Pandas刷题

Pandas刷题

Leecode刷题

176.第二高的薪水

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import pandas as pd

def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
    # 1. 删除所有重复的薪水.
    employee = employee.drop_duplicates(["salary"])
    
    # 2. 如果少于 2 个不同的薪水，返回 `np.NaN`。
    if len(employee["salary"].unique()) < 2:
        return pd.DataFrame({"SecondHighestSalary": [np.NaN]})
    
    # 3. 把表格按 `salary` 降序排序。
    employee = employee.sort_values(by="salary", ascending=False)
    
    # 4. 删除 `id` 列。
    employee.drop("id", axis=1, inplace=True)
    
    # 5. 重命名 `salary` 列。
    employee.rename({"salary": "SecondHighestSalary"}, axis=1, inplace=True)
    
    # 6, 7. 返回第 2 高的薪水
    return employee.head(2).tail(1)
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import pandas as pd

def second_highest_salary(employee: pd.DataFrame) -> pd.DataFrame:
    # 按照工资降序排序
    employee = employee.sort_values(by='salary', ascending=False)
    
    # 去除重复的工资
    employee = employee.drop_duplicates(subset=['salary'])
    
    # 选择第N高的工资，如果不存在则返回null
    if len(employee) >= 2:
        return pd.DataFrame({'SecondHighestSalary': [int(employee.iloc[1]['salary'])]})
    else:
        return pd.DataFrame({'SecondHighestSalary': [None]})

177.第N高薪水

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import pandas as pd

def nth_highest_salary(employee: pd.DataFrame, N: int) -> pd.DataFrame:
    
    dist = employee.drop_duplicates(subset='salary')
    dist['rnk'] = dist['salary'].rank(method='dense', ascending=False)
    ans = dist[dist.rnk == N][['salary']]
    if not len(ans):
        return pd.DataFrame({f'getNthHighestSalary({N})': [None]})
    ans = ans.rename(columns={'salary':f'getNthHighestSalary({N})'})
    return ans

178.分数排名

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import pandas as pd

def order_scores(scores: pd.DataFrame) -> pd.DataFrame:
    scores['rank'] = scores['score'].rank(method='dense', ascending=False)
    return scores[['score', 'rank']].sort_values('score', ascending=False)

180.连续出现的数字

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def consecutive_numbers(logs: pd.DataFrame) -> pd.DataFrame:
    ###类似sql的窗口函数 先将先窗口内排序
    logs['rank']=logs.groupby('num')['id'].rank(method='first')
    ##用id-rank 得到一个标签，同样的标签在同一个num下出现三次说明连续
    logs['is_']=logs['id']-logs['rank']
    ##分组count
    result_pre1=logs.groupby(by=['is_','num'])['id'].count().reset_index()
    result=result_pre1.loc[result_pre1['id']>=3,['num']]
    result.columns=['ConsecutiveNums']
    result.drop_duplicates(inplace=True)

    return result

182.查找重复的电子邮箱

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import pandas as pd

def duplicate_emails(person: pd.DataFrame) -> pd.DataFrame:
    person["duplicate"] = person["email"].duplicated(keep="first")
    df = person[person["duplicate"]].drop_duplicates(subset=["email"], keep="first")
    return df[["email"]]
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import pandas as pd

def duplicate_emails(person: pd.DataFrame) -> pd.DataFrame:
    df=person[person['email'].duplicated()]['email'].unique()
    return pd.DataFrame({'Email':df})
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import pandas as pd

def duplicate_emails(person: pd.DataFrame) -> pd.DataFrame:
    res = person['email'].value_counts()
    res = res[res>1]
    res = res.index
    print(res)
    # res.columns= ['email','value']
    # res = res[res['value']>1]
    return pd.DataFrame(res)