Regular expression used to split a unary operator

Question

Regular expression used to split a unary operator

Although there are many regex questions related to my issue, none of them address my specific problem as I do not have parentheses in my expressions.

Let's take a look at an example: 1*2+3-99/50

I am trying to tokenize this expression in order to convert it to post-fix notation. My first attempt was to use split with regex like this:

'1*2+3-99/50'.split(/([+\-*/])/)                   // operands have no unary (-)
// ['1', '*', '2', '+', '3', '-', '99', '/', '50']

It seems to work perfectly for binary operators but when dealing with unary operators, it falls short. I then attempted to utilize a lookahead to detect if a minus sign comes after another operator:

'1*-2+-3--99/-50'.split(/([+\-*/])(?=-)/               // all operands have unary (-)
// ['1', '*', '-2', '+', '-3', '-', '-99', '/', '-50']

This method also works, but the caveat is that all numbers already have negative signs attached to them.

I've tried capturing operands with unary (-) operators separately and appending them to the final tokens, but it messes up the order which is crucial for evaluation. I also looked into conditional regex but it doesn't seem applicable in my scenario.

Is it feasible to split tokens containing negative signs in one single split action?

javascript regex split

Answer 1

Answer №1

Consider using the \b delimiter for splitting

var numbers = '1*-2+-3-99/-50'.split(/(\b[-+*/])/);

console.log(JSON.stringify(numbers))

Answer 2

Consider using the \b delimiter for splitting

var numbers = '1*-2+-3-99/-50'.split(/(\b[-+*/])/);

console.log(JSON.stringify(numbers))

Answer 3

Answer №2

It is highly advisable to avoid using regular expressions for this task. The critical factor here is maintaining the correct order of operations, a feature that postfix notation can handle effortlessly but infix cannot.

Although this code snippet makes use of regular expressions, the core logic lies elsewhere:

function toPostfix(equation) {
  let output = "";
  let additions = equation.replace(/\b\s*-/g, "+-").split(/\+/);
  if (additions.length > 1) {
    output = "+)";
    for (let a = additions.length - 1; a >= 0; a--) {
      output = toPostfix(additions[a]) + " " + output;
    }
    return "( " + output;
  } else {
    if (equation.match(/^1\/-?[0-9]+$/)) {
       return "( 1 " + equation.substring(2) + " /)";
    }
    let multiplications = equation.replace(/\//g, "*1/").split(/\*/);
    if (multiplications.length > 1) {
      output = "*)";
      for (let m = multiplications.length - 1; m >= 0; m--) {
        output = toPostfix(multiplications[m]) + " " + output;
      }
      return "( " + output;
    } else {
      return equation;
    }
  }
}

console.log(toPostfix('1*2+3-99/50'));
//  "( ( 1 2 *) 3 ( -99 ( 1 50 /) *) +)"

To simplify the process, this function transforms subtractions into additions of negative numbers and divisions into multiplications by the inverse (1/n). To maintain the correct order of operations, we first split based on additions and subtractions, then proceed with each multiplication and division operation. The algorithm employs recursion to break down each segment and gradually build up the final answer from right to left.

This implementation solely handles addition, subtraction, multiplication, and division. There are no provisions for parentheses or exponents.

Answer 4

It is highly advisable to avoid using regular expressions for this task. The critical factor here is maintaining the correct order of operations, a feature that postfix notation can handle effortlessly but infix cannot.

Although this code snippet makes use of regular expressions, the core logic lies elsewhere:

function toPostfix(equation) {
  let output = "";
  let additions = equation.replace(/\b\s*-/g, "+-").split(/\+/);
  if (additions.length > 1) {
    output = "+)";
    for (let a = additions.length - 1; a >= 0; a--) {
      output = toPostfix(additions[a]) + " " + output;
    }
    return "( " + output;
  } else {
    if (equation.match(/^1\/-?[0-9]+$/)) {
       return "( 1 " + equation.substring(2) + " /)";
    }
    let multiplications = equation.replace(/\//g, "*1/").split(/\*/);
    if (multiplications.length > 1) {
      output = "*)";
      for (let m = multiplications.length - 1; m >= 0; m--) {
        output = toPostfix(multiplications[m]) + " " + output;
      }
      return "( " + output;
    } else {
      return equation;
    }
  }
}

console.log(toPostfix('1*2+3-99/50'));
//  "( ( 1 2 *) 3 ( -99 ( 1 50 /) *) +)"

To simplify the process, this function transforms subtractions into additions of negative numbers and divisions into multiplications by the inverse (1/n). To maintain the correct order of operations, we first split based on additions and subtractions, then proceed with each multiplication and division operation. The algorithm employs recursion to break down each segment and gradually build up the final answer from right to left.

This implementation solely handles addition, subtraction, multiplication, and division. There are no provisions for parentheses or exponents.

Answer 5

Answer №3

If you want to simplify the expression, consider making some modifications

this.expression = this.expression.replaceAll("&--", "+")
    .replaceAll("&++", "+")
    .replaceAll("&-+", "-")
    .replaceAll("&+-", "-");

Answer 6

If you want to simplify the expression, consider making some modifications

this.expression = this.expression.replaceAll("&--", "+")
    .replaceAll("&++", "+")
    .replaceAll("&-+", "-")
    .replaceAll("&+-", "-");

Regular expression used to split a unary operator

Answer №1

Answer №2

Answer №3

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