parse string into columns python pandas /xa0 in stead of white space

Question

How do I quickly make new columns that hold the three chunks contained in the column 'File'?

recieved messy data like this

d = {   'File' : pd.Series(['firstname lastname                   05/31/1996                     9999999999  ', 'FN SometimesMiddileInitial. LN                    05/31/1996                 9999999999  ']), 
    'Status' : pd.Series([0., 0.]), 
    'Error' : pd.Series([2., 2.])}
df=pd.DataFrame(d)

UPDATE In reality, i'm starting from a very messy excel file and my data has '\\xa0 \\xa0' between string characters. so my first attempt looks like

from pandas import DataFrame, ExcelFile
import pandas as pd
location = r'c:/users/meinzerc/Desktop/table.xlsx'
xls = ExcelFile(location)
table = xls.parse('Sheet1')
splitdf = df['File'].str.split('\s*)

My attempt doesn't work at all. WHY?

Answer 1

You could use a regex to pick up at least two spaces:

In [11]: df.File.str.split('\s\s+')
Out[11]: 
0       [firstname lastname, 05/31/1996, 9999999999, ]
1    [FN SometimesMiddileInitial. LN, 05/31/1996, 9...
Name: File, dtype: object

Perhaps a better option is to use extract (and perhaps there is a neater regex!!):

In [12]: df.File.str.extract('\s*(?P<name>.*?)\s+(?P<date>\d+/\d+/\d+)\s+(?P<number>\w+)\s*')
Out[12]: 
                             name        date      number
0              firstname lastname  05/31/1996  9999999999
1  FN SometimesMiddileInitial. LN  05/31/1996  9999999999

[2 rows x 3 columns]