从数组中删除重复的对象
从数组中删除重复的对象
提问于 2018-06-07 04:59:28
我一直致力于删除数组中的重复对象。在尝试读取filterListi+1.tagID时，我一直收到错误。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是个问题。我也想知道是否使用slicei，1比delete更好。
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 = 0; i < filterList.length; i++) {
if (filterList[i].tagId == filterList[i+1].tagID) {
delete filterList[i];
}
}
console.log(filterList)
回答 4
Stack Overflow用户
回答已采纳
发布于 2018-06-07 05:15:41
尝尝这个!使用filter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>
Stack Overflow用户
发布于 2018-06-07 05:15:28
您可以使用set对象来删除重复的ID，然后使用map和find函数来修改结果，以便只给出每个tagID的第一个外观。
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);
Stack Overflow用户
发布于 2018-06-07 05:06:49
您可以从索引1开始迭代，然后用实际元素检查最后一个元素。
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);
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原文链接：
https://stackoverflow.com/questions/50729508
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