CODEWITHSUNDEEP
python-3.xpandassplit
Dance Party
Dance Party

Reputation: 3713

Pandas Split Column

Given the following data frame:

import pandas as pd
import numpy as np
df = pd.DataFrame({
       'A' : ['a', 'b','c', 'd'],
       'B' : ['Y>`abcd', 'abcd','efgh', 'Y>`efgh']
    })
df

    A   B
0   a   Y>`abcd
1   b   abcd
2   c   efgh
3   d   Y>`efgh

I'd like to split column A on '>`' into 2 columns (C and D) so my data 

frame looks like this:
        A   C  D
    0   a   Y  abcd
    1   b      abcd
    2   c      efgh
    3   d   Y  efgh

Thanks in advance!

Upvotes: 1

Views: 760

Answers (4)

Mikko
Mikko

Reputation: 29

The simplest and most memory efficient way of doing it is:

df[['C', 'D']] = df.B.str.split('>`', expand=True)

Upvotes: 0

jezrael
jezrael

Reputation: 862481

You can use str.extract with fillna, last drop column B by drop:

df[['C','D']] = df['B'].str.extract('(.*)>`(.*)', expand=True)
df['D'] = df['D'].fillna(df['B'])
df['C'] = df['C'].fillna('')
df = df.drop('B', axis=1)

print df

   A  C     D
0  a  Y  abcd
1  b     abcd
2  c     efgh
3  d  Y  efgh

Next solution use str.split with mask and numpy.where:

df[['C','D']] =  df['B'].str.split('>`', expand=True) 
mask = pd.notnull(df['D'])
df['D'] = df['D'].fillna(df['C'])
df['C'] = np.where(mask, df['C'], '')
df = df.drop('B', axis=1)

Timings:

In large DataFrame is extract solution 100 times faster, in small 1.5 times:

len(df)=4:

In [438]: %timeit a(df)
100 loops, best of 3: 2.96 ms per loop

In [439]: %timeit b(df1)
1000 loops, best of 3: 1.86 ms per loop

In [440]: %timeit c(df2)
The slowest run took 4.44 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 1.89 ms per loop

In [441]: %timeit d(df3)
The slowest run took 4.62 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 1.82 ms per loop

len(df)=4k:

In [443]: %timeit a(df)
1 loops, best of 3: 799 ms per loop

In [444]: %timeit b(df1)
The slowest run took 4.19 times longer than the fastest. This could mean that an intermediate result is being cached 
100 loops, best of 3: 7.37 ms per loop

In [445]: %timeit c(df2)
1 loops, best of 3: 552 ms per loop

In [446]: %timeit d(df3)
100 loops, best of 3: 9.55 ms per loop

Code:

import pandas as pd
df = pd.DataFrame({
       'A' : ['a', 'b','c', 'd'],
       'B' : ['Y>`abcd', 'abcd','efgh', 'Y>`efgh']
    })
#for test 4k    
df = pd.concat([df]*1000).reset_index(drop=True)
df1,df2,df3 = df.copy(),df.copy(),df.copy()

def b(df):
    df[['C','D']] = df['B'].str.extract('(.*)>`(.*)', expand=True)
    df['D'] = df['D'].fillna(df['B'])
    df['C'] = df['C'].fillna('')
    df = df.drop('B', axis=1)
    return df

def a(df):
    df = pd.concat([df, df.B.str.split('>').apply(
    lambda l: pd.Series({'C': l[0], 'D': l[1][1: ]}) if len(l) == 2 else \
        pd.Series({'C': '', 'D': l[0]}))], axis=1)
    del df['B']
    return df

def c(df):
    df[['C','D']] = df['B'].str.split('>`').apply(lambda x: pd.Series(['']*(2-len(x)) + x))
    df = df.drop('B', axis=1)    
    return df   

def d(df):
    df[['C','D']] =  df['B'].str.split('>`', expand=True) 
    mask = pd.notnull(df['D'])
    df['D'] = df['D'].fillna(df['C'])
    df['C'] = np.where(mask, df['C'], '')
    df = df.drop('B', axis=1) 
    return df

Upvotes: 3

Colonel Beauvel
Colonel Beauvel

Reputation: 31161

I would use a one liner:

df['B'].str.split('>`').apply(lambda x: pd.Series(['']*(2-len(x)) + x))

#   0     1
#0  Y  abcd
#1     abcd
#2     efgh
#3  Y  efgh

Upvotes: 1

Ami Tavory
Ami Tavory

Reputation: 76297

Performing a str.split followed by an apply returning a pd.Series will create the new columns:

>>> df.B.str.split('>').apply(
    lambda l: pd.Series({'C': l[0], 'D': l[1][1: ]}) if len(l) == 2 else \
        pd.Series({'C': '', 'D': l[0]}))
    C   D
0   Y   abcd
1       abcd
2       efgh
3   Y   efgh

So you can concat this to the DataFrame, and del the original column:

df = pd.concat([df, df.B.str.split('>').apply(
    lambda l: pd.Series({'C': l[0], 'D': l[1][1: ]}) if len(l) == 2 else \
        pd.Series({'C': '', 'D': l[0]}))],
    axis=1)
del df['B']
>>> df
    A   C   D
0   a   Y   abcd
1   b       abcd
2   c       efgh
3   d   Y   efgh

Upvotes: 2

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