If I would like to split the string from the number of the sentence: "It was amazing in 2016"
I use:
re.split('\s*((?=\d+))
out: 'It was amazing in', '2016'
Now I would like to do the opposite, so if a sentence starts with a number, then followed by a string like: '2016 was amazing'
I would like the result to be: '2016', 'was amazing'
Using lookarounds you can use a single regex for both cases:
\s+(?=\d)|(?<=\d)\s+
Code:
>>> str = "It was amazing in 2016"
>>> re.split(r'\s+(?=\d)|(?<=\d)\s+', str)
['It was amazing in', '2016']
>>> str = "2016 was amazing"
>>> re.split(r'\s+(?=\d)|(?<=\d)\s+', str)
['2016', 'was amazing']
RegEx Breakup:
\s+ - Match 1 or more whitespaces(?=\d) - Lookbehind that asserts next character is a digit| - OR(?<=\d) - Lookbehind that asserts previous character is a digit\s+ - Match 1 or more whitespaces
filter(None, re.split(r'(\D+)(?=\d)|(?<=\d)(\D+)', str)) OR re.findall(r'\d+|\D+', str)In my opinion RegEx is an overkill for that task, so unless you already are using RegEx on your program or it's required (assignment or otherwise), I recommend some string manipulation functions to get what you want.
def ends_in_digit(my_string):
separated = my_string.rsplit(maxsplit=1)
return separated if separated[-1].isdigit() else False
def starts_with_digit(my_string):
separated = my_string.split(maxsplit=1)
return separated if separated[0].isdigit() else False
Another way to easily split into digits and non-digits is to match with \d+|\D+ regex. It will yield chunks with leading/trailing whitespaces though, but they can easily be removed (or kept if that is not important):
import re
r = re.compile(r'\d+|\D+')
ss = [ 'It was amazing in 2016', '2016 was amazing']
for s in ss:
print(r.findall(s)) # to get chunks with leading/trailing whitespace
print([x.strip() for x in r.findall(s)]) # no leading/trailing whitespace
See the Python demo.
re.split(r'(?<=\d)\s*', s)