如何在 pandas dataframe 中編寫以下 SQL 語句？

Question

幾天前我剛進入 pandas。 我想試試我以前用 SQL 做的事情，然后用 pandas dataframe 試試，不使用 pandasql，誰能給我一些線索？

> SELECT TOP 1 col FROM ( SELECT 
>     col AS col, 
>     LEN(col) AS length,
>     CAST(LEFT(col, 20) AS numeric(20,0)) AS col_p1,
>     (CASE
>         WHEN LEN(col) > 20 THEN CAST(RIGHT(col,LEN(col) - 20) AS numeric(20,0))
>     END) AS col_p2  FROM table_a WHERE SIGN(CAST(LEFT(col,2) AS SMALLINT)) = 1 ORDER BY length DESC, col_p1 DESC, col_p2 DESC )

Answer 1

您可以將查詢轉換為樹和 go 從葉子到根節點（又名關系代數樹）。 這就是我的意思：

1. 你從嵌套表開始

SELECT [COLUMNS]  FROM table_a WHERE [CONDITION] ORDER BY [CRITERIA]

2. 然后你 go 和 select 只有有趣的行：

WHERE SIGN(CAST(LEFT(col,2) AS SMALLINT)) = 1 

# equivalent to  mask = df['col'].apply(lambda val: int(val[:2]) > 0 )
# and then you apply the mask to the dataframe

3. 在此步驟之后，您 go 並創建新列，連接並重命名它們：

CAST(LEFT(col, 20) AS numeric(20,0)) AS col_p1,
     (CASE
         WHEN LEN(col) > 20 THEN CAST(RIGHT(col,LEN(col) - 20) AS numeric(20,0))
     END) AS col_p2


# for those 2, you can generate a numpy array and add it to the dataframe


# and then you rename your columns :
#      df.columns = [...]

4. 對於排序，您可以使用sort_values

ORDER BY length DESC, col_p1 DESC, col_p2 DESC 
# equivalent to df.sort_values(by =['length',...],ascending=False)

5. 最后：

SELECT TOP 1 col FROM T 
# equivalent to T[:1,'col']

其中 T = 是嵌套表。