Question

我一直在努力删除数组中的重复对象。我在尝试读取filterList [i + 1] .tagID时遇到错误。我可以手动输入[i + 1]值并获得正确的结果。我不确定为什么i + 1是一个问题。我还想知道是否使用切片[i，1]比删除更好。

＆＃13;

pandas

＆＃13;

Answer 1

您可以使用set对象删除重复的ID，然后使用map和find函数来改变结果，只为每个tagID提供第一个外观

 Array.from(new Set(filterList.map(_=> _.tagID)))
  .map(ID=> filterList.find(o=> o.tagID === ID)));

const filterList = [{
  tagID: 1,
  tagName: "Red"
}, {
  tagID: 1,
  tagName: "Red"
}, {
  tagID: 2,
  tagName: "Orange"
}, {
  tagID: 2,
  tagName: "Orange"
}, {
  tagID: 5,
  tagName: "Blue"
}, {
  tagID: 5,
  tagName: "Blue"
}, {
  tagID: 5,
  tagName: "Blue"
}, {
  tagID: 6,
  tagName: "Indigo"
}, {
  tagID: 6,
  tagName: "Indigo"
}, {
  tagID: 7,
  tagName: "Violet"
}, {
  tagID: 7,
  tagName: "Violet"
}, {
  tagID: 7,
  tagName: "Violet"
}]

let result = Array.from(new Set(filterList.map(_ => _.tagID))).map((ID, index) => filterList.find(o => o.tagID == ID));

console.log(result);

Answer 2

试试这个！使用过滤器和es2015

const filterListResult = filterList.filter((item, index, self) => index === self.findIndex((t) => (t.tagID === item.tagID && t.tagName === item.tagName)));
console.log(filterListResult);

<script>
const filterList = [{
  tagID: 1,
  tagName: "Red"
}, {
  tagID: 1,
  tagName: "Red"
}, {
  tagID: 2,
  tagName: "Orange"
}, {
  tagID: 2,
  tagName: "Orange"
}, {
  tagID: 5,
  tagName: "Blue"
}, {
  tagID: 5,
  tagName: "Blue"
}, {
  tagID: 5,
  tagName: "Blue"
}, {
  tagID: 6,
  tagName: "Indigo"
}, {
  tagID: 6,
  tagName: "Indigo"
}, {
  tagID: 7,
  tagName: "Violet"
}, {
  tagID: 7,
  tagName: "Violet"
}, {
  tagID: 7,
  tagName: "Violet"
}]
</script>

Answer 3

您可以从索引1开始迭代，并使用实际元素检查最后一个元素。

＆＃13;

const filterList = [{ tagID: 1, tagName: "Red" }, { tagID: 1, tagName: "Red" }, { tagID: 2, tagName: "Orange" }, { tagID: 2, tagName: "Orange" }, { tagID: 5, tagName: "Blue" }, { tagID: 5, tagName: "Blue" }, { tagID: 5, tagName: "Blue" }, { tagID: 6, tagName: "Indigo" }, { tagID: 6, tagName: "Indigo" }, { tagID: 7, tagName: "Violet" }, { tagID: 7, tagName: "Violet" }, { tagID: 7, tagName: "Violet" }]

filterList.sort(function(a, b) {
    return a.tagID - b.tagID;
});

for (let i = 1; i < filterList.length; i++) {
    if (filterList[i - 1].tagId === filterList[i].tagID) {
        delete filterList[i];
    }
}

console.log(filterList);

＆＃13;

Answer 4

使用reduce：

生成新列表

const filterList = [{ tagID: 1, tagName: "Red" }, { tagID: 1, tagName: "Red" }, { tagID: 2, tagName: "Orange" }, { tagID: 2, tagName: "Orange" }, { tagID: 5, tagName: "Blue" }, { tagID: 5, tagName: "Blue" }, { tagID: 5, tagName: "Blue" }, { tagID: 6, tagName: "Indigo" }, { tagID: 6, tagName: "Indigo" }, { tagID: 7, tagName: "Violet" }, { tagID: 7, tagName: "Violet" }, { tagID: 7, tagName: "Violet" }]

const newList = filterList.reduce((total, current) => {
    const exist = total.some(t => t.tagID === current.tagID)
    if (!exist) {
        total.push(current)
    }
    return total
}, [])

从数组中删除重复的对象

4 个答案: