Javascript，从对象中删除所有，真正的所有重复项

Question

有带有注释、时间条目的对象，想要删除所有（所有，抱歉重复这部分，我已经经历了很多例子，它们都只删除第二个、第三个等重复项，但它们包括第一个重复项）与它同时重复的笔记。

好吧，我需要将唯一项与重复项分开，我已经使用我在 Whosebug 上找到的这段代码成功地从中获取了所有重复项，

const allNotes = [
    {
        "note": 69,
        "time": 0
    },
    {
        "note": 57,
        "time": 0
    },
    {
        "note": 60,
        "time": 1.5
    },
    {
        "note": 64,
        "time": 2
    },
    {
        "note": 69,
        "time": 2.5
    },
    {
        "note": 71,
        "time": 3
    },
    {
        "note": 52,
        "time": 3
    },
    {
        "note": 64,
        "time": 4.5
    },
    {
        "note": 68,
        "time": 5
    },
    {
        "note": 71,
        "time": 5.5
    }
]

const getDuplicates = () => {
const values = allNotes;
const unique = new Object;
const lookup = values.reduce((a, e) => {
  a[e.time] = ++a[e.time] || 0;
  return a;
}, {});
const duplicates = values.filter(e => lookup[e.time]);
console.log(duplicates);
}

这段代码就像它产生的魅力一样工作

[
    {
        "note": 69,
        "time": 0
    },
    {
        "note": 57,
        "time": 0
    },
    {
        "note": 71,
        "time": 3
    },
    {
        "note": 52,
        "time": 3
    }
]

仍然需要能够只获取唯一的时间条目，需要这个结果

[
    {
        "note": 60,
        "time": 1.5
    },
    {
        "note": 64,
        "time": 2
    },
    {
        "note": 69,
        "time": 2.5
    },
    {
        "note": 64,
        "time": 4.5
    },
    {
        "note": 68,
        "time": 5
    },
    {
        "note": 71,
        "time": 5.5
    }
]

Answer 1

我猜你正在寻找这样的东西：

array.filter(a => array.filter(b => a.time === b.time).length === 1)
// filter array and sort for duplicates, just keep element if time-property is unique

请参阅下面的工作示例：

const allNotes=[{note:69,time:0},{note:57,time:0},{note:60,time:1.5},{note:64,time:2},{note:69,time:2.5},{note:71,time:3},{note:52,time:3},{note:64,time:4.5},{note:68,time:5},{note:71,time:5.5}];

const res = allNotes.filter(a => allNotes.filter(b => a.time === b.time).length === 1);
console.log(res);

Answer 2

您已经解决了您的问题，因为您能够找出重复项。过滤相反的就可以了。

const uniqueValues = values.filter(e => !lookup[e.time]);

Answer 3

const _dedupArr: any[] = [];

allNotes.forEach((v, i) => {
    const find = _dedupArr.find(d => d.time == v.time);

    if(!find) {
        _dedupArr.push(v);
    }
});

console.log('_dedupArr', _dedupArr);

Answer 4

不是一个高效的解决方案，但它绝对是一个简单的解决方案...将每个对象转换为基元，然后可以将这些基元分配给地图，并且只能检索值。

const allNotes = [
    {
        "note": 69,
        "time": 0
    },
    {
        "note": 57,
        "time": 0
    },
    {
        "note": 60,
        "time": 1.5
    },
    {
        "note": 64,
        "time": 2
    },
    {
        "note": 69,
        "time": 2.5
    },
    {
        "note": 71,
        "time": 3
    },
    {
        "note": 52,
        "time": 3
    },
    {
        "note": 64,
        "time": 4.5
    },
    {
        "note": 68,
        "time": 5
    },
    {
        "note": 71,
        "time": 5.5
    },
];

const uniqueMap = new Map();
allNotes.forEach(value => uniqueMap.set(
  // Any conversion to primitive is fine, JSON.stringify too
  `${value.note}-${value.time}`,
  value,
));
console.log(Array.from(uniqueMap.values()));

Javascript，从对象中删除所有，真正的所有重复项

Javasript, remove ALL, REALLY ALL duplicates from object

javascript

object

duplicates

unique