README_EN.md

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178. Rank Scores

中文文档

Description

Table: Scores

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| score       | decimal |
+-------------+---------+
id is the primary key (column with unique values) for this table.
Each row of this table contains the score of a game. Score is a floating point value with two decimal places.

 

Write a solution to find the rank of the scores. The ranking should be calculated according to the following rules:

• The scores should be ranked from the highest to the lowest.
• If there is a tie between two scores, both should have the same ranking.
• After a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no holes between ranks.

Return the result table ordered by score in descending order.

The result format is in the following example.

 

Example 1:

Input: 
Scores table:
+----+-------+
| id | score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+
Output: 
+-------+------+
| score | rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

Solutions

Solution 1

Python3

import pandas as pd


def order_scores(scores: pd.DataFrame) -> pd.DataFrame:
    # Use the rank method to assign ranks to the scores in descending order with no gaps
    scores["rank"] = scores["score"].rank(method="dense", ascending=False)

    # Drop id column & Sort the DataFrame by score in descending order
    result_df = scores.drop("id", axis=1).sort_values(by="score", ascending=False)

    return result_df

MySQL

# Write your MySQL query statement below
SELECT
    score,
    DENSE_RANK() OVER (ORDER BY score DESC) AS 'rank'
FROM Scores;

Solution 2

MySQL

SELECT
    Score,
    CONVERT(rk, SIGNED) `Rank`
FROM
    (
        SELECT
            Score,
            IF(@latest = Score, @rank, @rank := @rank + 1) rk,
            @latest := Score
        FROM
            Scores,
            (
                SELECT
                    @rank := 0,
                    @latest := NULL
            ) tmp
        ORDER BY
            Score DESC
    ) s;