Python 通过其他两列更新熊猫

我有一个大约3000个条目和10列的数据集，所以这是一个简单得多的版本

df =
   asset    tail    more_info
0   x         a     this is a long text field that is right
1   x         b     this is a long text field that is almost right
2   y         a     this is right
3   y         b     this is probably not right

期望结果

df =
   asset    tail    more_info
0   x         a     this is a long text field that is right
1   x         b     this is a long text field that is right
2   y         a     this is right
3   y         b     this is right

所以我试图更新我的more_info字段，其中资产匹配，尾部等于'a' 事实上，数据集更复杂，所以我需要以编程方式进行，这就是我在该逻辑中画空白的地方

def my_func(x):
    if x.asset == x.asset and x.tail =='b':
        '''
        this would be where I'd set it to x.more_info where tail = 'a' maybe numpy where ??
        '''
        
df['more_info'] = df['more_info'].apply(lambda x: my_func(x))

您可以尝试以下方法：

# reduce dataframe to contain only
df1 = df[['asset', 'tail']].copy()

# slice df to get only the ones you want to use for "more_info"
df2 = df[df['tail']=='a'][['asset', 'more_info']].copy()

df1.merge(df2, on=['asset'])

#   asset tail                                more_info
# 0     x    a  this_is_a_long_text_field_that_is_right
# 1     x    b  this_is_a_long_text_field_that_is_right
# 2     y    a                            this_is_right
# 3     y    b                            this_is_right

或者在一行中：

df[['asset', 'tail']].merge(df[df['tail']=='a'][['asset', 'more_info']], on=['asset'])

---

归档所有“a”记录，通过“资产”连接回原始数据