How to iterate over two lists with repeating elements

Question

The list_1 have repeated values that are repeated in list_2 I need to find the elements that match from list_1 and list_2 and save them in result_list_1 and result_list_2 respectively and also the ones that do not match and save them in not_equals_list_1 and not_equals_list_2. The result of the items that match must also match in position.

Note: I need to use only arrays and not sets.

list_1 = [1, 5, 2, 3, 2, 4, 1, 3]
list_2 = [1, 2, 2, 3, 1, 6]

It should result in:

result_list_1 = [1, 2, 3, 2, 1]
result_list_2 = [1, 2, 3, 2, 1]

not_equals_list_1 = [3, 4, 5]
not_equals_list_2 = [6]

Your help would be of great help to me.

Answer 1

collections.Counter is invaluable for things like this - like a set but with multiplicity. The algorithm is simple - we make a Counter for each list, and then for the opposite list we just check whether each element is in the other list's counter. If so, remove one occurrence and put it in the result_list. Otherwise, put it in the not_equals_list.

from collections import Counter

list_1 = [1, 5, 2, 3, 2, 4, 1, 3]
list_2 = [1, 2, 2, 3, 1, 6]

counter_1 = Counter(list_1)
counter_2 = Counter(list_2)

result_list_1 = []
result_list_2 = []
not_equals_list_1 = []
not_equals_list_2 = []

for item in list_1:
    if counter_2[item] > 0:
        result_list_1.append(item)
        counter_2[item] -= 1
    else:
        not_equals_list_1.append(item)

for item in list_2:
    if counter_1[item] > 0:
        result_list_2.append(item)
        counter_1[item] -= 1
    else:
        not_equals_list_2.append(item)

print(result_list_1)  # [1, 2, 3, 2, 1]
print(result_list_2)  # [1, 2, 2, 3, 1]
print(not_equals_list_1) # [5, 4, 3]
print(not_equals_list_2) # [6]

Order is preserved in not_equals_list from the order of the first list. If you desire it differently, you can use reversed() where necessary to change either order of iteration or to simply flip the result.

If using this type of solution with custom objects, you'll need to make sure that __hash__() is properly implemented for equality checking - since both sets and dicts are hashtable-based, and Counter is just a subclass of dict.

---

From a quick google search, multiset might provide a more efficient way of doing this by just converting your lists into sets and then doing set operations on them. I haven't tested this, though.

Answer 2

My method would decrease your time complexity at the cost of space complexity.

def taketwoarray(arr1,arr2):
my_dict={}
for i in arr1:# add all arr1 elements to dict
    my_dict[i]=0
for i in arr2:# match if they are in arr1 increase count else set to -1
    try:
        if my_dict[i]!=-1:
            my_dict[i]+=1
    except:
        my_dict[i]=-1
#now you have count of numbers that came in both lists and the numbers themselves
#and the numbers that dont match in either would be marked by zero and -1 count.
# count is set to -1 to avoid numbers coming again and again only in arr2.

Let me know if there is any issue!

Answer 3

result_list_1 = []
result_list_2 = []
not_equals_list_1 = []
not_equals_list_2 = []

for item in list_1:
    if item in list_2:
        result_list_1.append(item)
    else:
        not_equals_list_1.append(item)

for item in list_2:
    if item in list_1:
        result_list_2.append(item)
    else:
        not_equals_list_2.append(item)

Answer 4

from collections import Counter

list_1 = [1, 5, 2, 3, 2, 4, 1, 3]
list_2 = [1, 2, 2, 3, 1, 6]

c1 = Counter(list_1)
c2 = Counter(list_2)

result_list_1 = [*(c1 & c2).elements()]
result_list_2 = result_list_1[:]
not_equals_list_1 = [*(c1 - c2).elements()]
not_equals_list_2 = [*(c2 - c1).elements()]

print(result_list_1)      # [1, 1, 2, 2, 3]
print(result_list_2)      # [1, 1, 2, 2, 3]
print(not_equals_list_1)  # [5, 3, 4]
print(not_equals_list_2)  # [6]