Python Pandas：从数据框中删除不符合多个条件的行

Question

假设我有一个使用pandas.dataframe的列，如下所示：

index  fruits    origin      attribute
 1     apple     USA         tasty
 2     apple     France      yummy
 3     apple     USA         juicy
 4     apple     England     juicy
 5     apple     Japan       normal
 6     banana    Canada      nice
 7     banana    Italy       good
 .....

我想yummy apple from France(2)选择yummy apple from France(2)并从表中删除不匹配的apples ，如下所示：

index  fruits    origin      attribute
 1     apple     France      yummy
 2     banana    Canada      nice
 3     banana    Italy       good
 .....

我认为以下应该可行。 但事实并非如此：

df.drop(df[(df.fruits == "apple") & (df.origin != "France") | (df.fruits == "apple") & (df.attribute != "yummy")].index)

然后，我尝试了以下同样行不通的方法：

df = df[~df[(df.fruits == "apple") & (df.origin != "France") & (df.attribute != "yummy")]

伙计们，有什么帮助吗？

Answer 1

如果通过匹配条件选择：

df[(df.fruits != 'apple') | ((df.fruits == 'apple') & (df.origin == 'France') & (df.attribute == 'yummy'))]

#index  fruits  origin  attribute
#1  2    apple  France      yummy
#5  6   banana  Canada       nice
#6  7   banana   Italy       good

---

如果按不匹配条件删除：需要删除的是fruits是苹果但origin与France不匹配或attribute与yummy不匹配的行：

df[~((df.fruits == 'apple') & ((df.origin != 'France') | (df.attribute != 'yummy')))]

# index fruits  origin  attribute
#1    2  apple  France      yummy
#5    6 banana  Canada       nice
#6    7 banana   Italy       good

Answer 2

df.query(
    'fruits == "apple" & origin == "France" & attribute == "yummy"'
).append(df.query('fruits != "apple"'))

       fruits  origin attribute
index                          
2       apple  France     yummy
6      banana  Canada      nice
7      banana   Italy      good