How to 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 do this using pandas or python?
CustNum CustomerName ItemQty Item Seatblocks ItemExt
32363 McCartney, Paul 3 F04 2:218:10:4,6 60
31316 Lennon, John 25 F01 1:13:36:1,12 1:13:37:1,13 300
I want to split by the space (' ') and then the colon (':') in the Seatblocks column, but each cell would result in a different number of columns. I have a function to rearrange the columns so the Seatblocks column is at the end of the sheet, but I’m not sure what to do from there. I can do it in excel with the built in text-to-columns function and a quick macro, but my dataset has too many records for excel to handle.
Ultimately, I want to take records such John Lennon’s and create multiple lines, with the info from each set of seats on a separate line.
Answers:
This splits the Seatblocks by space and gives each its own row.
In [43]: df
Out[43]:
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
In [44]: s = df['Seatblocks'].str.split(' ').apply(Series, 1).stack()
In [45]: s.index = s.index.droplevel(-1) # to line up with df's index
In [46]: s.name = 'Seatblocks' # needs a name to join
In [47]: s
Out[47]:
0 2:218:10:4,6
1 1:13:36:1,12
1 1:13:37:1,13
Name: Seatblocks, dtype: object
In [48]: del df['Seatblocks']
In [49]: df.join(s)
Out[49]:
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
Or, to give each colon-separated string in its own column:
In [50]: df.join(s.apply(lambda x: Series(x.split(':'))))
Out[50]:
CustNum CustomerName ItemQty Item ItemExt 0 1 2 3
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
This is a little ugly, but maybe someone will chime in with a prettier solution.
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()"
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
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
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)
This seems a far easier method than those suggested elsewhere in this thread.
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
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 do this using pandas or python?
CustNum CustomerName ItemQty Item Seatblocks ItemExt
32363 McCartney, Paul 3 F04 2:218:10:4,6 60
31316 Lennon, John 25 F01 1:13:36:1,12 1:13:37:1,13 300
I want to split by the space (' ') and then the colon (':') in the Seatblocks column, but each cell would result in a different number of columns. I have a function to rearrange the columns so the Seatblocks column is at the end of the sheet, but I’m not sure what to do from there. I can do it in excel with the built in text-to-columns function and a quick macro, but my dataset has too many records for excel to handle.
Ultimately, I want to take records such John Lennon’s and create multiple lines, with the info from each set of seats on a separate line.
This splits the Seatblocks by space and gives each its own row.
In [43]: df
Out[43]:
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
In [44]: s = df['Seatblocks'].str.split(' ').apply(Series, 1).stack()
In [45]: s.index = s.index.droplevel(-1) # to line up with df's index
In [46]: s.name = 'Seatblocks' # needs a name to join
In [47]: s
Out[47]:
0 2:218:10:4,6
1 1:13:36:1,12
1 1:13:37:1,13
Name: Seatblocks, dtype: object
In [48]: del df['Seatblocks']
In [49]: df.join(s)
Out[49]:
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
Or, to give each colon-separated string in its own column:
In [50]: df.join(s.apply(lambda x: Series(x.split(':'))))
Out[50]:
CustNum CustomerName ItemQty Item ItemExt 0 1 2 3
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
This is a little ugly, but maybe someone will chime in with a prettier solution.
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()"
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
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
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)
This seems a far easier method than those suggested elsewhere in this thread.
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