JavaScript：根据数组项中的数字重新排序数组

Question

我正在尝试解读一个数组。该数组包含乱序的字符串项，并附有一个数字以指定顺序应该是什么。我想获取数组项中的数字并重新分配数组项索引以反映该数字。例如var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"]。我有一个函数可以在数组中搜索附加的数字并返回值。现在我想获取返回的值并将其转换为新数组中新项目的索引，例如var unscrambled = []。这是我目前所拥有的：

function unscramblePhrase() {
    var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"];   
    var unscrambled = [];
    for (var counter = 0; counter < scrambled.length; counter++) {
        numPosition = scrambled[counter].search('[0-9]');
        arrayIndex = scrambled[counter].substring(numPosition);
        console.log(arrayIndex);
        unscrambled.push(scrambled[arrayIndex]);
    }
 console.log(unscrambled)
}

我看到我的 arrayIndex 正在从加扰数组项的末尾提取数字，但我尝试根据此变量分配索引位置正在生成一个新加扰数组：["want1", "I0", "pizza4", "eat3", "to2"]。

Answer 1

这不是对数组进行排序的最佳方法，但您的问题是您使用Array.prototype.push，而您应该只是将值分配给unscrambled 中的索引。

unscrambled[arrayIndex] = scrambled[counter];

但真正的方法是使用Array.prototype.sort

function getNum(str){
    return Number(str.substring(str.length -1));
}


unscrambled.sort((a, b) => getNum(a) - getNum(b));

注意：此方法对您的数组in-place 进行排序。这可能好，也可能不好，具体取决于您的要求

但您始终可以在克隆上使用它：

[...unscrambled].sort((a, b) => getNum(a) - getNum(b));

Answer 2

您可以通过使用 RegEx 将数组拆分为值和索引，对它们进行排序，然后使用辅助 .map 删除附加信息以返回字符串数组来简化此操作。

scrambled.map(i => 
[i.replace(/\d/g,""), +i.replace(/\D/g,"")])
.sort((a, b) => a[1] - b[1])).map(i => i[0]);

var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"];   

var unscrambled = scrambled.map(i => 
  [i.replace(/\d/g,""), +i.replace(/\D/g,"")])
  .sort((a, b) => a[1] - b[1])
  .map( i => i[0]);

console.log(unscrambled);

Answer 3

试试这个（不知道它是否有效）：

function unscramblePhrase() { var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"];
var unscrambled = []; for (var counter = 0; counter < scrambled.length; counter++) { numPosition = scrambled[counter].search('[0-9]'); arrayIndex = scrambled[counter].substring(numPosition); console.log(arrayIndex); unscrambled[arrayIndex] = scrambled[counter]; } console.log(unscrambled) }

Answer 4

您也可以尝试使用“Array.sort”，如下所示

var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"]

let getNumberIndex = (d) => [...d].findIndex(v => Number.isInteger(+v))

let getNumber = (d) => d.slice(getNumberIndex(d))

let unscrambled = scrambled.slice(0).sort((a,b) => getNumber(a) - getNumber(b))

console.log(unscrambled)

Answer 5

const words = ["I0", "want1", "to2", "eat3", "pizza4"]

words
    .map (w => parseInt (w [--w.length]))
    .sort ()
    .map (i => words.filter (w => w [--w.length] == i) [0].replace (i, ''))

// ["I", "want", "to", "eat", "pizza"]

Answer 6

您可以使用单个循环并获取字符串和索引的部分，将值分配给索引并返回数组。

var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"],
    result = scrambled.reduce((r, string) => {
        var [s, i] = string.match(/\D+|\d+/g);
        r[i] = s;
        return r;
    }, []);
    
console.log(result);

对象更有趣。

var scrambled = ["pizza4", "to2", "I0", "eat3", "want1"],
    result = Object.assign(
        [],
        ...scrambled.map(s => (([v, k]) => ({ [k]: v }))(s.match(/\D+|\d+/g)))
    );
    
console.log(result);