Apply definition on multiple columns based on conditions- python

Question

i have dataframe that i want when one column equal to '02' apply the definition to 5 others columns. this is my code:

mask = (df['A']=='02')
z_valid = df[mask]
df.loc[((mask) & (df['B'] !=0)), 'B'] = z_valid['B'].apply(def)
df.loc[((mask) & (df['C'] !=0)), 'C'] = z_valid['C'].apply(def)
df.loc[((mask) & (df['D'] !=0)), 'D'] = z_valid['D'].apply(def)
df.loc[((mask) & (df['E'] !=0)), 'E'] = z_valid['E'].apply(def)
df.loc[((mask) & (df['F'] !=0)), 'F'] = z_valid['F'].apply(def)

i got this error: 'ValueError: Must have equal len keys and value when setting with an iterable'

but this code works when i apply def just on 2 columns:

mask = (df['A']=='02')
z_valid = df[mask]
df.loc[((mask) & (df['B'] !=0)), 'B'] = z_valid['B'].apply(def)
df.loc[((mask) & (df['C'] !=0)), 'C'] = z_valid['C'].apply(def)

how can i fix this error? thanks in advance.

Answer 1

It raised an exception because 'D' is an empty Series:

Suppose the following dataframe:

df = pd.DataFrame({'A': ['01', '02', '02', '02'], 'B': [9, 1, 2, 3], 'C': [9, 1, 2, 3], 'D': [9, 0, 0, 0]})
print(df)

# Output:
    A  B  C  D
0  01  9  9  9
1  02  1  1  0
2  02  2  2  0
3  02  3  3  0

Now if I try your code:

def f(sr):
    return 2*sr

mask = (df['A']=='02')
z_valid = df[mask]
df.loc[((mask) & (df['B'] !=0)), 'B'] = z_valid['B'].apply(f)
df.loc[((mask) & (df['C'] !=0)), 'C'] = z_valid['C'].apply(f)
df.loc[((mask) & (df['D'] !=0)), 'D'] = z_valid['D'].apply(f)

Output:

----> 8 df.loc[((mask) & (df['D'] !=0)), 'D'] = z_valid['D'].apply(f)
...
ValueError: Must have equal len keys and value when setting with an iterable

The exception is raised on 'D':

# Left side
>>> df.loc[((mask) & (df['D'] !=0)), 'D']
Series([], Name: E, dtype: int64)  # <- Empty Series

# Right side
>>> z_valid['D'].apply(f)
1    0
2    0
3    0
Name: E, dtype: int64