Ignore invalid value in pandas dataframe column

September 8, 2023

I have created the following pandas dataframe:

import pandas as pd
import numpy as np     

ds = {'col1' : 
['1489900119000',
'้้คค,1']
}

df = pd.DataFrame(data=ds)

which looks like this:

            col1
0  1489900119000
1         ้้คค,1

I am trying to build a new column (called col2) which contains the values in col1 arranged as list. Hence the code:

df['col2']  = [list(map(int, str(x))) for x in df.col1]

Since there is that unexpected / invalid value at row 1 (คค,1), the code fails.

Is there any way to by-pass that row and fill it in with a default value (e.g. [9,9])? For example, the resulting dataframe would look like this:

>Solution :

You can use a conditional with str.isdigit:

df['col2']  = [list(map(int, str(x))) if x.isdigit() else [9,9] for x in df.col1]

Or with a custom function and try/except:

def digitize(s):
    try:
        return [int(x) for x in s]
    except ValueError:
        return [9, 9]
    
df['col2'] = df['col1'].apply(digitize)

Output:

            col1                                     col2
0  1489900119000  [1, 4, 8, 9, 9, 0, 0, 1, 1, 9, 0, 0, 0]
1         ้้คค,1                                   [9, 9]