Splitting a pandas dataframe column by delimiter

Question:

i have a small sample data:

import pandas as pd

df = {'ID': [3009, 129, 119, 120, 121, 122, 130, 3014, 266, 849, 174, 844],
  'V': ['IGHV7-B*01', 'IGHV7-B*01', 'IGHV6-A*01', 'GHV6-A*01', 'IGHV6-A*01',
        'IGHV6-A*01', 'IGHV4-L*03', 'IGHV4-L*03', 'IGHV5-A*01', 'IGHV5-A*04',
        'IGHV6-A*02','IGHV6-A*02'],
  'Prob': [1, 1, 0.8, 0.8056, 0.9, 0.805, 1, 1, 0.997, 0.401, 1, 1]}

df = pd.DataFrame(df)

looks like

df    

Out[25]: 
      ID    Prob           V
0    3009  1.0000  IGHV7-B*01
1     129  1.0000  IGHV7-B*01
2     119  0.8000  IGHV6-A*01
3     120  0.8056  IGHV6-A*01
4     121  0.9000  IGHV6-A*01
5     122  0.8050  IGHV6-A*01
6     130  1.0000  IGHV4-L*03
7    3014  1.0000  IGHV4-L*03
8     266  0.9970  IGHV5-A*01
9     849  0.4010  IGHV5-A*04
10    174  1.0000  IGHV6-A*02
11    844  1.0000  IGHV6-A*02

I want to split the column ‘V’ by the ‘-‘ delimiter and move it to another column named ‘allele’

    Out[25]: 
      ID    Prob      V    allele
0    3009  1.0000  IGHV7    B*01
1     129  1.0000  IGHV7    B*01
2     119  0.8000  IGHV6    A*01
3     120  0.8056  IGHV6    A*01
4     121  0.9000  IGHV6    A*01
5     122  0.8050  IGHV6    A*01
6     130  1.0000  IGHV4    L*03
7    3014  1.0000  IGHV4    L*03
8     266  0.9970  IGHV5    A*01
9     849  0.4010  IGHV5    A*04
10    174  1.0000  IGHV6    A*02
11    844  1.0000  IGHV6    A*02

the code i have tried so far is incomplete and didn’t work:

df1 = pd.DataFrame()
df1[['V']] = pd.DataFrame([ x.split('-') for x in df['V'].tolist() ])

or

df.add(Series, axis='columns', level = None, fill_value = None)
newdata = df.DataFrame({'V':df['V'].iloc[::2].values, 
                        'Allele': df['V'].iloc[1::2].values})
Asked By: Jessica

||

Answers:

Use vectoried str.split with expand=True:

In [42]:
df[['V','allele']] = df['V'].str.split('-',expand=True)
df

Out[42]:
      ID    Prob      V allele
0   3009  1.0000  IGHV7   B*01
1    129  1.0000  IGHV7   B*01
2    119  0.8000  IGHV6   A*01
3    120  0.8056   GHV6   A*01
4    121  0.9000  IGHV6   A*01
5    122  0.8050  IGHV6   A*01
6    130  1.0000  IGHV4   L*03
7   3014  1.0000  IGHV4   L*03
8    266  0.9970  IGHV5   A*01
9    849  0.4010  IGHV5   A*04
10   174  1.0000  IGHV6   A*02
11   844  1.0000  IGHV6   A*02
Answered By: EdChum

For storing data into a new dataframe use the same approach, just with the new dataframe:

tmpDF = pd.DataFrame(columns=['A','B'])
tmpDF[['A','B']] = df['V'].str.split('-', expand=True)

Eventually (and more usefull for my purposes) if you would need get only a part of the string value (i.e. text before ‘-‘), you could use .str.split(…).str[idx] like:

df['V'] = df['V'].str.split('-').str[0]
df
    ID      V       Prob
0   3009    IGHV7   1.0000
1   129     IGHV7   1.0000
2   119     IGHV6   0.8000
3   120     GHV6    0.8056

– splits ‘V’ values into list according to separator ‘-‘ and stores 1st item back to the column

Answered By: Lukas

Use the below:

df['allele'] = [x.split('-')[-1] for x in df['V']]

The above first part retains any values after the ‘-‘ sign

df['V'] = [x.split('-')[-0] for x in df['V']]

The above second part retains any values before the ‘-‘ sign and automatically replaces the main column

df.head(3)
Answered By: Ariel Jumba
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