Python SQL 到 pandas DataFrame

Question

我有一个 SQL 查询:

pd.read_sql_query("""SELECT UvTab.A, UvTab.Uv,
                            IFNULL(DvTab.Dv, 0) AS Dv 
                                FROM
                                    (
                                        SELECT A, COUNT(*) AS Uv FROM B
                                            WHERE Vtype = 2 GROUP BY A
                                    ) AS UvTab
                                LEFT JOIN 
                                    (
                                        SELECT A, COUNT(*) AS Dv FROM B
                                            WHERE Vtype = 3 GROUP BY A
                                    ) AS DvTab
                                ON UvTab.A = DvTab.A
                            """, conn)

我的目标是获得相同的结果，但仅使用 pandas 的方法。我得到的是:

UvTab = B.loc[B.Vtype == 2].groupby("A").size()
UvTab = pd.DataFrame({'A' : UvTab.index, 'Uv' : UvTab.values})
DvTab = B.loc[B.Vtype == 3].groupby("A").size()
DvTab = pd.DataFrame({'A' : DvTab.index, 'Dv' : DvTab.values})
df = pd.merge(UvTab, DvTab, how='left', on='A')
df['Dv'] = df['Dv'].fillna(0)

而且看起来还不错。但这是表示查询的最简单且最好的方式吗？

Answer 1

一个想法是聚合 sum 进行计数匹配，然后使用 DataFrame.join :

UvTab = (B.Vtype == 2).astype(int).groupby(B["A"]).sum().reset_index(name='Uv')
DvTab = (B.Vtype == 3).astype(int).groupby(B["A"]).sum().to_frame('Dv')
df = UvTab.join(DvTab, on='A').fillna({'DV':0})

或者使用merge替代:

UvTab = (B.Vtype == 2).astype(int).groupby(B["A"]).sum().reset_index(name='Uv')
DvTab = (B.Vtype == 3).astype(int).groupby(B["A"]).sum().reset_index(name='Dv')
df = UvTab.merge(DvTab, on='A', how='left').fillna({'DV':0})