根据每个列表的子集从列表列表中删除重复项

Question

I wrote a function to remove "duplicates" from a list of list.

The elements of my list are:

[ip, email, phone number].

I would like to remove the sublists that got the same EMAIL and PHONE NUMBER, I don't really care about the IP address.

The solution that I currently use is :

def remove_duplicate_email_phone(data):
    for i in range(len(data)):
        for j in reversed(range(i+1,len(data))):
            if data[i][1] == data[j][1] and data[i][2] == data[j][2] :
                data.pop(j)
    return data

I would like to optimize this. It took more than 30 minutes to get the result.

Answer 1

Your approach does a full scan for each and every element in the list, making it take O(N**2) (quadratic) time. The list.pop(index) is also expensive as everything following index is moved up, making your solution approach O(N**3) cubic time.

Use a set and add (email, phonenumber) tuples to it to check if you already have seen that pair; testing containment against a set takes O(1) constant time, so you can clean out dupes in O(N) total time:

def remove_duplicate_email_phone(data):
    seen = set()
    cleaned = []
    for ip, email, phone in data:
        if (email, phone) in seen:
            continue
        cleaned.append([ip, email, phone])
        seen.add((email, phone))
    return cleaned

This produces a new list, the old list is left untouched.

Answer 2

Another solution might be to use groupby.

from itertools import groupby
from operator import itemgetter

deduped = []

data.sort(key=itemgetter(1,2))
for k, v in groupby(data, key=itemgetter(1,2):
    deduped.append(list(v)[0])

or using a list comprehension:

deduped = [next(v) for k, v in groupby(data, key=itemgetter(1,2))]

Answer 3

Another approach could be to use a Counter

from collections import Counter

data = [(1, "a@b.com", 1234), (1, "a@b.com", 1234), (2, "a@b.com", 1234)]
counts = Counter([i[:2] for i in data])

print [i for i in data if counts[i[:2]] == 1]  # Get unique