Python: Count and replace regular expression in same pass?

Question

I can globally replace a regular expression with re.sub(), and I can count matches with

for match in re.finditer(): count++

Is there a way to combine these two, so that I can count my substitutions without making two passes through the source string?

Note: I'm not interested in whether the substitution matched, I'm interested in the exact count of matches in the same call, avoiding one call to count and one call to substitute.

Answer 1

You can use re.subn.

re.subn(pattern, repl, string, count=0, flags=0)

it returns (new_string, number_of_subs_made)

For example purposes, I'm using the same example as @Shubham Sharma used.

text = "Jack 10, Lana 11, Tom 12, Arthur, Mark"
out_str, count = re.subn(r"(\d+)", repl='repl', string=text)

# out_str--> 'Jack repl, Lana repl, Tom repl, Arthur, Mark'
# count---> 3

Answer 2

You can pass a repl function while calling the re.sub function. The function takes a single match object argument, and returns the replacement string. The repl function is called for every non-overlapping occurrence of pattern.

Try this:

count = 0
def count_repl(mobj): # --> mobj is of type re.Match
    global count
    count += 1 # --> count the substitutions
    return "your_replacement_string" # --> return the replacement string

text = "The original text" # --> source string
new_text = re.sub(r"pattern", repl=count_repl, string=text) # count and replace the matching occurrences in one pass.

OR,

You can use re.subn which performs the same operation as re.sub, but return a tuple (new_string, number_of_subs_made).

new_text, count = re.sub(r"pattern", repl="replacement", string=text)

---

Example:

count = 0
def count_repl(mobj):
    global count
    count += 1
    return f"ID: {mobj.group(1)}"

text = "Jack 10, Lana 11, Tom 12, Arthur, Mark"
new_text = re.sub(r"(\d+)", repl=count_repl, string=text)

print(new_text)
print("Number of substitutions:", count)

Output:

Jack ID: 10, Lana ID: 11, Tom ID: 12
Number of substitutions: 3

Answer 3

import re


text = "Jack 10, Lana 11, Tom 12"
count = len([x for x in re.finditer(r"(\d+)", text)])
print(count)

# Output: 3

Ok, there's a better way

import re


text = "Jack 10, Lana 11, Tom 12"
count = re.subn(r"(\d+)", repl="replacement", string=text)[1]
print(count)

# Output: 3