Python正则表达式循环跳过每三个项目

Question

I'm doing a tokenizer and I want to separate strings like "word-bound-with-hyphen" into "word xxsep bound xxsep with xxsep hyphen".

I tried this:

import re

s = "words-bound-with-hyphen"
reg_m = re.compile("[\w\d]+-[\w\d]+")
reg = re.compile("([\w\d]+)-([\w\d]+)")
while(reg_m.match(s)):
    s = reg.sub(r"\1 xxsep \2", s)
print(s) #prints "words xxsep bound-with xxsep hyphen"

But this leaves every third hyphen-bound word.

Answer 1

You could just replace the hyphens with a regex:

In [4]: re.sub("-", " xxsep ", "word-bound-with-hyphen")
Out[4]: 'word xxsep bound xxsep with xxsep hyphen'

or with string substitution:

In [7]: "word-bound-with-hyphen".replace("-", " xxsep ")
Out[7]: 'word xxsep bound xxsep with xxsep hyphen'

The reason your current approach doesn't work is that re.sub() returns non-overlapping groups whereas word-bound overlaps with bound-with overlaps with with-hyphen .

Answer 2

If you don't want to just replace all hyphens but only those that are preceded and followed by certain characters than use regex lookbacks and lookaheads.

import re
s = "words-bound-with-hyphen"
re.sub('(?<=[\w\d])-(?=[\w\d])',' xxsep ', s)
# result: 'words xxsep bound xxsep with xxsep hyphen'

Answer 3

import re
s = "words-bound-with-hyphen"
re.sub('-',' xxsep ',s)

or without using regular expressions

" xxsep ".join(x.split('-'))

here, the list will be separated taking - as delimiter and then joined using "xxsep"

Answer 4

Why not use word boundaries . Search for \\b-\\b and replace with xxsep .