如何将 Dataframe 列中值的最后 3 位拆分为两个新的数据帧？

Question

df=pd.DataFrame({'col1':[25021,25002,8002,40211,2232,""]})

    col1    
0   25021    
1   25002
2   8002
3   40211
4   2232
5   

I would like to get the following, not too sure how to split based on last 3 digits into col3 and whatever preceding into col1

    col2   col3    
0   25     021       
1   25     002       
2   8      002      
3   40     211       
4   2      232 
5

Answer 1

This is my approach:

df['col2'] = df['col1'].astype(str).str[-3:]

df['col1'] = df['col1'].astype(str).str[:-3]

Output:

   col1 col2
0    25  021
1    25  002
2     8  002
3    40  211
4     2  232

Answer 2

Try this.

df=pd.DataFrame({'col1':[25021,25002,8002,40211,2232]})
df['col2'] = df['col1'].astype(str).apply(lambda x:x[-3:]).astype(int)
df['col1'] = df['col1'].astype(str).apply(lambda x:x[:-3]).astype(int)

Answer 3

Just a play on Pandas' string split method; you can wrap the delimiter(in a regex), so that it is included in the output:

(df
 .astype(str)
 .col1
 .str.split(r'(\d{3}$)', n=1, expand=True)
 .drop(2,axis=1)
 .set_axis(['col1','col2'],axis='columns')
)

    col1    col2
0    25     021
1    25     002
2    8      002
3    40     211
4    2      232