Salaries - 腾讯云开发者社区

Department Top Three Salaries Desicription The Employee table holds all employees....| | 2 | Sales | +----+----------+ Write a SQL query to find employees who earn the top three salaries

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MySQL练习三：查找薪水记录超过15次的员工号emp_no以及其对应的记录次数t

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什么是Python中的匿名函数

f=lambda x,n:x ** n print(f(2,3)) lambda匿名函数的应用：**max,min,sorted,map,reduce,filter** 求工资最高的人：max salaries...={ 'egon':3000, 'alex':100000000, 'wupeiqi':10000, 'yuanhao':2000 } def get(k): return salaries...[k] print(max(salaries,key=get)) #'alex' print(max(salaries,key=lambda x:salaries[x])) info = [ {'...# 默认按照字典的键排序 print(salaries) # salaries=sorted(salaries,key=lambda x:salaries[x]) #默认是升序排 alaries=sorted...(salaries,key=lambda x:salaries[x],reverse=True) #降序 print(salaries) info=[ {'name':'egon','age

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SQL练习八

有一个全部员工的薪水表salaries简况如下: 有一个各个部门的领导表dept_manager简况如下: 请你查找各个部门当前领导的薪水详情以及其对应部门编号dept_no，输出结果以salaries.emp_no...升序排序，并且请注意输出结果里面dept_no列是最后一列，以上例子输出如下: 表结构 drop table if exists `salaries` ; drop table if exists...`dept_manager` ; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `...'9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries...'9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries

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MySQL练习是一：查找当前薪水详情以及部门编号dept_no

表结构：数据表 drop table if exists `salaries` ; drop table if exists `dept_manager` ; CREATE TABLE `salaries...VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02',...'9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries...VALUES(10004,74057,'2001-11-27','9999-01-01'); INSERT INTO salaries VALUES(10005,94692,'2001-09-09',...'9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries

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8.scala高阶函数

在第3行，函数doubleSalary被应用在列表salaries中的每一个元素。...为了简化压缩代码，我们可以使用匿名函数，直接作为参数传递给map: val salaries = Seq(20000, 70000, 40000) val newSalaries = salaries.map...对于上述代码，一种更惯用的写法为： val salaries = Seq(20000, 70000, 40000) val newSalaries = salaries.map(_ * 2) 既然Scala...[Double]): List[Double] = salaries.map(salary => salary * 1.1) def greatPromotion(salaries: List...: Double => Double): List[Double] = salaries.map(promotionFunction) def smallPromotion(salaries

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数据可视化-Matplotlib在线图上填充区域

rcParams['font.sans-serif'] = 'Microsoft YaHei' data = pd.read_csv('data.csv') ages = data['Age'] dev_salaries...= data['All_Devs'] py_salaries = data['Python'] js_salaries = data['JavaScript'] plt.plot(ages,dev_salaries...,color='#444444', linestyle='--',label='所有开发人员薪水' ) plt.plot(ages,py_salaries,label='Python开发薪水..., dev_salaries, where=(py_salaries>dev_salaries), interpolate=True,color='yellow',alpha=0.25, label...,dev_salaries, where=(py_salaries<=dev_salaries), interpolate=True,color='red',alpha=0.25, label

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MySQL练习三：查找当前有薪水记录的员工以及部门信息

] 表结构：数据表 drop table if exists `salaries` ; drop table if exists `dept_manager` ; CREATE TABLE `...salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date...VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02',...'9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries...'9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries

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数据可视化-Matplotlib使用子图绘制数据

import font_manager plt.style.use('seaborn') data = pd.read_csv('data.csv') ages = data['Age'] dev_salaries...= data['All_Devs'] py_salaries = data['Python'] js_salaries = data['JavaScript'] #创建子图 fig, (ax1,ax2...) = plt.subplots(nrows=2,ncols=1) ax1.plot(ages, dev_salaries, color='#444444', linestyle='-...-', label='All Devs') ax2.plot(ages, py_salaries, label='Python') ax2.plot(ages, js_salaries, label='

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牛客网数据库实战题解题思路及答案

'9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries...'9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries...salaries.to_date,dept_no from salaries inner join dept_manager on salaries.emp_no = dept_manager.emp_no...-25'); INSERT INTO salaries VALUES(10001,66074,'1988-06-25','1989-06-25'); INSERT INTO salaries VALUES...情况 drop table if exists `salaries` ; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary`

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入门MySQL——查询语法练习

salaries ：薪资表，记录每个员工每段时期的薪资。 titles ：职称表，记录每个员工每段时期的职位名称。...-+5 rows in set (0.11 sec) 总结：limit限定显示前多少行，可与order by联合使用 5.聚合函数 #查找某员工薪水总和SELECT SUM(salary) FROM salaries...INNER JOIN dept_empON salaries.emp_no = dept_emp.emp_noWHERE salaries.emp_no = 10010; SELECT *FROM salaries...LEFT JOIN dept_empON salaries.emp_no = dept_emp.emp_noWHERE salaries.emp_no = 10010; SELECT *FROM salaries...RIGHT JOIN dept_empON salaries.emp_no = dept_emp.emp_noWHERE salaries.emp_no = 10010; 总结：a left join

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MySQL 调优 | OPTIMIZER_TRACE 详解

`emp_no` AS `emp_no`,`salaries`.`salary` AS `salary`,`salaries`....`from_date` AS `from_date`,`salaries`.`to_date` AS `to_date` from `salaries` where ((`salaries`....`salary` AS `salary`,`salaries`.`from_date` AS `from_date`,`salaries`....`to_date` AS `to_date` from `salaries` where ((`salaries`....": "((`salaries`.

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MySQL练习十三：找出所有员工当前薪水salary情况

题目：有一个薪水表，salaries简况如下: 请你找出所有员工具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示，以上例子输出如下: D272D88115F2A8870C9D588A098CDD57...数据表：表结构 drop table if exists `salaries` ; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary...date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); 数据：insert语句 INSERT INTO salaries...VALUES(10001,72527,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02',...参考答案： select salary from salaries group by salary order by salary desc;

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剑指offer数据库语句

from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `salaries...,salaries.salary,salaries.from_date,salaries.to_date,dept_manager.dept_no from salaries inner join dept_manager...on dept_manager.emp_no = salaries.emp_no and dept_manager.to_date = '9999-01-01' and salaries.to_date...查找所有员工入职时候的薪水情况，给出emp_no以及salary，并按照emp_no进行逆序 SELECT e.emp_no,s.salary FROM employees e LEFT JOIN salaries...s.salary)>15; 找出所有员工当前(to_date='9999-01-01')具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示 SELECT s.salary FROM salaries

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MySQL练习十二：查找薪水记录超过15次的员工号emp_no以及其对应的记录次数t

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等值连接和非等值连接

以下是一个使用非等值连接的示例：假设我们有两个表employees和salaries，它们分别存储了员工和薪资的信息。...employees表包含id、name和department三个列，salaries表包含id、employee_id和salary三个列。现在，我们想查询每个部门中薪资高于平均水平的员工。...employees.id = salaries.employee_idWHERE salaries.salary > (SELECT AVG(salary) FROM salaries);执行结果如下...departmentnamesalaryAAlice80000ABob85000BCharlie90000BDavid95000在这个查询中，我们使用了INNER JOIN关键字将employees表和salaries...表连接起来，并使用ON子句指定连接条件为employees.id = salaries.employee_id。

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通过一条简单的SQL 来理解MYSQL的解析SQL的过程

首先一个SQL 语句，会经历二个阶段， 1 解析， 2 执行计划生成 select * from employees as e left join salaries as s on e.emp_no...salaries 有 2835359 条记录，并且走的是 salaries 表的主键，（emp_no from_date） ?...—————————————————————————— | select * from employees as e left join salaries as s on e.emp_no = s.emp_no...`to_date` AS `to_date` from (`employees` `e` left join `salaries` `s` on((`e`.`emp_no` = `s`....`to_date` AS `to_date` from `employees` `e` left join `salaries` `s` on((`e`.`emp_no` = `s`.

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Python基础10-函数的递归

[name] res=max(salaries,key=func) print(res) # 使用匿名函数 salaries={ 'zls':10000000, 'qls':10000..., 'lls':11111, 'zhang3':100, 'li4':1 } res=max(salaries,key=lambda name:salaries[name])...':100, 'li4':1 } res=min(salaries,key=lambda name:salaries[name]) print(res) # 需求三：使用 sorted 排序...'zhang3':100, 'li4':1 } res=sorted(salaries,key=lambda name:salaries[name]) print(res) res=sorted...(salaries,key=lambda name:salaries[name],reverse=True) print(res) # 需求四：使用 map 映射，把每一个人名加上DSB names=

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「R」ggplot2数据可视化

我们依旧以Salaries数据集来进行相关探索。...首先，查看薪水是如何随学术等级变化的： data(Salaries, package='car') library(ggplot2) ggplot(data=Salaries, aes(x=salary...Salaries density by Rank.png 接下来，我们通过性别和学术等级分组，绘制获得博士学位年数和薪水的关系： ggplot(Salaries, aes(x=yrs.since.phd...data(Salaries, package='car') library(ggplot2) ggplot(data=Salaries, aes(x=yrs.since.phd, y=salary))...data(Salaries, package = 'car') library(ggplot2) p1 <- ggplot(data=Salaries, aes(x=rank)) + geom_bar(

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数据科学家赚多少？基于pandasql和plotly的薪资分析与可视化 ⛵

实战数据集下载（百度网盘）：公众号『ShowMeAI研究中心』回复『实战』，或者点击这里获取本文 [37]基于pandasql和plotly的数据科学家薪资分析与可视化『ds_salaries数据集...# Loading data salaries = pd.read_csv('ds_salaries.csv') 要查看前五个记录，我们可以使用 salaries.head() 方法。...LIMIT 5 """) 输出：图片数据预处理我们数据集中的第1列“Unnamed: 0”是没有用的，在分析之前我们把它剔除： salaries = salaries.drop...(names=salaries["employee_residence"], to="name") salaries["company_location"] = coco.convert(names=salaries...df = salaries df["company_country"] = coco.convert(names = salaries["company_location"], to = 'name_short

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LeetCode 0185 - Department Top Three Salaries

MySQL练习三：查找薪水记录超过15次的员工号emp_no以及其对应的记录次数t

什么是Python中的匿名函数

SQL练习八

MySQL练习是一：查找当前薪水详情以及部门编号dept_no

8.scala高阶函数

数据可视化-Matplotlib在线图上填充区域

MySQL练习三：查找当前有薪水记录的员工以及部门信息

数据可视化-Matplotlib使用子图绘制数据

牛客网数据库实战题解题思路及答案

入门MySQL——查询语法练习

MySQL 调优 | OPTIMIZER_TRACE 详解

MySQL练习十三：找出所有员工当前薪水salary情况

剑指offer数据库语句

MySQL练习十二：查找薪水记录超过15次的员工号emp_no以及其对应的记录次数t

等值连接和非等值连接

通过一条简单的SQL 来理解MYSQL的解析SQL的过程

Python基础10-函数的递归

「R」ggplot2数据可视化

数据科学家赚多少？基于pandasql和plotly的薪资分析与可视化 ⛵

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