Javasript，從物件中洗掉所有，真正的所有重復項

有帶有注釋、時間條目的物件，想要洗掉所有（所有，抱歉重復這部分，我經歷了很多例子，它們都只洗掉了第二個、第三個等重復項，但它們包括第一個重復項）具有相同時間的筆記重復從中。

好吧，我需要將唯一的與重復的分開，我已經設法使用我在 StackOverflow 上找到的這段代碼從中獲取所有重復，

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);
}

這段代碼就像它產生的魅力一樣作業

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

仍然需要能夠獲得唯一的時間條目，需要這個結果

[
    {
        "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
    }
]

uj5u.com熱心網友回復：

我猜你正在尋找這樣的東西：

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

請參閱下面的作業示例：

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);

uj5u.com熱心網友回復：

您已經解決了您的問題，因為您能夠挑選出重復項。只過濾相反的。

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

uj5u.com熱心網友回復：

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);

uj5u.com熱心網友回復：

不是一個高性能的解決方案，但它絕對是一個簡單的解決方案......將每個物件轉換為一個基元，然后可以將這些基元分配給一個映射，并且只能檢索值。

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()));

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