5

我有一个类似于这样的数组设置：

var ary1 = new Array("d", "a", "b", "c");
var ary2 = new Array("ee", "rr", "yy", "mm");

var mdAry = new Array(ary1, ary2);

ary1 和 ary2 索引是宏观事物中彼此相关的信息。

d ee
a rr
b yy
c mm

我可以 sort() ary1 并得到：

a
b
c
d

但如果我独立排序 ary2 我会得到：

ee
mm
rr
yy

列出时会在视觉上破坏 ary1 和 ary2 连接。我可以检索 ary1 的排序解决方案并将其应用于 ary2 吗？我想得到这个：

a rr
b yy
c mm
d ee

如果不是，是否可以对 mdAry 进行排序，以便将 mdAry[0] 排序的解决方案应用于剩余的索引？

4

5 回答 5

3

如果您的数组项是相关的，则将它们存储在一起：

var arr = [
  {x: 'd', y: 'ee'}, 
  {x: 'a', y: 'rr'}, 
  {x: 'b', y: 'yy'},
  {x: 'c', y: 'mm'}
];

arr.sort(function(a, b) {
  if (a.x != b.x) {
      return a.x < b.x ? -1 : 1;
  }
  return 0;
});
于 2013-05-23T04:26:45.240 回答
1

只需在其中添加另一种方法，您就可以从第一个数组中获得排序“结果”并将其应用于任何其他相关列表：

function getSorter(model) {
    var clone = model.slice(0).sort();
    var sortResult = model.map(function(item) { return clone.indexOf(item); });

    return function(anyOtherArray) {
        result = [];
        sortResult.forEach(function(idx, i) {
            result[idx] = anyOtherArray[i];
        });
        return result;
    }
}

然后，

var arr = ["d", "a", "b", "c"];
var arr2 = ["ee", "rr", "yy", "mm"];

var preparedSorter = getSorter(arr);
preparedSorter(arr2); 
//=> ["rr", "yy", "mm", "ee"];

或者，

multidimensional = [arr, arr2];
multidimensional.map(getSorter(arr)); 
// => [["a", "b", "c", "d"], ["rr", "yy", "mm", "ee"]]
于 2013-05-23T04:18:18.023 回答
1

一种方法是将数据结构转换为更容易排序的数据结构，然后再将其转换回来

var ary1  = ["d", "a", "b", "c"],
    ary2  = ["ee", "rr", "mm", "yy"]
    mdAry = [ary1, ary2];

// convert to form [[d, ee], [a, rr], ..]
var tmp = mdAry[0].map(function (e, i) {
    return [e, mdAry[1][i]];
});
// sort this
tmp.sort(function (a, b) {return a[0] > b[0];});
// revert to [[a, b, ..], [rr, mm, ..]]
tmp.forEach(function (e, i) {
    mdAry[0][i] = e[0];
    mdAry[1][i] = e[1];
});
// output
mdAry;
// [["a", "b", "c", "d"], ["rr", "mm", "yy", "ee"]]
于 2013-05-23T01:58:02.977 回答
0

在您的示例中，结果应该是（如果我理解正确的话）

a rr
b mm
c yy
d ee

所以这个应该做的工作：

Array.prototype.sortRelated = function(related) {
    var clone = this.slice(0), 
        sortedRelated = [];
    clone.sort();

    for(var i = 0; i < this.length; i ++) {
        sortedRelated[clone.indexOf(this[i])] = related[i];
    }

    return sortedRelated;
}

var ary1 = new Array("d", "a", "b", "c");
var ary2 = new Array("ee", "rr", "mm", "yy");

var sorted = ary1.sortRelated(ary2);

在这里工作演示：http: //jsfiddle.net/cwgN8/

于 2013-05-23T01:51:17.397 回答
0

您可以将它们“合并”为具有两个属性的单个对象，按第一个属性排序，然后最后分开（参见此处的演示）：

function sortBoth(ary1, ary2) {
    var merged = [];
    for (var i=0; i < ary1.length; i++) merged.push({'ary1': ary1[i], 'ary2': ary2[i]});
    merged.sort(function(o1, o2) { return ((o1.ary1 < o2.ary1) ? -1 : ((o1.ary1 == o2.ary1) ? 0 : 1)); });
    for (var i=0; i < merged.length; i++) { ary1[i] = merged[i].ary1; ary2[i] = merged[i].ary2; }
}

var ary1 = new Array("d", "a", "b", "c");
var ary2 = new Array("ee", "rr", "mm", "yy");

console.log(ary1);
console.log(ary2);

sortBoth(ary1, ary2);

console.log(ary1);
console.log(ary2);

输出：

[ "d",  "a",  "b",  "c"]
["ee", "rr", "mm", "yy"]
[ "a",  "b",  "c",  "d"]
["rr", "mm", "yy", "ee"]
于 2013-05-23T01:52:42.993 回答