我有一个类似的数组设置:
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索引在宏观方案中是彼此相关的信息。
d ee
a rr
b yy
c mm
我可以排序()ary1并获取:
a
b
c
d
但如果我独立排序ary2,我会得到:
ee
mm
rr
yy
在列出时可视地破坏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
如果没有,可以对mdAry进行排序,以便将mdAry [0]排序解决方案应用于剩余的指标吗?
答案 0 :(得分:3)
如果你的数组项是相关的,那么将它们存储在一起:
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;
});
答案 1 :(得分:1)
这样做的一种方法是将数据结构转换为可以更容易排序的内容,然后在
之后将其转换回来。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"]]
答案 2 :(得分:1)
只是在那里添加另一个方法,你可以从第一个数组中得到一个排序“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"]]
答案 3 :(得分: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
所以这个人应该接受这个工作:
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);
此处的演示演示:http://jsfiddle.net/cwgN8/
答案 4 :(得分: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,然后在最后分开( see demo here ):
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"]