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简单 |
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表: Weather
+---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | recordDate | date | | temperature | int | +---------------+---------+ id 是该表具有唯一值的列。 没有具有相同 recordDate 的不同行。 该表包含特定日期的温度信息
编写解决方案,找出与之前(昨天的)日期相比温度更高的所有日期的 id
。
返回结果 无顺序要求 。
结果格式如下例子所示。
示例 1:
输入: Weather 表: +----+------------+-------------+ | id | recordDate | Temperature | +----+------------+-------------+ | 1 | 2015-01-01 | 10 | | 2 | 2015-01-02 | 25 | | 3 | 2015-01-03 | 20 | | 4 | 2015-01-04 | 30 | +----+------------+-------------+ 输出: +----+ | id | +----+ | 2 | | 4 | +----+ 解释: 2015-01-02 的温度比前一天高(10 -> 25) 2015-01-04 的温度比前一天高(20 -> 30)
我们可以通过自连接的方式,将 Weather
表中的每一行与它的前一行进行比较,如果温度更高,并且日期相差一天,那么就是我们要找的结果。
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;
# 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;