I want to replace string like
'''1 2 3 4 5 6 abcde fghij klmno pqrst 7 8 9 10 uvwxyz abcdef 11 12 13'''
to
'''1 2 3 4 5 6
abcde fghij klmno pqrst
7 8 9 10
uvwxyz abcdef
11 12 13'''
that is my method:
s = re.sub(r'(\d) ([a-z])', r'\1\n\2', s)
s = re.sub(r'([a-z]) (\d)', r'\1\n\2', s)
how can I do this in one regular expression? and I know I can do it use re.findall and groups but I want to find a more easy way?
I really think the easiest way would be to match using findall instead of splitting or sub-ing:
result = re.findall(r"\d+(?:\s+\d+)*|[a-z]+(?:\s+[a-z]+)*", text)
print('\n'.join(result))
or in one line:
result = '\n'.join(re.findall(r"\d+(?:\s+\d+)*|[a-z]+(?:\s+[a-z]+)*", text))
Gives:
1 2 3 4 5 6
abcde fghij klmno pqrst
7 8 9 10
uvwxyz abcdef
11 12 13
\d+(?:\s+\d+)* matches the parts with digits and spaces.
[a-z]+(?:\s+[a-z]+)* matches the parts with letters and spaces.
Here are two ways to do it with a single regex:
Use a conditional pattern. Capture \1 is straightforward. Capture \4 checks whether we grabbed \2 or \3, and then defines the rest of the pattern accordingly.
re.sub(r'((\d)|([a-z])) ((?(2)[a-z]|\d))', r'\1\n\4', s)
Replace only the space, and surround it with look-behind and look-ahead assertions.
re.sub(r'(?<=\d) (?=[a-z])|(?<=[a-z]) (?=\d)', '\n', s)
But your two simple regexes are better than all of this nonsense.
You can use the regular expression or command:
s = re.sub(r'((\d) ([a-z])|([a-z]) (\d))', r'\2\4\n\3\5', s)
It'll match or group 2 & 3 or group 4 & 5. =]
error: unmatched group.You could use re.split
>>> s = '''1 2 3 4 5 6 abcde fghij klmno pqrst 7 8 9 10 uvwxyz abcdef 11 12 13'''
>>> for i in re.split(r'(?<=\d)\s+(?=[A-Za-z])|(?<=[A-Za-z])\s+(?=\d)', s):
print(i)
1 2 3 4 5 6
abcde fghij klmno pqrst
7 8 9 10
uvwxyz abcdef
11 12 13
>>> print('\n'.join(re.split(r'(?<=\d)\s+(?=[A-Za-z])|(?<=[A-Za-z])\s+(?=\d)', s)))
OR
re.sub
>>> print(re.sub(r'(?<=\d)\s+(?=[A-Za-z])|(?<=[A-Za-z])\s+(?=\d)', r'\n', s))
1 2 3 4 5 6
abcde fghij klmno pqrst
7 8 9 10
uvwxyz abcdef
11 12 13
The above re.sub command will replace one or more spaces which exists between digit and a letter or between a letter and a digit with newline character.
You can use a replacement:
re.sub(r'(\d[\d\s]*|[a-z][a-z\s]*)', r'\1\n', s)
To be more rigorous with trailing whitespaces, you can do that:
re.sub(r'(\d(?:[\d\s]*\d)?|[a-z](?:[a-z\s]*[a-z])?)\s*', r'\1\n', s).rstrip()
