Question

我正在尝试根据另一列中的信息填充空白列

我的数据框

   A        B                                    C
0  F    House                     Are you at home?
1  E    House    description: to deliver tomorrow
2  F    Apt                 Here is some exemples 
3  F    House          description: a brown table
4  E    Apt               description: in the bus
5  F    House                 Hello, how are you?
6  E    Apt                     description: keys

因此，我创建了一个D列，如果列C以'description'开头，则填写'fuzzy'，如果不是'buzzy'，则填写

。

new_column['D'] = ''

然后我尝试填充它们

def fill_column(delete_column):
    if new_column['D'].loc[new_column['D'].str.startswith('description:'):
        new_column['D'] == 'fuzzy'
    else:
        new_column['D'] == 'buzzy'

    return new_column

我的输出：

  File "<ipython-input-41-ec3c1407168c>", line 6
    else:
       ^
SyntaxError: invalid syntax

好的输出：

   A        B                                   C       D
0  F    House                    Are you at home?   buzzy
1  E    House    description: to deliver tomorrow   fuzzy
2  F    Apt                 Here is some exemples   buzzy
3  F    House          description: a brown table   fuzzy
4  E    Apt               description: in the bus   fuzzy
5  F    House                 Hello, how are you?   buzzy
6  E    Apt                     description: keys   fuzzy

Answer 1

您在这里不需要if-else语句，可以使用np.where在一行中简洁地完成此操作：

df['D'] = np.where(
    df['C'].str.startswith('description:'), 'fuzzy', 'buzzy')

由于您仅分配了两个值，因此可以通过一个loc调用来完成此操作。

df['D'] = 'buzzy'
df.loc[df['C'].str.startswith('description:'), 'D'] = 'fuzzy'

或者使用df.mask / df.where，例如注释中建议的@jpp：

df['D'] = 'buzzy'
df['D'] = df['D'].mask(df['C'].str.startswith('description:'), 'fuzzy')

最后，使用map：

m = {True: 'fuzzy', False: 'buzzy'}
df['D'] = df['C'].str.startswith('description:').map(m)

print(df)
   A      B                                 C      D
0  F  House                  Are you at home?  buzzy
1  E  House  description: to deliver tomorrow  fuzzy
2  F    Apt             Here is some exemples  buzzy
3  F  House        description: a brown table  fuzzy
4  E    Apt           description: in the bus  fuzzy
5  F  House               Hello, how are you?  buzzy
6  E    Apt                 description: keys  fuzzy

Answer 2

new_column.loc[new_column['C'].str.startswith('description:'), 'D'] = 'fuzzy'
new_column.loc[~new_column['C'].str.startswith('description:'), 'D'] = 'buzzy'

根据来自另一列的信息用熊猫填充一个空列

2 个答案: