Two list comparison in Python -

i have 2 lists:

list1=[[['b', 10], 1], [['c', 15], 1], [['f', 30], 1]] list2=[[['g', 20], 2], [['d', 25], 1]] 

now [['b', 10], 1]] in list1 matches [['d', 25], 1]] in list2 because have equal second element ([1] , [1], first match).

i want ['b', 10] , ['d', 25] modified , [['c', 15], 1] , [['f', 30], 1] deleted list1 [['g', 20], 2] added list2.

can me here? i've tried sets , iteration of lists , comparing both doesn't work.

note: not homework, wondering how list works.

you can try this:

list1=[[['b', 10], 1], [['c', 15], 1], [['f', 30], 1]] list2=[[['g', 20], 2], [['d', 25], 1]]  list2_iter = iter(list2) item1 in list1:     in_list2 = false     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 

note inner loop list2 using iterator, not start beginning of list2 each time, last element stopped in previous iteration of outer loop. else rather straightforward.

output:

inserted [['g', 20], 2] modified [['b', 10], 1] [['d', 25], 1] deleted [['c', 15], 1] deleted [['f', 30], 1] 

note not find best match, 1 pass through both, list1 , list2. more complete algorithm, take @ diff , longest common subsequence.

---

update: realized how problem in fact virtually same finding minimum edit distance of 2 strings, happened have code lying around. after generalization:

import operator  def diff(s1, s2, match=operator.eq, neutral="*", subst=2):     """s1, s2: 2 strings or lists match     match: function determine whether 2 elements match     neutral: 'neutral' element padding     subst: substitution costs     """     s1, s2 = neutral + s1, neutral + s2      # calculate edit distance / sequence match dp     = [[0] * len(s2) in range(len(s1))]     in range(len(s1)):         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 or k:         if a[i][k] == a[i-1][k] + 1:             path, = [-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)     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)  # strings word1, word2 = "intention", "execution" x, path = diff(word1, word2) print_match(word1, word2, path)  # 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) 

note resulting match can still different expect, since there many equally optimal ways match elements, should optimal nonetheless.