Get difference between two lists

Question

I have two lists in Python, like these:

temp1 = [\'One\', \'Two\', \'Three\', \'Four\']
temp2 = [\'One\', \'Two\']

I need to create a third

Answer 1

Let's say we have two lists

list1 = [1, 3, 5, 7, 9]
list2 = [1, 2, 3, 4, 5]

we can see from the above two lists that items 1, 3, 5 exist in list2 and items 7, 9 do not. On the other hand, items 1, 3, 5 exist in list1 and items 2, 4 do not.

What is the best solution to return a new list containing items 7, 9 and 2, 4?

All answers above find the solution, now whats the most optimal?

def difference(list1, list2):
    new_list = []
    for i in list1:
        if i not in list2:
            new_list.append(i)

    for j in list2:
        if j not in list1:
            new_list.append(j)
    return new_list

versus

def sym_diff(list1, list2):
    return list(set(list1).symmetric_difference(set(list2)))

Using timeit we can see the results

t1 = timeit.Timer("difference(list1, list2)", "from __main__ import difference, 
list1, list2")
t2 = timeit.Timer("sym_diff(list1, list2)", "from __main__ import sym_diff, 
list1, list2")

print('Using two for loops', t1.timeit(number=100000), 'Milliseconds')
print('Using two for loops', t2.timeit(number=100000), 'Milliseconds')

returns

[7, 9, 2, 4]
Using two for loops 0.11572412995155901 Milliseconds
Using symmetric_difference 0.11285737506113946 Milliseconds

Process finished with exit code 0

Answer 2

I prefer to use converting to sets and then using the "difference()" function. The full code is :

temp1 = ['One', 'Two', 'Three', 'Four'  ]                   
temp2 = ['One', 'Two']
set1 = set(temp1)
set2 = set(temp2)
set3 = set1.difference(set2)
temp3 = list(set3)
print(temp3)

Output:

>>>print(temp3)
['Three', 'Four']

It's the easiest to undersand, and morover in future if you work with large data, converting it to sets will remove duplicates if duplicates are not required. Hope it helps ;-)

Answer 3

Can be done using python XOR operator.

• This will remove the duplicates in each list
• This will show difference of temp1 from temp2 and temp2 from temp1.

---

set(temp1) ^ set(temp2)

Answer 4

def diffList(list1, list2):     # returns the difference between two lists.
    if len(list1) > len(list2):
        return (list(set(list1) - set(list2)))
    else:
        return (list(set(list2) - set(list1)))

e.g. if list1 = [10, 15, 20, 25, 30, 35, 40] and list2 = [25, 40, 35] then the returned list will be output = [10, 20, 30, 15]

Answer 5

most simple way,

use set().difference(set())

list_a = [1,2,3]
list_b = [2,3]
print set(list_a).difference(set(list_b))

answer is set([1])

can print as a list,

print list(set(list_a).difference(set(list_b)))

Answer 6

if you want something more like a changeset... could use Counter

from collections import Counter

def diff(a, b):
  """ more verbose than needs to be, for clarity """
  ca, cb = Counter(a), Counter(b)
  to_add = cb - ca
  to_remove = ca - cb
  changes = Counter(to_add)
  changes.subtract(to_remove)
  return changes

lista = ['one', 'three', 'four', 'four', 'one']
listb = ['one', 'two', 'three']

In [127]: diff(lista, listb)
Out[127]: Counter({'two': 1, 'one': -1, 'four': -2})
# in order to go from lista to list b, you need to add a "two", remove a "one", and remove two "four"s

In [128]: diff(listb, lista)
Out[128]: Counter({'four': 2, 'one': 1, 'two': -1})
# in order to go from listb to lista, you must add two "four"s, add a "one", and remove a "two"