How to add the occurance number after the selected character in string?

Question

I have a string which contains multiple Q = and the goal is to add the number of occurance after each Q.

For example, 'Q = 1 t h \n Q = 2 t h \n Q = 3 t h' should be 'Q1 = 1 t h \n Q2 = 2 t h \n Q3 = 3 t h'

Here's my method:

import re

test = 'Q = 1 t h \n Q = 2 t h \n Q = 3 t h'
num = test.count('Q =')
pattern = re.compile('[Q]')

for n in range(num):
    where = [m for m in pattern.finditer(test)]
    test = test[:where[n].start()+1]+f'{n+1}'+test[where[n].start()+1:]

print(test)

Is there any better solution?

Answer 1

Possible approaches:

1. Use the .sub method of regular expressions to replace each matched occurrence. (I wanted to include a reference link for that, but I couldn't find a decent question that directly asks about using regex for substitution where OP didn't already know about re.sub.) It can accept a function that accept the matched string and returns the replacement string.

   The function we need doesn't actually need the matched string for its logic (because it's a constant and because taking it apart would be harder than just re-creating the parts we need), but needs to give a different result each time. We can create those results with logic that simply interpolates numbers sourced from itertools.count (which produces an unlimited sequence of integers counting up, on demand). Since the function only needs to consume that iterator as a one-off, we can declare it locally, and (carefully) use it as a closure in a lambda.

   import re
   from itertools import count
   from functools import partial
   
   pattern = re.compile('Q =')
   test = 'Q = 1 t h \n Q = 2 t h \n Q = 3 t h'
   c = count(1)
   test = pattern.sub(lambda _: f'Q{next(c)} =', test)
   

2. Instead of regular expressions, since the searched-for substring is a constant, just .split() the string on that value, generate and collect the replacement strings, interweave the replacements with the parts in between and join the results again.

   from itertools import count
   
   test = 'Q = 1 t h \n Q = 2 t h \n Q = 3 t h'
   
   between = test.split('Q =')
   parts = [None] * (len(between) * 2 - 1)
   parts[::2] = between
   parts[1::2] = [f'Q{i} =' for i in range(1, len(between))]
   test = ''.join(parts)
   

3. Supposing that we can assume that the string is also split into lines (i.e., there is \n at the end of each "question" part): split the string into those lines, replace the start of each line using similar techniques, and join the lines back together:

   test = 'Q = 1 t h \n Q = 2 t h \n Q = 3 t h'
   lines = test.split('\n')
   test = '\n'.join(
       line.replace('Q =', f'Q{i} =') for i, line in enumerate(lines, 1)
   )
   

   Here, enumerate is a built-in function that replaces the count logic - it automatically matches up values like the ones from itertools.count, with values taken from another sequence (here, the lines of the original text). See also Accessing the index in 'for' loops.

4. If we can't make that assumption, use regex to split the string instead. Use a lookahead assertion so that the actually matched pattern doesn't include any text, but matches right before each Q =, allowing the string to be split in those places. Then make the same replacements as before. This time, we join with an empty string, because we did not remove newlines from the original data.

   import re
   
   pattern = re.compile('(?=Q =)')
   test = 'Q = 1 t h \n Q = 2 t h \n Q = 3 t h'
   lines = pattern.split(test)
   test = ''.join(
       line.replace('Q =', f'Q{i} =') for i, line in enumerate(lines)
   )
   

   Note that this time, we start the enumeration from 0, because splitting this way will produce an empty string at the beginning of the lines.

Answer 2

I would use a regex approach this way :

out = re.sub(r"Q\s*=\s*(\d+)", r"Q\1 = \1", test)

Or simply use replace (to follow your approach) :

out = "\n".join([l.replace("Q =", f"Q{i+1} =") for i, l in enumerate(test.split("\n"))])

Output :

>>> print(out)

'Q1 = 1 t h \n Q2 = 2 t h \n Q3 = 3 t h'

Answer 3

If your strings don't contain curly braces, you could use the format function to inject the sequence numbers:

test = "Q = 1 t h \n Q = 2 t h \n Q = 3 t h"

result = test.replace("Q =","Q{} =").format(*range(1,test.count("Q =")+1))

print(result)

Q1 = 1 t h 
 Q2 = 2 t h 
 Q3 = 3 t h

If there could be curly braces, you could use split instead, separating the string using "Q =" as the separator and reassembling it with the injected numbering:

result = "".join(f"Q{i} ="*(i>0)+s for i,s in enumerate(test.split("Q =")))

Or you could just use a loop that replaces the "Q =" one at a time:

for i in range(test.count("Q =")):
    test = test.replace("Q =",f"Q{i+1} =",1)

print(test)
    Q1 = 1 t h 
     Q2 = 2 t h 
     Q3 = 3 t h