您必须过滤df 2帧，检查df 2中的元素是否不在df 1中：

df2[~df2["product"].isin(df1['product'])]

• ~对布尔值Series的值求反。
• ser1.isin（ser 2）是一个布尔级数，它为ser 1的每个元素给出是否可以在ser 2中找到该值。

我猜您正在尝试执行左反连接，这意味着您只想保留df2中不存在于df1中的行。在这种情况下：

df1 = pd.DataFrame({"product":['apples', 'bananas', 'oranges', 'kiwi']})
df2 = pd.DataFrame({"product":['apples', 'aples', 'appples', 'banans', 'oranges', 'kiwki'], "key":[1, 2, 3, 4, 5, 6]})

# left join
joined_df = df2.merge(df1, on='product', how='left', indicator=True)
# keeping products that were only present in df2
products_only_in_df2 = joined_df.loc[joined_df['_merge'] == 'left_only', 'product']
# filtering df2 using the above df so we have the keys as well
result = df2[df2['product'].isin(products_only_in_df2)]