compare 2 lists in python

 a = [u'120 devon 15.61', u'126 korn05000000000000 12.96', u'128 itswat 20.58',] b = [u'120', u'devon', u'15.61', u'126', u'korn05', u'12.96', u'128', u'itswat', u'20.58'] x = u' '.join(a) y = u' '.join(b) if(x == y): pass else: a1 = x.split(' ') b1 = y.split(' ') for index, value in enumerate(a1): if (b1[index] != value): print(index) 

 a = [u'120 devon 15.61', u'126 korn05000000000000 12.96', u'128 itswat 20.58',] b = [u'120', u'devon', u'15.61', u'126', u'korn05', u'12.96', u'128', u'itswat', u'20.58'] x = u' '.join(a).split() y = u' '.join(b).split() for index, value in enumerate(x): if (y[index] != value): print(index) 

 a = ' '.join(a).split() b = ' '.join(b).split() index_list = [a.index(x) for x, y in zip(a, b) if not x == y] 

if the length of the lists is different

 a = iter(' '.join(a).split()) b = iter(' '.join(b).split()) for i, x in enumerate(a): try: if not x == next(b): print(i) except StopIteration: # len(a) > len(b) print(i) try: while next(b): # len(a) < len(b) i += 1 print(i) except StopIteration:pass 

To find positions in which items with the same index differ in a pair of lists:

 a = [1, 2, 3, 4] b = [1, 20, 30] indices_uniq = [i for i, (x, y) in enumerate(zip(a, b)) if x != y] # -> [1, 2] 

Result [1, 2] as the second and third elements are different. Only i < min(len(a), len(b)) itertools.zip_longest(a, b, fill_value=sentinel) are itertools.zip_longest(a, b, fill_value=sentinel) , since there is no match for the shorter list of elements (otherwise, itertools.zip_longest(a, b, fill_value=sentinel) be used instead of zip(a,b) ).

For the literal input in the question, the result is [0,1,2] , that is, all elements are distinguished in pairs in the given lists. If you want to compare the input as a sequence of words separated by a space, you first need to normalize the input ( get_words() ), and then compare the corresponding words ( unequal_indicies(a, b) ):

 def get_words(chunks): return (word for chunk in chunks for word in chunk.split()) def unequal_indices(a, b): return (i for i, (x, y) in enumerate(zip(a, b)) if x != y) a = ["120 devon 15.61", "126 korn05000000000000 12.96", "128 itswat 20.58"] b = ["120", "devon", "15.61", "126", "korn05", "12.96", "128", "itswat", "20.58"] print(list(unequal_indices(get_words(a), get_words(b)))) # -> [4] 

In this case (when comparing by words) the result [4] , that is, only the fifth words in a given pair of texts differ.

In tests , you can use self.assertEqual(a, b) , which will show the differences between lists (Python 3), if any:

 Traceback (most recent call last): File "./__main__.py", line 13, in test self.assertEqual(a, b) AssertionError: Lists differ: [1, 2, 3, 4] != [1, 20, 30] First differing element 1: 2 20 First list contains 1 additional elements. First extra element 3: 4 - [1, 2, 3, 4] ? ^^^ + [1, 20, 30] ? + ^ 

To compare sequences containing text, you can difflib to try :

 >>> print(*difflib.ndiff(a, b), sep='\n') - 120 devon 15.61 - 126 korn05000000000000 12.96 - 128 itswat 20.58 + 120 + devon + 15.61 + 126 + korn05 + 12.96 + 128 + itswat + 20.58 

The output shows that all elements are different.

If you normalize input, then only one word is expected to be different:

 >>> print(*difflib.ndiff(list(get_words(a)), list(get_words(b))), sep='\n') 120 devon 15.61 126 - korn05000000000000 + korn05 12.96 128 itswat 20.58