CODEWITHSUNDEEP
pythonregex
Subhayan Bhattacharya
Subhayan Bhattacharya

Reputation: 5705

Python regular expressions substitution

I have the following string : "n && && && && && &&n" and I want to substitiute && with and but only those &&s which have a space before and after them.

I am using the below code:

import re
str = input()
change = re.sub(r' (&&) ', ' and ', str)
print (change)

But the output is coming as: n and && and && and &&n

The correct output should be: n and and and and and &&n

Can someone please tell me where I am going wrong?

Upvotes: 0

Views: 52

Answers (3)

TigerhawkT3
TigerhawkT3

Reputation: 49320

In your expression, you are expecting the spaces to be counted twice, but a space after an && will make it unavailable to be counted as a space before the next &&. You are looking for "lookarounds":

>>> import re
>>> s = 'n && && && && && &&n'
>>> change = re.sub(r'(?<= )(&&)(?= )', 'and', s)
>>> print (change)
n and and and and and &&n

These "look" for the given matches without actually consuming them. You're actually only interested in the ands; they just have to be surrounded by the spaces.

Also, don't name your variable str or you make it difficult to access the built-in function str().

Upvotes: 1

meuh
meuh

Reputation: 12255

Your regular expression will not look at the ending space again after doing one substitution, so the next character checked will be "&& ...". Instead "lookahead" for a space instead of substituting for it:

change = re.sub(r' (&&)(?= )',' and',str)

Upvotes: 0

Sniggerfardimungus
Sniggerfardimungus

Reputation: 11831

The trouble is that the 'n && ', at the beginning, is picked up by the first substitution match. Once that space after the && is removed (albeit, replaced by a space in the ' and ',) it can't be used in another match.

The good news is that you'll never end up with consecutive &&s after your sub pass and - for the simplest solution - you can just run sub twice:

In [7]: input_str = 'n && && && && && &&n'

In [8]: first_sub = re.sub(r' (&&) ', ' and ', input_str)

In [9]: second_sub = re.sub(r' (&&) ', ' and ', first_sub)

In [10]: print second_sub
n and and and and and &&n

Upvotes: 0

Related Questions