Question

我有两个列表，oldPanodatas和newPanodatas。我想只获取newPanodatas中oldPanodatas中不存在的var filteredPanodatas = _.difference(newPanodatas, oldPanodatas)中的对象。我这样做了：

_.difference

但是我收到了所有项目，OLD: Object {roomModelId: "56a9e0088ac247005538d6d3", x: 262, index: 1, y: 211, panoDataRotate: 0…} OLD: Object {roomModelId: "56a9e0088ac247005538d6d3", x: 177, index: 0, y: 182, panoDataRotate: 0…} NEW: Object {index: 0, x: 177, y: 182, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0} NEW: Object {index: 1, x: 262, y: 211, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0} NEW: Object {index: 2, x: 200, y: 200, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0} FILTERED: Object {index: 0, x: 177, y: 182, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0} FILTERED: Object {index: 1, x: 262, y: 211, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0} FILTERED: Object {index: 2, x: 200, y: 200, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0}并未过滤任何内容：

{{1}}

这是为什么？什么是实现我想要的正确方法？

Answer 1

尝试使用differceWith，调用isEqual比较器，将数组元素与值进行比较

_.differenceWith(newPanodatas, oldPanodatas, _.isEqual);

代码示例，

var old = [{roomModelId: "56a9e0088ac247005538d6d3", x: 262, index: 1, y: 211, panoDataRotate: 0},  {roomModelId: "56a9e0088ac247005538d6d3", x: 177, index: 0, y: 182, panoDataRotate: 0}];
newobj = [{index: 0, x: 177, y: 182, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0}, {index: 1, x: 262, y: 211, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0}, {index: 2, x: 200, y: 200, roomModelId: "56a9e0088ac247005538d6d3", panoDataRotate: 0}];
r = _.differenceWith(newobj, old, _.isEqual);

r

的结果

[{index: 2, panoDataRotate: 0, roomModelId: "56a9e0088ac247005538d6d3", x: 200, y: 200}]

为什么_.difference会返回所有项目？

1 个答案: