1. 知识点

1.27 左连接

data=pd.merge(employees,employee_uni,on='id',how='left')

1.28 数据填充与交叉连接

• 交叉合并

  df = pd.merge(students,subjects,how='cross')
  

• 空值填充

  result['attended_exams'] = result['attended_exams'].fillna(0)
  

1.29 获取列值列表

data_max_id=employee.groupby('managerId').agg(counts=('managerId','count')).reset_index().query('`counts`>=5')['managerId'].tolist()

题目

2.26 合作过至少三次的演员和导演

import pandas as pd

def actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:
    actor_director=actor_director.groupby(['actor_id','director_id']).agg(counts=('director_id','count')).reset_index()
    actor_director=actor_director.query('`counts`>=3')[['actor_id','director_id']]
    actor_director.drop_duplicates(inplace=True)
    return actor_director

2.27 使用唯一标识码替换员工ID

import pandas as pd

def replace_employee_id(employees: pd.DataFrame, employee_uni: pd.DataFrame) -> pd.DataFrame:
    data=pd.merge(employees,employee_uni,on='id',how='left')
    data=data[['unique_id','name']]
    return data

2.28 学生们参加各科测试的次数

import pandas as pd

def students_and_examinations(students: pd.DataFrame, subjects: pd.DataFrame, examinations: pd.DataFrame) -> pd.DataFrame:
    # 交叉合并
    df = pd.merge(students,subjects,how='cross')
    # 计算参加的同学
    group = examinations.groupby(['student_id','subject_name']).agg(attended_exams=('subject_name','count')).reset_index()
    # 左连接
    result = pd.merge(df,group,on=['student_id','subject_name'],how='left')
    # 为空的未参加，填充为0
    result['attended_exams'] = result['attended_exams'].fillna(0)
    result = result.sort_values(by=['student_id','subject_name'])
    return result[['student_id', 'student_name', 'subject_name', 'attended_exams']]

2.29 至少有5名直接下属的经理

import pandas as pd

def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
    data_max_id=employee.groupby('managerId').agg(counts=('managerId','count')).reset_index().query('`counts`>=5')['managerId'].tolist()
    results=employee.query(f'`id` in {data_max_id}')[['name']]
    return results

2.30 销售员

import pandas as pd

def sales_person(sales_person: pd.DataFrame, company: pd.DataFrame, orders: pd.DataFrame) -> pd.DataFrame:
    # 找到与RED有关的，剩余的就是没有关的
    # 方法一：连表
    data_c_o=pd.merge(orders,company,on='com_id',how='left')
    data_c_o=data_c_o[data_c_o['name']=='RED']
    data_s_o=pd.merge(data_c_o,sales_person,on='sales_id',how='left')
    un=data_s_o['sales_id'].unique().tolist()
    results=sales_person.query(f'`sales_id` not in {un}')[['name']]
    return results

    # 方法二：不连表
    all_com_id=company.query('`name`=="RED"')['com_id'].values.tolist()
    all_sales_id=orders.query(f'`com_id` in {all_com_id}')['sales_id'].values.tolist()
    results=sales_person.query(f'`sales_id` not in {all_sales_id}')[['name']]
    return results