python split a string with at least 2 whitespaces

Question

I would like to split a string only where there are at least two or more whitespaces.

For example

str = '10DEUTSCH        GGS Neue Heide 25-27     Wahn-Heide   -1      -1' print str.split() 

Results:

['10DEUTSCH', 'GGS', 'Neue', 'Heide', '25-27', 'Wahn-Heide', '-1', '-1'] 

I would like it to look like this:

['10DEUTSCH', 'GGS Neue Heide 25-27', 'Wahn-Heide', '-1', '-1'] 

Answer 1

In [4]: import re     In [5]: text = '10DEUTSCH        GGS Neue Heide 25-27     Wahn-Heide   -1      -1' In [7]: re.split(r'\s{2,}', text) Out[7]: ['10DEUTSCH', 'GGS Neue Heide 25-27', 'Wahn-Heide', '-1', '-1'] 

Update 2021+ answer.

str.split now accepts regular expressions to split on.

read more here

row = '10DEUTSCH        GGS Neue Heide 25-27     Wahn-Heide   -1      -1' df = pd.DataFrame({'string' : row},index=[0]) 

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print(df)                                               string 0  10DEUTSCH        GGS Neue Heide 25-27     Wahn... 

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df1 = df['string'].str.split('\s{2,}',expand=True) print(df1)             0                     1           2   3   4 0  10DEUTSCH  GGS Neue Heide 25-27  Wahn-Heide  -1  -1 

Answer 2

As has been pointed out, str is not a good name for your string, so using words instead:

output = [s.strip() for s in words.split('  ') if s] 

The .split(' ') -- with two spaces -- will give you a list that includes empty strings, and items with trailing/leading whitespace. The list comprehension iterates through that list, keeps any non-blank items (if s), and .strip() takes care of any leading/trailing whitespace.