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196. Delete Duplicate Emails


Table: Person

| Column Name | Type    |
| id          | int     |
| email       | varchar |
id is the primary key (column with unique values) for this table.
Each row of this table contains an email. The emails will not contain uppercase letters.


Write a solution to delete all duplicate emails, keeping only one unique email with the smallest id.

For SQL users, please note that you are supposed to write a DELETE statement and not a SELECT one.

For Pandas users, please note that you are supposed to modify Person in place.

After running your script, the answer shown is the Person table. The driver will first compile and run your piece of code and then show the Person table. The final order of the Person table does not matter.

The result format is in the following example.


Example 1:

Person table:
| id | email            |
| 1  | john@example.com |
| 2  | bob@example.com  |
| 3  | john@example.com |
| id | email            |
| 1  | john@example.com |
| 2  | bob@example.com  |
Explanation: john@example.com is repeated two times. We keep the row with the smallest Id = 1.


Delete the same email with a large Id.

Or, group according to the email, then use the Min keyword to pick out the smaller ones, and then delete the complement set.

Possible Pitfalls

p1.Id > p2.Id 

here is >, not !=, since

  1. we are keeping smaller-id of duplicates in result
  2. if !=, then all duplicates will be removed, but we are expecting smallest-id duplicate being kept
  • import pandas as pd
    # Modify Person in place
    def delete_duplicate_emails(person: pd.DataFrame) -> None:
        # Sort the rows based on id (Ascending order)
        person.sort_values(by="id", ascending=True, inplace=True)
        # Drop the duplicates based on email.
        person.drop_duplicates(subset="email", keep="first", inplace=True)
  • # Write your MySQL query statement below
    DELETE FROM Person
        SELECT Id
        FROM (SELECT MIN(Id) Id FROM Person GROUP BY Email) p
    DELETE p1.*
        Person p1,
        Person p2
        p1.Email = p2.Email AND p1.Id > p2.Id

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