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)

What have you tried, and what exactly is the problem with it? — jonrsharpe
– jonrsharpe, Commented Apr 18, 2021 at 19:44
@jonrsharpe I edited my post with functioning code, but if you have any suggestions how to make it better anything is appreciated — Craig Nathan
– Craig Nathan, Commented Apr 18, 2021 at 20:10
@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 — Craig Nathan
– Craig Nathan, Commented Apr 18, 2021 at 20:10
"better" how, exactly? Elegance isn't an objective measure. If you have working code that you think could be improved, see Code Review. — jonrsharpe
– jonrsharpe, Commented Apr 18, 2021 at 20:12
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. — Craig Nathan
– Craig Nathan, Commented Apr 18, 2021 at 20:19

Marcel Preda · Accepted Answer · 2021-04-18 20:25:21Z

2

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.

edited Apr 18, 2021 at 20:25

answered Apr 18, 2021 at 20:03

Marcel Preda

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2 Comments

Craig Nathan Over a year ago

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 Over a year ago

@CraigNathan, just edited my initial answer, the if condition now is shorter.

Andrej Kesely · Accepted Answer · 2021-04-18 20:35:04Z

1

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))

edited Apr 18, 2021 at 20:35

answered Apr 18, 2021 at 20:20

Andrej Kesely

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Comments

Thierry Lathuille · Accepted Answer · 2021-04-18 20:42:49Z

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.

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