Python Pandas write_frame删除sqlite表

当尝试用Pandas

dataframe

的内容覆盖sqlite表时，Pandas

会删除该表，但在尝试插入之前不会重新创建该表

下面是一个最低限度的工作示例：

import sqlite3 as sq
import pandas as pd
import pandas.io.sql as pd_sql

d = pd.DataFrame({'year':[2008,2008,2008,2008,2009,2009,2009,2009], 
                  'flavour':['strawberry','strawberry','banana','banana',
                  'strawberry','strawberry','banana','banana'],
                  'day':['sat','sun','sat','sun','sat','sun','sat','sun'],
                  'sales':[10,12,22,23,11,13,23,24]})

# Connect to the database (create if necessary)                  
conn = sq.connect('mydb')

# Create the table 'mytable' if necessary
if not pd_sql.table_exists('mytable', conn, 'sqlite'):
    pd_sql.write_frame(d, 'mytable', conn)

# Change some data    
d['sales'][d.sales==24] = 25

# Confirm the table exists
print "Table 'mytable' exists:"
print pd_sql.table_exists('mytable', conn, 'sqlite')

# Get some data from the table
cur = pd_sql.execute("SELECT DISTINCT flavour FROM mytable", conn)
print "Here's the data to prove the table exists:"
print cur.fetchall()

try:
    print "Attempting write_frame..."
    pd_sql.write_frame(d, 'mytable', conn, if_exists='replace')
except sq.OperationalError as e:
    print "sq.OperationalError is: " + str(e)
    print pd_sql.table_exists('mytable', conn, 'sqlite')
    conn.close()

运行此脚本将产生以下输出：

Table 'mytable' exists:
True
Here's the data to prove the table exists:
[(u'banana',), (u'strawberry',)]
Attempting write_frame...
sq.OperationalError is: no such table: mytable
Table 'mytable' exists after write_frame:
False

这看起来像熊猫身上的虫子。有人能证实这一点吗

谢谢你一如既往，
Rob
在对原始帖子的评论中，这已经被证实是熊猫中已知的一个bug

在我看来，这一解决办法还可以：

pd_sql.uquery("DELETE FROM mytable", conn)
pd_sql.write_frame(d, 'mytable', conn, if_exists='append')

这个问题似乎离题了，因为它是一个bug报告（所以应该发布）。@AndyHayden这是一样的吗@Jeff看起来是的，不要认为它被合并了，会在SQLAlchemy的东西完成后解决。对不起，先生们，我不相信我能找到一个真正的bug！我将发布我一直用作此问题“答案”的变通方法。