Insert operators from string into expression string

Question

Given a string with variables and parentheses:

'a((bc)((de)f))'

and a string of operators:

'+-+-+'

I would like to insert each operator (in order) into the first string between the following patterns (where char is defined as a character that is not an open or close parenthesis):

• char followed by char
• char followed by '('
• ')' followed by '('
• ')' followed by char

To give the result:

'a+((b-c)+((d-e)+f))'

Edit: I got it to work with the following code, but is there a more elegant way to do this, i.e. without a for loop?

 x = 'a((bc)((de)f))'
 operators = '+-+-+'
 y = x
 z = 0
 for i in range(len(x)):
 if i < len(x)-1:
 xx = x[i]
 isChar = True if x[i] != '(' and x[i] != ')' else False
 isPO = True if x[i] == '(' else False
 isPC = True if x[i] == ')' else False
 isNxtChar = True if x[i+1] != '(' and x[i+1] != ')' else False
 isNxtPO = True if x[i+1] == '(' else False
 isNxtPC = True if x[i+1] == ')' else False
 if (isChar and (isNxtChar or isNxtPO)) or (isPC and (isNxtPO or isNxtChar)):
 aa = operators[z]
 split1 = x[:i+1]
 split2 = x[i+1:]
 y = y[:i+z+1] + operators[z] + x[i+1:]
 if z+1 < len(operators):
 z+=1
 
 print (y)

• jonrsharpe: What have you tried, and what exactly is the problem with it?
• Craig Nathan: @jonrsharpe I edited my post with functioning code, but if you have any suggestions how to make it better anything is appreciated
• Craig Nathan: @JohnColeman sorry, I didn't mean it like that. I'm just looking for the most efficient way to do this with the least amount of code
• jonrsharpe: "better" how, exactly? Elegance isn't an objective measure. If you have working code that you think could be improved, see Code Review.
• Craig Nathan: My code wasn't working at the time of my post. I got it working after asking. Do you or anyone have any suggestions on how to achieve the result without using a for loop? I wasn't aware of code review, thank you for pointing that out. I can post there as well.

Answer 1

initialExpr = 'a((bc)((de)f))'
operators = '+-+-+'
countOp = 0
countChar = 0
for char in initialExpr:
 countChar += 1
 print(char,end='')
 if countChar < len(initialExpr) and (char == ')' or char.isalpha()) and (initialExpr[countChar] == '(' or initialExpr[countChar].isalpha()):
 print(operators[countOp], end='')
 countOp += 1

This should do the job. Assumption is the the variables, parenthesis and operators are in the right order and number.

• Craig Nathan: Thank you! This is better than what I came up with. The only thing I have modified is instead of the print statements, I've appended to a string so I can use it elsewhere
• Marcel Preda: @CraigNathan, just edited my initial answer, the if condition now is shorter.

Answer 2

One-liner using re:

import re
s = "a((bc)((de)f))"
o = "+-+-+"
print(
 re.sub(
 r"(?:[a-z](?:\(|[a-z]))|(?:\)(?:\(|[a-z]))",
 lambda g, i=iter(o): next(i).join(g.group()),
 s,
 )
)

Prints:

a+((b-c)+((d-e)+f))

Answer 3

You can use a regex matching the pairs of characters inside which you want to insert an operator.

Then, you can use re.sub with a replacement function that joins the two characters with the next operator.

We can use a class with a __call__ method, that uses an iterator on the operators:

import re
rules = re.compile(r'[a-z]{2}|[a-z]\(|\)\(|\)[a-z]')
class Replace:
 def __init__(self, operators):
 self.it_operators = iter(operators)
 
 def __call__(self, match):
 return next(self.it_operators).join(match.group())
variables = 'a((bc)((de)f))'
operators = '+-+-+'
print(rules.sub(Replace(operators), variables))
# a+((b-c)+((d-e)+f))

Replace(operators) returns a callable Replace instance with an it_operators attribute that is an iterator, ready to iterate on the operators. For each matching pair of characters, sub calls this instance, and its __call__ method returns the replacement for the two characters, that it builds by joining them with the next operator.