Comparing two un-ordered lists for differences in individual elements

Question

Given two lists:

listA = ['A' , 'B' , 'C, D' , 'E, F, G', 'H' , 'I']

listB = ['C' , 'E, G' , 'A' , 'B' , 'I']

I want to compare each element and highlight the differences where it is applicable.

listA['E,F,G'] and listB['E,G'].

The difference would be ['F']

There are several differences between the individual list elements and ideally would like to flag them all. Is this possible with python? Is the below thinking correct?

set(listA).intersection(listB)

Answer 1

What you are looking for is symmetric difference. In python you can achieve it by using symmetric_difference function or using short hand s ^ t.

s.symmetric_difference(t)

---

This will give you the difference elements. Now, what you can do is

def split_words(element):
   if len(element) > 1:
      element = element.split(',')
   return element

result = []
for e1, e2 in zip(sorted(list_a), sorted(list_b)):
    if e1 not in list_b:
       e1 = split_words(e1)
       e2 = split_words(e2)

       diff = set(e1) ^ set(e2)
       result.append(diff)

Answer 2

From comments, it seems like you want to "unpack" the nested pseudo-string-lists before calculating the difference. You could define a simple helper function for that.

>>> listA = ['A' , 'B' , 'C, D' , 'E, F, G', 'H' , 'I']
>>> listB = ['C' , 'E, G' , 'A' , 'B' , 'I']
>>> elements = lambda s: set(x for y in s for x in y.split(", "))
>>> elements(listA)
{'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I'}

Then, you can use set operations like -, ^, &, or | to get what you want.

>>> elements(listA) - elements(listB) # difference
{'D', 'F', 'H'}
>>> elements(listA) ^ elements(listB) # sym. diff.
{'D', 'F', 'H'}
>>> elements(listA) & elements(listB) # intersection
{'A', 'B', 'C', 'E', 'G', 'I'}
>>> elements(listA) | elements(listB) # union
{'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I'}