python re.sub如何使用

Question

Considering a string django-1.8.0.dist-info . I wish to transform it to django==1.8.0 using re.sub .

I could do it using the combination of re.search and .format but I wish to know if it could be achieved with re.sub .

I tried this:

re.sub(r"(?P<name>\\w+)-(?P<version>[\\d\\.]+)", "\\g<name>==\\g<version>", "django-1.8.0.dist-info") which returns django==1.8.0.dist-info .

I wish to take this opportunity to learn something new on regex.

Answer 1

(?P<name>\w+)-(?P<version>[\d\.]+)\..*

Use this.Just capture the last part as well.See demo.

https://regex101.com/r/sJ9gM7/23#python

Answer 2

Not sure you really need to use regexp, but what you are after you can simply achieve as follows:

s="django-1.8.0.dist-info"

s=s.replace(".dist-info", "").replace('-','==')    

print(s)    # django==1.8.0  

One regex for this could be:

import re
s="django-1.8.0.dist-info"
s = re.sub(r'^(\w+)-([\d\.]{5})(.+)', r'\1==\2', s)
print(s) # django==1.8.0

The expression make 3 groups.

• group one are characters till -
• group two are 5 characters (number or dot) with version number
• group three is rest

Answer 3

Your regex is almost ok, but you don't want to replace - you should be using re.match or re.search

>>> "==".join(re.match(r"(?P<name>\w+)-(?P<version>[\d\.]+)", "django-1.8.0.dist-info").groups())
'django==1.8.0.'

As @Avinash pointed out - there is an extra . at the end

Here is one way to fix the regex (and an example of using re.findall ):

>>> s = "django-1.8.0.dist-info"
>>> "==".join(*re.findall(r"(\w+)-(\d+(?:\.\d+)*)", s))
'django==1.8.0'

I left out the group names for (hopefully) clarity