Group unique users with two changing IDs

Can you think of a faster algorithm for this problem? Or improve the code?

Problem:

I have two customer IDs:

• ID1 (e.g. phone number)
• ID2 (e.g. email address)

A user sometimes change their ID1 and sometimes ID2. How can I find unique users?

Example:

ID1 = [7, 7, 8, 9]

ID2 = [a, b, b, c]

Desired result:

ID3 = [Anna, Anna, Anna, Paul]

The real world scenario has ca. 600 000 items per list.

There is already an SQL idea here: How can I match Employee IDs from 2 columns and group them into an array?

And I got help from a friend which had this idea with TypeScript: https://stackblitz.com/edit/typescript-leet-rewkmh?file=index.ts

A second friend of mine helped me with some pseudo-code, and I was able to create this:

Fastest working code so far:

ID1 = [7, 7, 8, 9]
ID2 = ["a", "b", "b", "c"]

def timeit_function(ID1, ID2):
    
    def find_user_addresses():
        phone_i = []
        email_i = []
        
        tmp1 = [ID1[0]]
        tmp2 = []
        tmp_index = []

        while len(tmp1) != 0 or len(tmp2) != 0:
            while len(tmp1) != 0:
                tmp_index = []  
                for index, value in enumerate(ID1):
                    if value == tmp1[0]:
                        tmp2.append(ID2[index])
                        tmp_index.insert(-1, index)

                for i in tmp_index: 
                    del ID1[i]
                    del ID2[i]
                tmp1 = list(dict.fromkeys(tmp1))
                phone_i.append(tmp1.pop(0))

            while len(tmp2) != 0:
                tmp_index = [] 
                for index, value in enumerate(ID2):
                    if value == tmp2[0]:
                        tmp1.append(ID1[index])
                        tmp_index.insert(0, index)

                for i in tmp_index: 
                    del ID1[i]
                    del ID2[i]
                tmp2 = list(dict.fromkeys(tmp2))
                email_i.append(tmp2.pop(0))

        return phone_i, email_i
    
    users = {}
    i = 0
    while len(ID1) != 0:
        phone_i, email_i = find_user_addresses()
        users[i] = [phone_i, email_i]
        i += 1
    return users

Output:

{0: [[7, 8], ['a', 'b']], 1: [[9], ['c']]}

Meaning: {User_0: [[phone1, phone2], [email1, email2]], User_1: [phone3, email3]}

%timeit timeit_function(ID1, ID2)

575 ns ± 3.86 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

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