I have a string s, a pattern p and a replacement r, i need to get the list of strings in which only one match with p has been replaced with r.
Example:
s = 'AbcAbAcc'
p = 'A'
r = '_'
// Output:
['_bcAbAcc', 'Abc_bAcc', 'AbcAb_cc']
I have tried with re.finditer(p, s) but i couldn't figure out how to replace each match with r.
You can replace them manually after finding all the matches:
[s[:m.start()] + r + s[m.end():] for m in re.finditer(p,s)]
The result is:
['_bcAbAcc', 'Abc_bAcc', 'AbcAb_cc']
How does it work?
re.finditer(p,s) will find all matches (each will be a re.Match
object)re.Match objects have start() and end() method which return the location of the matchs[:begin] + replacement + s[end:]You don't need regex for this, it's as simple as
[s[:i]+r+s[i+1:] for i,c in enumerate(s) if c==p]
Full code: See it working here
s = 'AbcAbAcc'
p = 'A'
r = '_'
x = [s[:i]+r+s[i+1:] for i,c in enumerate(s) if c==p]
print(x)
Outputs:
['_bcAbAcc', 'Abc_bAcc', 'AbcAb_cc']
As mentioned, this only works on one character, for anything longer than one character or requiring a regex, use zvone's answer.
For a performance comparison between mine and zvone's answer (plus a third method of doing this without regex), see here or test it yourself with the code below:
import timeit,re
s = 'AbcAbAcc'
p = 'A'
r = '_'
def x1():
return [s[:i]+r+s[i+1:] for i,c in enumerate(s) if c==p]
def x2():
return [s[:i]+r+s[i+1:] for i in range(len(s)) if s[i]==p]
def x3():
return [s[:m.start()] + r + s[m.end():] for m in re.finditer(p,s)]
print(x1())
print(timeit.timeit(x1, number=100000))
print(x2())
print(timeit.timeit(x2, number=100000))
print(x3())
print(timeit.timeit(x3, number=100000))
p is always one character long, but for what i need p is going to be an actual pattern. But yes with the example I provided that's going to work perfectly!