將嵌套回圈轉換為forEach();

我正在嘗試學習 forEach() 方法，但找不到更高級的示例。所以我考慮重構我的 Codewars 代碼以從中學習。我不知道在嵌套回圈中正確使用 forEach 方法。希望你能幫助我從這個例子中學習:)

6 kyu - 替換為字母位置 https://www.codewars.com/kata/546f922b54af40e1e90001da/train/javascript

function alphabetPosition(text) {
    let textToArray = text.replace(/[^a-zA-Z]/gi,'').toUpperCase().split(''); //Eliminate anything thats not a letter 
    const alphabet = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"];
    let pointsHolder = [];                            //empty array for score
    for (let i = 0; i < textToArray.length; i  ){
        for  (let j = 0; j < alphabet.length; j  ) {
            if (textToArray[i] == alphabet[j] ) {     //We check the index of given string letter in alphabet
                pointsHolder.push(j 1)                //give it a score based on place in alphabet( 1 for 0 as 1st index)
            }
        }
    }
    return pointsHolder.join(' ');                    //return scored array as a string with spaces
}

uj5u.com熱心網友回復：

確實沒有必要使用計算量大的嵌套回圈。這樣，您也不必手動創建 AZ 陣列。

您可以使用 . 輕松地將字母轉換為任意數字String.charCodeAt()。a字符代碼為 97，b字符代碼為 98，等等...要獲得從 1 開始的索引（a=1，b=2，...），您只需從數字中減去 96。

function alphabetPosition(text) {
  const alphabets = text.toLowerCase().replace(/[^a-z]/g, '').split('');
  return alphabets.map(alphabet => alphabet.charCodeAt(0) - 96).join(' ');
}

或者，您也可以使用for...of回圈，但這需要在回傳之前將陣列存盤在另一個變數中：

function alphabetPosition(text) {
  const alphabets = text.toLowerCase().replace(/[^a-z]/g, '');
  
  const codes = [];
  for (const alphabet of alphabets) {
    codes.push(alphabet.charCodeAt() - 96);
  }
  
  return codes.join(' ');
}

uj5u.com熱心網友回復：

（注意：@Terry 的解決方案仍然是解決您的代碼挑戰的更有效的解決方案）

您可以通過以下方式替換它：

function alphabetPosition(text) {
  let textToArray = text.replace(/[^a-zA-Z]/gi, '').toUpperCase().split(''); 
  const alphabet = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"];
  let pointsHolder = []; 
  textToArray.forEach(t2a => {
    alphabet.forEach((a, j) => {
      if (t2a == a) { pointsHolder.push(j   1) }
    })
  })
  return pointsHolder.join(' '); 
}

console.log(alphabetPosition("ABCSTU"))

uj5u.com熱心網友回復：

Terry's answercharCode中提出的解決方案的替代方案，但也避免了嵌套回圈，是創建一個您想要得分的字符，然后從傳遞的字串中按字符訪問它。Map

請記住，字串是可迭代的，無需轉換為陣列。

function alphabetPosition(text) {
  text = text.toUpperCase().replace(/[^A-Z]/gi, '');

  const alphabet = new Map(
    ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]
      .map((v, i) => [v, i   1])
  );

  const pointsHolder = [];
  for (const char of text) {
    pointsHolder.push(alphabet.get(char))
  }

  return pointsHolder.join(' ');
}

console.log(alphabetPosition("AB??CS??TU"))

這也允許您使用不一定具有連續字符代碼的 Map

function alphabetPosition(text) {
  text = text.toUpperCase().replace(/[^??????]/gi, '');

  const alphabet = new Map(
    ["??", "??", "??"]
      .map((v, i) => [v, i   1])
  );

  const pointsHolder = [];
  for (const char of text) {
    pointsHolder.push(alphabet.get(char))
  }

  return pointsHolder.join(' ');
}

console.log(alphabetPosition("AB??CS??TU"))

標籤：javascript 数组 功能 循环

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