pandas: How do I split text in a column into multiple rows?

Question

I\'m working with a large csv file and the next to last column has a string of text that I want to split by a specific delimiter. I was wondering if there is a simple way to

Answer 1

Another approach would be like this:

temp = df['Seatblocks'].str.split(' ')
data = data.reindex(data.index.repeat(temp.apply(len)))
data['new_Seatblocks'] = np.hstack(temp)

Answer 2

Differently from Dan, I consider his answer quite elegant... but unfortunately it is also very very inefficient. So, since the question mentioned "a large csv file", let me suggest to try in a shell Dan's solution:

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print df['col'].apply(lambda x : pd.Series(x.split(' '))).head()"

... compared to this alternative:

time python -c "import pandas as pd;
from scipy import array, concatenate;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(concatenate(df['col'].apply( lambda x : [x.split(' ')]))).head()"

... and this:

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(dict(zip(range(3), [df['col'].apply(lambda x : x.split(' ')[i]) for i in range(3)]))).head()"

The second simply refrains from allocating 100 000 Series, and this is enough to make it around 10 times faster. But the third solution, which somewhat ironically wastes a lot of calls to str.split() (it is called once per column per row, so three times more than for the others two solutions), is around 40 times faster than the first, because it even avoids to instance the 100 000 lists. And yes, it is certainly a little ugly...

EDIT: this answer suggests how to use "to_list()" and to avoid the need for a lambda. The result is something like

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(df.col.str.split().tolist()).head()"

which is even more efficient than the third solution, and certainly much more elegant.

EDIT: the even simpler

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print pd.DataFrame(list(df.col.str.split())).head()"

works too, and is almost as efficient.

EDIT: even simpler! And handles NaNs (but less efficient):

time python -c "import pandas as pd;
df = pd.DataFrame(['a b c']*100000, columns=['col']);
print df.col.str.split(expand=True).head()"

Answer 3

import pandas as pd
import numpy as np

df = pd.DataFrame({'ItemQty': {0: 3, 1: 25}, 
                   'Seatblocks': {0: '2:218:10:4,6', 1: '1:13:36:1,12 1:13:37:1,13'}, 
                   'ItemExt': {0: 60, 1: 300}, 
                   'CustomerName': {0: 'McCartney, Paul', 1: 'Lennon, John'}, 
                   'CustNum': {0: 32363, 1: 31316}, 
                   'Item': {0: 'F04', 1: 'F01'}}, 
                    columns=['CustNum','CustomerName','ItemQty','Item','Seatblocks','ItemExt'])

print (df)
   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0    32363  McCartney, Paul        3  F04               2:218:10:4,6       60
1    31316     Lennon, John       25  F01  1:13:36:1,12 1:13:37:1,13      300

Another similar solution with chaining is use reset_index and rename:

print (df.drop('Seatblocks', axis=1)
             .join
             (
             df.Seatblocks
             .str
             .split(expand=True)
             .stack()
             .reset_index(drop=True, level=1)
             .rename('Seatblocks')           
             ))

   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    32363  McCartney, Paul        3  F04       60  2:218:10:4,6
1    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13

---

If in column are NOT NaN values, the fastest solution is use list comprehension with DataFrame constructor:

df = pd.DataFrame(['a b c']*100000, columns=['col'])

In [141]: %timeit (pd.DataFrame(dict(zip(range(3), [df['col'].apply(lambda x : x.split(' ')[i]) for i in range(3)]))))
1 loop, best of 3: 211 ms per loop

In [142]: %timeit (pd.DataFrame(df.col.str.split().tolist()))
10 loops, best of 3: 87.8 ms per loop

In [143]: %timeit (pd.DataFrame(list(df.col.str.split())))
10 loops, best of 3: 86.1 ms per loop

In [144]: %timeit (df.col.str.split(expand=True))
10 loops, best of 3: 156 ms per loop

In [145]: %timeit (pd.DataFrame([ x.split() for x in df['col'].tolist()]))
10 loops, best of 3: 54.1 ms per loop

But if column contains NaN only works str.split with parameter expand=True which return DataFrame (documentation), and it explain why it is slowier:

df = pd.DataFrame(['a b c']*10, columns=['col'])
df.loc[0] = np.nan
print (df.head())
     col
0    NaN
1  a b c
2  a b c
3  a b c
4  a b c

print (df.col.str.split(expand=True))
     0     1     2
0  NaN  None  None
1    a     b     c
2    a     b     c
3    a     b     c
4    a     b     c
5    a     b     c
6    a     b     c
7    a     b     c
8    a     b     c
9    a     b     c

Answer 4

This seems a far easier method than those suggested elsewhere in this thread.

split rows in pandas dataframe

Answer 5

Can also use groupby() with no need to join and stack().

Use above example data:

import pandas as pd
import numpy as np


df = pd.DataFrame({'ItemQty': {0: 3, 1: 25}, 
                   'Seatblocks': {0: '2:218:10:4,6', 1: '1:13:36:1,12 1:13:37:1,13'}, 
                   'ItemExt': {0: 60, 1: 300}, 
                   'CustomerName': {0: 'McCartney, Paul', 1: 'Lennon, John'}, 
                   'CustNum': {0: 32363, 1: 31316}, 
                   'Item': {0: 'F04', 1: 'F01'}}, 
                    columns=['CustNum','CustomerName','ItemQty','Item','Seatblocks','ItemExt']) 
print(df)

   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0  32363    McCartney, Paul  3        F04  2:218:10:4,6               60     
1  31316    Lennon, John     25       F01  1:13:36:1,12 1:13:37:1,13  300  


#first define a function: given a Series of string, split each element into a new series
def split_series(ser,sep):
    return pd.Series(ser.str.cat(sep=sep).split(sep=sep)) 
#test the function, 
split_series(pd.Series(['a b','c']),sep=' ')
0    a
1    b
2    c
dtype: object

df2=(df.groupby(df.columns.drop('Seatblocks').tolist()) #group by all but one column
          ['Seatblocks'] #select the column to be split
          .apply(split_series,sep=' ') # split 'Seatblocks' in each group
         .reset_index(drop=True,level=-1).reset_index()) #remove extra index created

print(df2)
   CustNum     CustomerName  ItemQty Item  ItemExt    Seatblocks
0    31316     Lennon, John       25  F01      300  1:13:36:1,12
1    31316     Lennon, John       25  F01      300  1:13:37:1,13
2    32363  McCartney, Paul        3  F04       60  2:218:10:4,6

Answer 6

It may be late to answer this question but I hope to document 2 good features from Pandas: pandas.Series.str.split() with regular expression and pandas.Series.explode().

import pandas as pd
import numpy as np

df = pd.DataFrame(
    {'CustNum': [32363, 31316],
     'CustomerName': ['McCartney, Paul', 'Lennon, John'],
     'ItemQty': [3, 25],
     'Item': ['F04', 'F01'],
     'Seatblocks': ['2:218:10:4,6', '1:13:36:1,12 1:13:37:1,13'],
     'ItemExt': [60, 360]
    }
)

print(df)
print('-'*80+'\n')

df['Seatblocks'] = df['Seatblocks'].str.split('[ :]')
df = df.explode('Seatblocks').reset_index(drop=True)
cols = list(df.columns)
cols.append(cols.pop(cols.index('CustomerName')))
df = df[cols]


print(df)
print('='*80+'\n')
print(df[df['CustomerName'] == 'Lennon, John'])

The output is:

   CustNum     CustomerName  ItemQty Item                 Seatblocks  ItemExt
0    32363  McCartney, Paul        3  F04               2:218:10:4,6       60
1    31316     Lennon, John       25  F01  1:13:36:1,12 1:13:37:1,13      360
--------------------------------------------------------------------------------

    CustNum  ItemQty Item Seatblocks  ItemExt     CustomerName
0     32363        3  F04          2       60  McCartney, Paul
1     32363        3  F04        218       60  McCartney, Paul
2     32363        3  F04         10       60  McCartney, Paul
3     32363        3  F04        4,6       60  McCartney, Paul
4     31316       25  F01          1      360     Lennon, John
5     31316       25  F01         13      360     Lennon, John
6     31316       25  F01         36      360     Lennon, John
7     31316       25  F01       1,12      360     Lennon, John
8     31316       25  F01          1      360     Lennon, John
9     31316       25  F01         13      360     Lennon, John
10    31316       25  F01         37      360     Lennon, John
11    31316       25  F01       1,13      360     Lennon, John
================================================================================

    CustNum  ItemQty Item Seatblocks  ItemExt  CustomerName
4     31316       25  F01          1      360  Lennon, John
5     31316       25  F01         13      360  Lennon, John
6     31316       25  F01         36      360  Lennon, John
7     31316       25  F01       1,12      360  Lennon, John
8     31316       25  F01          1      360  Lennon, John
9     31316       25  F01         13      360  Lennon, John
10    31316       25  F01         37      360  Lennon, John
11    31316       25  F01       1,13      360  Lennon, John