Question

我有下面提到的2个阵列。使用此我需要根据dimentionFilterKeyOrder中需要与dimentionFilterKeys值进行比较的值创建一个新数组。此外，dimentionFilterKeys中不匹配的项目需要在新数组的最后一个列表中列出。

注意：新数组Order应与dimentionFilterKeyOrder类似。

dimentionFilterKeyOrder= ["21","22","23","14","10312","81","10315","10314"];

dimentionFilterKeys = [
  {"filterKey": "81"},
  {"filterKey": "82"},
  {"filterKey": "10312"},
  {"filterKey": "10313"},
  {"filterKey": "10315"},
  {"filterKey": "10314"},
  {"filterKey": "20"},
  {"filterKey": "21"},
  {"filterKey": "22"},
  {"filterKey": "23"},
  {"filterKey": "24"},
  {"filterKey": "14"}
]

OutPut寻找类似于下面的内容。

newArryaOutPut = [
  {"filterKey": "21"},
  {"filterKey": "22"},
  {"filterKey": "23"},
  {"filterKey": "14"},
  {"filterKey": "10312"},
  {"filterKey": "81"},
  {"filterKey": "10315"},
  {"filterKey": "10314"},
  {"filterKey": "82"},
  {"filterKey": "10313"},
  {"filterKey": "20"},
  {"filterKey": "24"}
]

Answer 1

您可以使用自定义排序方法。使用filterKey索引，您可以对对象数组进行排序。

注意：不会对dimentionFilterKeyOrder中未移动到最后的值进行排序。 Array#sort()执行就地排序，如果您希望取消修改原始数组，可以使用array#slice()然后对其进行排序。

＆＃13;

const dimentionFilterKeyOrder= ["21","22","23","14","10312","81","10315","10314"],
      dimentionFilterKeys = [{"filterKey":"81"},{"filterKey":"82"},{"filterKey":"10312"},{"filterKey":"10313"},{"filterKey":"10315"},{"filterKey":"10314"},{"filterKey":"20"},{"filterKey":"21"},{"filterKey":"22"},{"filterKey":"23"},{"filterKey":"24"},{"filterKey":"14"}];

dimentionFilterKeys.sort((a,b) => {
  let indexA = dimentionFilterKeyOrder.indexOf(a.filterKey),
      indexB = dimentionFilterKeyOrder.indexOf(b.filterKey);
  indexA = indexA == -1 ? Infinity : indexA + 1;
  indexB = indexB == -1 ? Infinity : indexB + 1;
  return indexA - indexB;
});

console.log(dimentionFilterKeys);

＆＃13;

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＆＃13;

Answer 2

这应该有效，顺序数组中不存在的键的顺序无法预测。

var compareFunc = function(a, b) {
    let indexA = dimentionFilterKeyOrder.indexOf(a['filterKey']);
  let indexB = dimentionFilterKeyOrder.indexOf(b['filterKey']);
  if(indexA==-1)
    return 1;
  if(indexB==-1)
    return -1;
  return indexA-indexB;
}

var sorted = dimentionFilterKeys.sort(compareFunc)

Answer 3

您可以使用sorting with map并将已知位置的索引和Infinity用于排序，并保留数组的原始顺序。

var order = ["21","22","23","14","10312","81","10315","10314"],
    array = [{ filterKey: "81" }, { filterKey: "82" }, { filterKey: "10312" }, { filterKey: "10313" }, { filterKey: "10315" }, { filterKey: "10314" }, { filterKey: "20" }, { filterKey: "21" }, { filterKey: "22" }, { filterKey: "23" }, { filterKey: "24" }, { filterKey: "14" }],
    result = array.map(function (o, i) {
        return { index: i, order: (order.indexOf(o.filterKey) + 1) || Infinity };
    })
    .sort(function (a, b) {
        return a.order - b.order || a.index - b.index;
    })
    .map(function (o) {
        return array[o.index];
    });
    
console.log(result);

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Javascript比较2个数组并根据订单创建一个新数组

3 个答案: