Python: pandas DataFrame new column based on other columns

Question

I have a df with 2 columns like below:

        A       B
0  100-00     nan
1  200-00     nan
2   other  300-00
3  100-00    text
4   other     nan

I need to create column C which will apply logic as below:

if B is nan, then A
if B starts with numeric, then B
else A

I've got code as below, which is working just fine, but I believe there might be a better, more efficent way to do it:

C = []
for r in range(df.shape[0]):
    if df['B'].iloc[r] == 'nan':
        C.append(df['A'].iloc[r])
    elif df['B'].iloc[r][:3].isnumeric():
        C.append(df['B'].iloc[r])
    else:
        C.append(df['A'].iloc[r])
df['C'] = C

df
        A       B       C
0  100-00     nan  100-00
1  200-00     nan  200-00
2   other  300-00  300-00
3  100-00    text  100-00
4   other     nan   other

Thanks in advance for all your help.

jezrael · Accepted Answer · 2020-04-27 09:48:31Z

2

I simplify solution for df.B if first character in df.B is numeric tested with Series.str.contains with regex ^ for start of string and \d for number in numpy.where:

df['C'] = np.where(df['B'].str.contains(r'^\d', na=False), df.B, df.A)
#alternative
#df['C'] = df.B.where(df['B'].str.contains(r'^\d', na=False), df.A)
print (df)
        A       B       C
0  100-00     NaN  100-00
1  200-00     NaN  200-00
2   other  300-00  300-00
3  100-00    text  100-00
4   other     NaN   other

answered Apr 27, 2020 at 9:48

jezrael

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Pani · Accepted Answer · 2020-04-27 09:57:47Z

1

Not necessarily more efficient, but a more pythonic way to do it

import pandas as pd

df = pd.DataFrame({'A': ['100-00', '200-00', 'other', '100-00', 'other'], 'B': ['nan', 'nan', '300-00', 'text', 'nan']})

def label_columnC(row):
    if row['B'] == 'nan':
        return row['A']
    elif row['B'][:3].isnumeric():
        return row['B']
    else:
        return row['A']

df['C'] = df.apply(lambda row: label_columnC(row), axis=1)

edited Apr 27, 2020 at 9:57

answered Apr 27, 2020 at 9:51

Pani

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