很久没有写AI或者Python相关的代码，毕竟现在已经不是一个算法工程师了。所以就用白嫖版的leetcode练练手。

先丢个代码，再慢慢填坑把

import pandas as pd# 1.大的国家，选出面基至少为300万或者人口至少为2500万的国家
# 简单的pandas过滤
def big_countries(world: pd.DataFrame) -> pd.DataFrame:df = world[(world['area'] >= 3000000) | (world['population'] >= 25000000)]return df[['name', 'population', 'area']]# 1757 可回收且低脂的产品，过滤出可回收且低脂
def find_products(products: pd.DataFrame) -> pd.DataFrame:df = products[(products['low_fats'] == 'Y') & (products['recyclable'] == 'Y')]return df[['product_id']]# 183 从不订购的客户，考察且或非中的非，用~表示，而不是用Not表示
def find_customers(customers: pd.DataFrame, orders: pd.DataFrame) -> pd.DataFrame:df = customers[~customers['id'].isin(orders['customerId'])]df = df[['name']].rename(columns={'name': 'Customers'})return df# 1148 文章浏览1， 考察的不同列相等
def article_views(views: pd.DataFrame) -> pd.DataFrame:df = views[views['author_id'] == views['viewer_id']][['author_id']]df = df.rename(columns={'author_id': 'id'})df = df.drop_duplicates()df = df.sort_values(by='id')return df# 1683. 无效的推文，对比不同列的值进行过滤
def article_views(views: pd.DataFrame) -> pd.DataFrame:views = views[views['author_id'] == views['viewer_id']]['author_id']views = views.rename({'author_id': 'id'})views.drop_duplicates(inplace=True)views = views.sort_values(by='id')return views['id']# 1683. 无效的推文，列转str，使用str的方法进行过滤
def invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:return tweets[tweets['content'].str.len() > 15][['tweet_id']]# 1873. 计算特殊奖金, 通过apply的方式进行复杂的过滤
def calculate_special_bonus(employees: pd.DataFrame) -> pd.DataFrame:employees['bonus'] = employees.apply(lambda x: x['salary'] if x['employee_id'] % 2 and not x['name'].startswith('M') else 0,axis=1)df = employees[['employee_id', 'bonus']].sort_values('employee_id')return df# 1667. 修复表中的名字 这边注意，使用apply的话，本身就是str，不用强制转
def fix_names(users: pd.DataFrame) -> pd.DataFrame:users['name'] = users.apply(lambda x: x['name'][0].upper() + x["name"][1:].lower(),axis=1)return users.sort_values('user_id')# 1517. 查找拥有有效邮箱的用户 正则表达式的应用
def valid_emails(users: pd.DataFrame) -> pd.DataFrame:return users[users["mail"].str.match(r"^[a-zA-Z][a-zA-Z0-9_.-]*\@leetcode\.com$")]# 1527. 患某种疾病的患者 字符串contains正则表达
def find_patients(patients: pd.DataFrame) -> pd.DataFrame:return patients[patients['conditions'].str.contains(r'\bDIAB1')]# 177. 第N高的薪水
def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:df = employee[['salary']].drop_duplicates()key = 'getNthHighestSalary(' + str(N) + ')'if len(df) < N:#这边不能直接返回Nonereturn pd.DataFrame({key: [None]})# 这边类似数据库的count N limit 1df = df.sort_values(by='salary', ascending=False).head(N).tail(1)return df.rename(columns={'salary': key})# 176. 第二高的薪水
def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:df = employee[['salary']].drop_duplicates()if len(df) == 1:return pd.DataFrame({'SecondHighestSalary': None})df = df.sort_values(by='salary', ascending=False).head(2).tail(1)# 这边就换种写法把，不用rename了return pd.DataFrame({'SecondHighestSalary': df['salary'][0]})def department_highest_salary(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:# 类似sql的left join on a.departmentId = b.iddf = employee.merge(department, left_on='departmentId', right_on='id', how='left')df.rename(columns={'name_x': 'Employee', 'name_y': 'Department', 'salary': 'Salary'}, inplace=True)# 先找出最高工资，groupby分组, transform是执行某个函数，可以用lambda自己定义，也可以输入约定速成的几个max_salary = df.groupby('Department')['Salary'].transform('max')# 过滤工资等于最高工资的员工df = df[df['Salary'] == max_salary]return df[['Department', 'Employee', 'Salary']]# 178. 分数排名
def order_scores(scores: pd.DataFrame) -> pd.DataFrame:# 类似sql的 dense_rank over (partition score order by score desk)scores['rank'] = scores['score'].rank(method='dense', ascending=False)return scores[['score', 'rank']].sort_values('score', ascending=False)# 196. 删除重复的电子邮箱
def delete_duplicate_emails(person: pd.DataFrame) -> None:person.sort_values('id', inplace=True)person.drop_duplicates(['email'], keep='first', inplace=True)return# 1795. 每个产品在不同商店的价格
def rearrange_products_table(products: pd.DataFrame) -> pd.DataFrame:df =products.set_index(['product_id']).stack(dropna=True)df = df.reset_index()df.columns = ['product_id', 'store', 'price']return df# 1907. 按分类统计薪水
def count_salary_categories(accounts: pd.DataFrame) -> pd.DataFrame:low = (accounts['income'] < 20000).sum()avg = ((accounts['income'] <= 50000) & (accounts['income'] >= 20000)).sum()high = (accounts['income'] > 50000).sum()return pd.DataFrame({'category': ['Low Salary', 'Average Salary', 'High Salary'],'accounts_count': [low, avg, high]})# 1741. 查找每个员工花费的总时间
def total_time(employees: pd.DataFrame) -> pd.DataFrame:df = employees.groupby(by=['event_day', 'emp_id']).sum().reset_index().rename(columns={'event_day': 'day'})df['total_time'] = df['out_time'] - df['in_time']df.drop(columns=['in_time', 'out_time'], inplace=True)return df.sort_values(by='emp_id')# 511. 游戏玩法分析 I
def game_analysis(activity: pd.DataFrame) -> pd.DataFrame:df = activity.sort_values(by=['player_id', 'event_date'])[['player_id', 'event_date']].rename(columns={'event_date': 'first_login'})print(df)return df.drop_duplicates('player_id')# 2356. 每位教师所教授的科目种类的数量
def count_unique_subjects(teacher: pd.DataFrame) -> pd.DataFrame:df = teacher.drop_duplicates(['teacher_id', 'subject_id'])df['cnt'] = df.groupby('teacher_id')['subject_id'].transform('count')return df[['teacher_id', 'cnt']].drop_duplicates(['teacher_id', 'cnt'])# 596. 超过5名学生的课
def find_classes(courses: pd.DataFrame) -> pd.DataFrame:courses['cnt'] = courses.groupby('class')['student'].transform('count')df = courses[courses['cnt'] >= 5].drop('student', axis=1)[['class']]return df.drop_duplicates()# 586. 订单最多的客户
def largest_orders(orders: pd.DataFrame) -> pd.DataFrame:if orders.empty:return pd.DataFrame({'customer_number': []})df = orders.groupby('customer_number').size().reset_index(name='count')df.sort_values(by='count', ascending=False, inplace=True)return df[['customer_number']][:1]# 1484. 按日期分组销售产品
def categorize_products(activities: pd.DataFrame) -> pd.DataFrame:groups = activities.groupby('sell_date')stats = groups.agg(num_sold=('product', 'nunique'),products=('product', lambda x: ','.join(sorted(set(x))))).reset_index()stats.sort_values('sell_date', inplace=True)return stats# 1693. 每天的领导和合伙人
def daily_leads_and_partners(daily_sales: pd.DataFrame) -> pd.DataFrame:df = daily_sales.groupby(['date_id', 'make_name']).nunique().reset_index()df = df.rename(columns={'lead_id':'unique_leads', 'partner_id': 'unique_partners'})return df# 1050. 合作过至少三次的演员和导演
def actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:df = actor_director.groupby(['actor_id', 'director_id']).size().reset_index(name='cnt')df = df[df['cnt'] >= 3]return df[['actor_id', 'director_id']]# 1378. 使用唯一标识码替换员工ID
def replace_employee_id(employees: pd.DataFrame, employee_uni: pd.DataFrame) -> pd.DataFrame:df = employees.merge(employee_uni, how='left', left_on='id', right_on='id')return df[['unique_id', 'name']]# 1280. 学生们参加各科测试的次数
def students_and_examinations(students: pd.DataFrame, subjects: pd.DataFrame,examinations: pd.DataFrame) -> pd.DataFrame:# 按 id 和科目分组，并计算考试次数。grouped = examinations.groupby(['student_id', 'subject_name']).size().reset_index(name='attended_exams')all_id_subjects = pd.merge(students, subjects, how='cross')id_subjects_count = pd.merge(all_id_subjects, grouped, on=['student_id', 'subject_name'], how='left')id_subjects_count['attended_exams'] = id_subjects_count['attended_exams'].fillna(0).astype(int)id_subjects_count.sort_values(['student_id', 'subject_name'], inplace=True)return id_subjects_count[['student_id', 'student_name', 'subject_name', 'attended_exams']]# 570. 至少有5名直接下属的经理
def find_managers(employee: pd.DataFrame) -> pd.DataFrame:df = employee.groupby(['managerId']).size().reset_index(name='cnt')df = df[df['cnt'] >= 5]df = df.merge(employee, how='inner', right_on='id', left_on='managerId')return df[['name']]# 607. 销售员
def sales_person(sales_person: pd.DataFrame, company: pd.DataFrame, orders: pd.DataFrame) -> pd.DataFrame:company = company[company['name'] == 'RED']df = orders.merge(company, how='inner', on='com_id')sales_person = sales_person[~sales_person['sales_id'].isin(df['sales_id'])]return sales_person[['name']]