Retain intermediate string using regex

Question

I want to append a substring to a string pattern at all occurrences in multiple python code files. However the original string follows a pattern and is not an exact same string each time. Below are some examples of the variation:

Original Code:  a.b();
Want Code:      a.b().c();

Original Code:  a.b(param1=1);
Want Code:      a.b(param1=1, param2=2).c();

Original Code:  a.b(param1=1, param2=2);
Want Code:      a.b(param1=1, param2=2).c();

Original Code:  a.b(param1=D());
Want Code:      a.b(param1==D()).c();

Original Code:  X(a.b(param1=D()));
Want Code:      X(a.b(param1==D()).c());

---

Update: Since I am attempting to replace code in a file, the file contains indention and new lines for better readability: e.g

Original Code:  X(a.b(
                     param1=D()
                     )
                 );

Want Code:      X(a.b(
                     param1=D()
                     ).c()
                 );

Original Code:  X(a.b(
                     param1=D(),
                     param2="qwerty"
                     )
                 );

Want Code:      X(a.b(
                     param1=D(),
                     param2="qwerty"
                     ).c()
                 );

Original Code:  X(a.b(
                       newObj())
                 );

Want Code:      X(a.b(
                       newObj()).c()
                 );

I am not really concerned about parameters passed in function b. I simply need to append invocation of c() every time a.b() is invoked.

I am using the regex 'a.b(.*?)' to detect the appropriate original code. I tried using the following solution regexes: a.b($1).c() or a.b(\1).c() but to no avail.

Answer 1

You can use

a\.b\([^()]*\)(?=;)

• a\.b Match literally and escape the dot
• \([^()]*\) Match from an opening parenthesis till closing parenthesis using a negated character class
• (?=;) Positive lookahead, assert a ; to the right

Regex demo | Python demo

And replace with the full match \g<0> followed by .c()

\g<0>.c()

For example:

import re

regex = r"a\.b\([^()]*\)(?=;)"

s = ("a.b();\n"
    "a.b(param1=1);\n"
    "a.b(param1=1, param2=2);")

result = re.sub(regex, r"\g<0>.c()", s)

if result:
    print (result)

Output

a.b().c();
a.b(param1=1).c();
a.b(param1=1, param2=2).c();

---

Matching balanced parenthesis using the PyPi regex module:

a\.b(\((?>[^()]++|(?1))*\))

The pattern matches:

• a\.b Match .b
• ( Capture group 1
  • \( Match (
  • (?> Atomic group (no backtracking)
    • [^()]++ Match 1+ occurrences of any char except ( or )
    • | Or
    • (?1) Recurse the first subpattern (group 1)
  • )* Close the group and optionally repeat
  • \) Match )
• ) Close group 1

Regex demo | Python demo

import regex

pattern = r'a\.b(\((?>[^()]++|(?1))*\))'
strings = [
    "a.b();",
    "a.b(param1=1);",
    "a.b(param1=1, param2=2);",
    "a.b(param1=d(abc=\"123\"));"
]

for s in strings:
    m = regex.match(pattern, s)
    if m:
        print(f"{m.group()}.c()")

Output

a.b().c()
a.b(param1=1).c()
a.b(param1=1, param2=2).c()
a.b(param1=d(abc="123")).c()