按字母順序對字符串數組進行排序，然后按特殊字符對數字進行排序

Question

給定以下字符串數組，我一直嘗試自然地對az字符串進行排序，然后是數字字符串，最后是特殊字符。

    nextSearchTerms = ["T","D","I","C","Y","O","4","K","N","800","S","1","V","(","10","'","`","B","M","[","-"," ","J","U","H","G","R","E","P"];
    console.log(nextSearchTerms);

    Array.prototype.naturalSort = function(){
        var a, b, a1, b1, rx=/(\d+)|(\D+)/g, rd=/\d+/;
        return this.sort(function(as, bs){
            a= String(as).toLowerCase().match(rx);
            b= String(bs).toLowerCase().match(rx);
            while(a.length && b.length){
                a1= a.shift();
                b1= b.shift();
                if (rd.test(a1) || rd.test(b1)){
                    if(!rd.test(a1)) return -1;
                    if(!rd.test(b1)) return 1;
                    if(a1 != b1) return a1-b1;
                }
                else if (a1 != b1) {
                    var aIsLetter = a1[0].charAt(0).match(/[a-z]/i),
                        bIsLetter = b1[0].charAt(0).match(/[a-z]/i);
                    if (aIsLetter && !bIsLetter) return -1;
                    if (!aIsLetter && bIsLetter) return 1;
                    return (a1[0] == b1[0] ? 0 : (a1[0] < b1[0] ? -1 : 1));
                }
            }
            return a.length - b.length;
        });
    }
    console.log(nextSearchTerms.naturalSort());

我一直試圖修改的函數返回。

[“B”，“C”，“D”，“E”，“G”，“H”，“I”，“J”，“K”，“M”，“N”，“O”，“ P“，”R“，”S“，”T“，”U“，”V“，”Y“，”“，”'“，”（“，” - “，”[“，”`“， “1”，“4”，“10”，“800”]

我想最終的數組輸出。

[“B”，“C”，“D”，“E”，“G”，“H”，“I”，“J”，“K”，“M”，“N”，“O”，“ P“，”R“，”S“，”T“，”U“，”V“，”Y“，”1“，”4“，”10“，”800“，”'“，”“ ，“ - ”，“[”，“`”，“”]

關於我缺少什么的任何建議？

Answer 1

這是我對你所尋找的東西的刺激。 我認為它比你擁有的更清潔：

Array.prototype.naturalSort = function() {
  var stringRE = /^[A-Za-z]+$/
  var numberRE = /^[\d]+$/
  return this.sort(function(a, b) {
    var aIsString = stringRE.test(a);
    var bIsString = stringRE.test(b)
    var aIsNumeric = numberRE.test(a);
    var bIsNumeric = numberRE.test(b);
    if (aIsString && bIsString) {
      return a.localeCompare(b);
    } else if (aIsNumeric && bIsNumeric) {
      return parseInt(a, 10) - parseInt(b, 10);
    } else if (aIsString && bIsNumeric) {
      return -1;
    } else if (aIsNumeric && bIsString) {
      return 1;
    } else if (aIsString || aIsNumeric) {
      return -1;
    } else if (bIsString || bIsNumeric) {
      return 1;
    } else {
      return a.localeCompare(b);
    }
  })
};

var chars = ["T","D","I","C","Y","O","4","K","N","800","S","1","V","(","10","'","`","B","M","[","-"," ","J","U","H","G","R","E","P"];

console.log(chars.naturalSort());
// ["B", "C", "D", "E", "G", "H", "I", "J", "K", "M", "N", "O", "P", "R", "S", "T", "U", "V", "Y", "4", "1", "10", "800", " ", "-", "'", "(", "[", "`"]

Answer 2

這個怎么樣？ 什么東西與數字相比......

            if (rd.test(a1) || rd.test(b1)){
                var aIsLetter = a1[0].charAt(0).match(/[a-z]/i),
                    bIsLetter = b1[0].charAt(0).match(/[a-z]/i);
                if(!rd.test(a1) && aIsLetter) return -1;
                if(!rd.test(a1) && !aIsLetter) return 1;
                if(!rd.test(b1) && bIsLetter) return 1;
                if(!rd.test(b1) && !bIsLetter) return -1;
                if(a1 != b1) return a1-b1;
            }

Answer 3

這是一個更短的解決方案：

function naturalSort(a, b) {
  var aPriority = /[a-z]/i.test(a) * 3 + /\d+/i.test(a) * 2; 
  var bPriority = /[a-z]/i.test(b) * 3 + /\d+/i.test(b) * 2; 

  if (aPriority === bPriority) return a.localeCompare(b, 'en', {numeric: true});
  return aPriority < bPriority ? 1: -1;
}

nextSearchTerms.sort(naturalSort)
// ["B", "C", "D", "E", "G", "H", "I", "J", "K", "M", "N", "O", "P", "R", "S", "T", "U", "V", "Y", "1", "4", "10", "800", " ", "-", "'", "(", "[", "`"]

如果我理解正確，這應該回答你的要求。 它很簡單易懂。 字母具有更高的優先級（3），數字具有2，其余的具有1.並且數組按優先級順序排序。

LE：修正了數字排序;

Answer 4

我想以下方法將完成所請求的任務。 實際上，該功能只是一個單線程，它利用了可鏈接的數組功能。 為了更好地理解，我將其分為三行。 按以下步驟操作;

1. 將數組減少為3個不同類型的數組。 這樣做時會將小數轉換為數字，因為下一步將需要它。
2. 通過對類型數組進行排序來減少數組，如果它們是數字類型並且與前一個數據類型相連，則將它們映射回字符串。

我相信它可以進一步簡化，但到目前為止，這是我能想到的。 請指教...

 var testar = ["T","D","I","C","Y","O","4","K","N","800","S","1","V","(","10","'","`","B","M","[","-"," ","J","U","H","G","R","E","P", "0", "-17"]; function naturalSort(a){ return a.reduce((p,c) => { /[AZ]/i.test(c) ? p[0].push(c) : /\\d+/.test(c) ? p[1].push(c*1) : p[2].push(c); return p }, [[],[],[]]) .reduce((p,c) => p.concat(c.sort((p,c) => p < c ? -1:1) .map(e => typeof e == "number" ? e.toString() : e)),[]); } document.write('<pre>' + JSON.stringify(naturalSort(testar), 0, 2) + '</pre>');