pandas series name conflict with column name

September 2, 2023

So I was going over pandas leetcode and I came across this question. Where I am getting errors int not subscriptable. And then I found out because the column name and the series name is the problem where I cant get the column properly, is there any way to avoid this kinda problem?

data = [[2, 'Meir', 3000], [3, 'Michael', 3800], [7, 'Addilyn', 7400], [8, 'Juan', 6100], [9, 'Kannon', 7700]]
employees = pd.DataFrame(data, columns=['employee_id', 'name', 'salary']).astype({'employee_id':'int64', 'name':'object', 'salary':'int64'})

def get_bonus(row):
    if (row.name[0] == "M") and (row.employee_id %2 == 0)
        return row.salary
    else:
        return 0
    
def calculate_special_bonus(employees: pd.DataFrame) -> pd.DataFrame:
    employees["bonus"] = employees.apply(get_bonus, axis="columns")
    return employees[["employee_id", "bonus"]]

calculate_special_bonus(employees)


employee_id       2
name           Meir
salary         3000
bonus          None
Name: 0, dtype: object
employee_id          3
name           Michael
salary            3800
bonus             None
Name: 1, dtype: object
employee_id          7
name           Addilyn
salary            7400
bonus             None
Name: 2, dtype: object
employee_id       8
name           Juan
salary         6100
bonus          None
Name: 3, dtype: object
employee_id         9
name           Kannon
salary           7700
bonus            None
Name: 4, dtype: object

this is the row/series going into the apply function

>Solution :

Don’t use dot notation here, prefer bracket notation to use the standard indexing.

From the documentation:

You can use this access only if the index element is a valid Python identifier, e.g. s.1 is not allowed.
The attribute will not be available if it conflicts with an existing method name, e.g. s.min is not allowed, but s['min'] is possible.
Similarly, the attribute will not be available if it conflicts with any of the following list: index, major_axis, minor_axis, items.
In any of these cases, standard indexing will still work, e.g. s['1'], s['min'], and s['index'] will access the corresponding element or column.

So you should replace:

if (row.name[0] == "M") and ...

with:

if (row[name][0] == "M") and ...

Obviously, you can also do (vectorized code):

m1 = employees.name.str[0] == 'M'
m2 = employees.employee_id.mod(2) == 0
employees['bonus'] = employees['salary'].where(m1 & m2, other=0)
print(employees)

# Output
   employee_id     name  salary  bonus
0            2     Meir    3000   3000
1            3  Michael    3800      0
2            7  Addilyn    7400      0
3            8     Juan    6100      0
4            9   Kannon    7700      0