熊猫:如何将一列中的文本分成多行?-Python 实用宝典

熊猫:如何将一列中的文本分成多行?

我正在处理一个较大的csv文件,并且最后一列的旁边是一串文本,我想用一个特定的分隔符来分割它。我想知道是否有使用pandas或python的简单方法? 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 我想先按空格(' ')再(':')在Seatblocks列中按冒号分开,但每个单元格将导致列数不同。我具有重新排列列的功能,因此Seatblocks列位于工作表的末尾,但是我不确定从那里开始如何做。我可以使用内置text-to-columns函数和快速宏在excel中完成此操作,但是我的数据集记录太多,无法处理excel。 最终,我想记录约翰·列侬的记录并创建多行,并将每组座位的信息放在单独的行上。

问题:熊猫:如何将一列中的文本分成多行?

我正在处理一个较大的csv文件,并且最后一列的旁边是一串文本,我想用一个特定的分隔符来分割它。我想知道是否有使用pandas或python的简单方法?

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

我想先按空格(' ')(':')Seatblocks列中按冒号分开,但每个单元格将导致列数不同。我具有重新排列列的功能,因此Seatblocks列位于工作表的末尾,但是我不确定从那里开始如何做。我可以使用内置text-to-columns函数和快速宏在excel中完成此操作,但是我的数据集记录太多,无法处理excel。

最终,我想记录约翰·列侬的记录并创建多行,并将每组座位的信息放在单独的行上。

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.


回答 0

这将座垫按空间划分,并给每个单独的行。

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

或者,将每个冒号分隔的字符串放在自己的列中:

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

这有点丑陋,但也许有人会用更漂亮的解决方案。

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.


回答 1

与Dan不同的是,我认为他的回答相当优雅……但是不幸的是,它的效率也非常低下。因此,由于问题提到“大的csv文件”,因此我建议尝试使用Shell Dan的解决方案:

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()"

...与这种替代方案相比:

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()"

... 还有这个:

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()"

第二个简单地避免了分配10万个序列,这足以使它快10倍左右。但是,第三种解决方案有点讽刺地浪费了对str.split()的调用(每行每列调用一次,因此比其他两种解决方案多三倍),它比第一种解决方案快40倍,因为它甚至避免实例化100000个列表。是的,这确实有点丑陋...

编辑: 此答案建议如何使用“ to_list()”并避免使用lambda。结果是像

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()"

这比第三个解决方案更有效,而且肯定更优雅。

编辑:更简单

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()"

也可以,并且几乎一样有效。

编辑: 更简单!并处理NaN(但效率较低):

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

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()"

回答 2

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

链接的另一个类似解决方案是use reset_indexrename

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

如果in列中不是NOT NaN值,则最快的解决方案是listDataFrame构造函数使用理解:

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

但是如果列NaN只包含str.splitexpand=True返回的参数一起使用DataFrame值为(document)的,那么它解释了为什么它比较慢:

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
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

回答 3

另一种方法是这样的:

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

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)

回答 4

也可以使用groupby()而不需要加入和stack()。

使用上面的示例数据:

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

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

回答 5

这似乎比该线程其他地方建议的方法容易得多。

在熊猫数据框中拆分行

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

split rows in pandas dataframe


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