如何使用Snowflake SQL查询的结果填充 pandas DataFrame？ [英] How can I populate a pandas DataFrame with the result of a Snowflake sql query?

问题描述

使用Python Connector我可以查询雪花：

import snowflake.connector

# Gets the version
ctx = snowflake.connector.connect(
    user=USER,
    password=PASSWORD,
    account=ACCOUNT,
    authenticator='https://XXXX.okta.com',
    )
ctx.cursor().execute('USE warehouse MY_WH')
ctx.cursor().execute('USE MYDB.MYSCHEMA')


query = '''
select * from MYDB.MYSCHEMA.MYTABLE
LIMIT 10;
'''

cur = ctx.cursor().execute(query)

结果是snowflake.connector.cursor.SnowflakeCursor。如何将其转换为 pandas DataFrame？

推荐答案

您可以将DataFrame.from_records()或pandas.read_sql()与snowflake-sqlalchemy配合使用。雪花炼金术选项有一个更简单的API

pd.DataFrame.from_records(iter(cur), columns=[x[0] for x in cur.description])

将返回一个DataFrame，其中包含取自SQL结果的正确列名。iter(cur)会将游标转换为迭代器，cur.description提供列的名称和类型。

所以完整的代码将是

import snowflake.connector
import pandas as pd

# Gets the version
ctx = snowflake.connector.connect(
    user=USER,
    password=PASSWORD,
    account=ACCOUNT,
    authenticator='https://XXXX.okta.com',
    )
ctx.cursor().execute('USE warehouse MY_WH')
ctx.cursor().execute('USE MYDB.MYSCHEMA')


query = '''
select * from MYDB.MYSCHEMA.MYTABLE
LIMIT 10;
'''

cur = ctx.cursor().execute(query)
df = pd.DataFrame.from_records(iter(cur), columns=[x[0] for x in cur.description])

如果您喜欢使用pandas.read_sql，则可以

import pandas as pd

from sqlalchemy import create_engine
from snowflake.sqlalchemy import URL


url = URL(
    account = 'xxxx',
    user = 'xxxx',
    password = 'xxxx',
    database = 'xxx',
    schema = 'xxxx',
    warehouse = 'xxx',
    role='xxxxx',
    authenticator='https://xxxxx.okta.com',
)
engine = create_engine(url)


connection = engine.connect()

query = '''
select * from MYDB.MYSCHEMA.MYTABLE
LIMIT 10;
'''

df = pd.read_sql(query, connection)

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