python，re.sub每'x'个字符添加一个空格

Question

string = 'ABCDEFGHIJLM'

I'm trying to achieve the following result using

re.sub(SOME CODE HERE):
'ABC DEF GHI JLM'

Is it possible to do that?

Answer 1

Just match three letters "(\\w\\w\\w)" and replace them with themselves plus a space ("\\1 "):

print re.sub(r'(\w\w\w)', r'\1 ', 'ABCDEFGHIJLM')

Prints:

'ABC DEF GHI JLM '

To get rid of the trailing space, you can put a negative look-ahead of "Not end of string (?!$)":

print re.sub(r'(\w\w\w)(?!$)', r'\1 ', 'ABCDEFGHIJLM')

Prints:

'ABC DEF GHI JLM'

If you want to make the group size a parameter, you can specify the letter count as a quantifier ({n}) after the \\w to define the size. Eg:

group_size = 2
print re.sub(r'(\w{%d})(?!$)' % group_size, r'\1 ', 'ABCDEFGHIJLM')

Prints:

'AB CD EF GH IJ LM'

Answer 2

Use re findall and join this way:

 s='ABCDEFGHIJLM'
 n=3
 pattern= '.'*n 
 ' '.join(re.findall(pattern, s))
'ABC DEF GHI JLM'