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发布于 2024-06-17 01:04:03 字数 2853 浏览 0 评论 0 收藏 0

197. Rising Temperature

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Description

Table: Weather

+---------------+---------+
| Column Name   | Type  |
+---------------+---------+
| id      | int   |
| recordDate  | date  |
| temperature   | int   |
+---------------+---------+
id is the column with unique values for this table.
There are no different rows with the same recordDate.
This table contains information about the temperature on a certain day.

 

Write a solution to find all dates' Id with higher temperatures compared to its previous dates (yesterday).

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Weather table:
+----+------------+-------------+
| id | recordDate | temperature |
+----+------------+-------------+
| 1  | 2015-01-01 | 10      |
| 2  | 2015-01-02 | 25      |
| 3  | 2015-01-03 | 20      |
| 4  | 2015-01-04 | 30      |
+----+------------+-------------+
Output: 
+----+
| id |
+----+
| 2  |
| 4  |
+----+
Explanation: 
In 2015-01-02, the temperature was higher than the previous day (10 -> 25).
In 2015-01-04, the temperature was higher than the previous day (20 -> 30).

Solutions

Solution 1: Self-Join + DATEDIFF/SUBDATE Function

We can use self-join to compare each row in the Weather table with its previous row. If the temperature is higher and the date difference is one day, then it is the result we are looking for.

import pandas as pd


def rising_temperature(weather: pd.DataFrame) -> pd.DataFrame:
  weather.sort_values(by="recordDate", inplace=True)
  return weather[
    (weather.temperature.diff() > 0) & (weather.recordDate.diff().dt.days == 1)
  ][["id"]]
# Write your MySQL query statement below
SELECT w1.id
FROM
  Weather AS w1
  JOIN Weather AS w2
    ON DATEDIFF(w1.recordDate, w2.recordDate) = 1 AND w1.temperature > w2.temperature;

Solution 2

# Write your MySQL query statement below
SELECT w1.id
FROM
  Weather AS w1
  JOIN Weather AS w2
    ON SUBDATE(w1.recordDate, 1) = w2.recordDate AND w1.temperature > w2.temperature;

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