The following code is functional but I would like to refactor the !== part that allows my ternary to run only on values that are not white space so I can include edge case test's. This would include any non letter value as well as white space and I know that regex may play a part but I can't find a niftty way to incorporate it into the if() statement that precedes the ternary operations.

const letterPositions = function(strPos) {
  if (typeof strPos !== 'string') {
    return console.log('Sorry your input is not a string');
  }

  const result = {};
  for (let i = 0; i < strPos.length; i  ) {
    if (strPos[i] !== ' ') {
      result[strPos[i]] ? result[strPos[i]].push(i) : (result[strPos[i]] = [i]);
    }
  }

  return result;
};

console.log(letterPositions('aa bb cc'));

CodePudding user response：

mainly two options, regex and charcode, welcome to edit if more ways available

const codes = ['A', 'Z', 'a', 'z'].map(x => x.charCodeAt(0))
console.log(codes)

const letterPositions = function(strPos) {
  if (typeof strPos !== 'string') {
    return console.log('Sorry your input is not a string');
  }

  const result = {};
  for (let i = 0; i < strPos.length; i  ) {

    //if (strPos[i] !== ' ') {                       // old

    //if (/[A-Za-z]/.test(strPos[i])) {                 // regex

    let code = strPos.charCodeAt(i)
    if ((code >= 65 && code <= 90) || (code >= 97 && code <= 122)) { // charcode
      result[strPos[i]] ? result[strPos[i]].push(i) : (result[strPos[i]] = [i]);
    }
  }

  return result;
};

console.log(letterPositions('aa bb cc'));

CodePudding user response：

you can also do that...

const letterPositions = str =>
  {
  if (typeof str !== 'string') 
    return console.log('Sorry your input is not a string' )
 
  return [...str].reduce((r,l,i)=>((l===' ')?null:(r[l]??=[],r[l].push(i)),r),{})
  }

console.log( letterPositions('aa bb cc') )

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