=operator.eq, neutral="*", subst=2): """s1, s2: the two strings or lists to match match: function to determine whether two elements match up neutral: 'neutral' element for padding subst: substitution costs """ s1, s2 = neutral + s1, neutral + s2 # calculate edit distance / sequence match with DP A = [[0] * len(s2) for i in range(len(s1))] for i in range(len(s1)): for k in range(len(s2)): if min(i, k) == 0: A[i][k] = max(i, k) else: diag = 0 if match(s1[i], s2[k]) else subst A[i][k] = min(A[i-1][k-1] + diag, A[i ][k-1] + 1, A[i-1][k ] + 1) # reconstruct path path, i, k = [], len(s1)-1, len(s2)-1 while i or k: if A[i][k] == A[i-1][k] + 1: path, i = [-1] + path, i-1 elif A[i][k] == A[i][k-1] + 1: path, k = [1] + path, k-1 else: path, i, k = [0] + path, i-1, k-1 return A[len(s1)-1][len(s2)-1], path def print_match(list1, list2, path): i1, i2 = iter(list1), iter(list2) for p in path: if p == -1: print "DEL %20r" % next(i1) if p == 0: print "EQ %20r %20r" % (next(i1), next(i2)) if p == +1: print "INS %41r" % next(i2) # with strings word1, word2 = "INTENTION", "EXECUTION" x, path = diff(word1, word2) print_match(word1, word2, path) # with your lists of lists list1 = [[['B', 10], 1], [['C', 15], 1], [['F', 30], 1]] list2 = [[['G', 20], 2], [['D', 25], 1]] x, path = diff(list1, list2, match=lambda x, y: x[1] == y[1], neutral=[[None, -1]]) print_match(list1, list2, path) </code></pre> <p>Note that the resulting match can <em>still</em> be different from what you expect, since often there are many equally optimal ways to match the elements, but it should be optimal nonetheless.</p>
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    1. CO@ tobias:Thanks that's what I wanted to do.
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    2. CONow when I am using list1=[[['B', 10], 2], [['C', 15], 1]] list2=[[['B', 10], 1], [['C', 15], 1]] Output is: Added [['B', 10], 1] Added [['C', 15], 1] Deleted [['B', 10], 2] Deleted [['C', 15], 1] According to me it should be: Modified: ['C', 15] ['B', 10] Deleted ['B', 10] Added: ['B', 10]
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    3. COI am sorry but this doesn't work for every case
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