根据另一个对象数组的顺序对对象数组进行排序

Question

我有两个对象数组，我需要将其中一个数组的顺序与另一个数组匹配。 这是一个例子：

const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}];

const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}];

如何使outOfOrderArray匹配outOfOrderArray中对象的arrayInProperOrder ？

Answer 1

这是比@Faly解决方案少一个循环的方法。 无需对数组进行排序。 您可以循环遍历arrayInProperOrder并从outOfOrderArray找到匹配的元素，以便后者变得有序。

 const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}]; const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}]; const newArray = []; arrayInProperOrder.forEach(item => { const original = outOfOrderArray.find(i => i.field === item.field); if (original) { newArray.push(original); } }); console.log(newArray); 

如果需要它，只需将newArray分配给outOfOrderArray 。

Answer 2

您可以创建一个临时对象来存储订单并将其用作sort()的基础

如果在临时对象上找不到某些字段，请使用Infinity作为默认值。

 const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}]; const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}]; //Make a temp object - This will store the order of the field. This nessasary so that no need to search every reiteration on sort() const tempObj = arrayInProperOrder.reduce((c, v, i) => Object.assign(c, {[v.field]: i + 1}), {}); //Sort the array. outOfOrderArray.sort((a, b) => (tempObj[a.field] || Infinity) - (tempObj[b.field] || Infinity)); console.log(outOfOrderArray); 

Doc： sort（） ， Infinity

Answer 3

您可以使用array.prototype.sort和array.prototype.findIndex ：

 const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}]; const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}]; outOfOrderArray.sort( (a, b) => arrayInProperOrder.findIndex(e => e.field === a.field) - arrayInProperOrder.findIndex(e => e.field === b.field) ); console.log(outOfOrderArray); 

如果您关心性能，则可以构建一个对象，该对象允许您从arrayInProperOrder数组中按字段值获取索引：

 const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}]; const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}]; const indexes = arrayInProperOrder.reduce((m, o, i) => (m[o.field] = i, m), {}); outOfOrderArray.sort( (a, b) => indexes[a.field] - indexes[b.field] ); console.log(outOfOrderArray);