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By convention, ES6 code

const result = arr.reduce((r, t) => {
    // 在結(jié)果中查找 index，
    // 如果找到，更新該位置的對象引用
    // 找到則加一個
    var i = r.findIndex(m => m.id === t.id);
    if (i >= 0) {
        r[i] = t;
    } else {
        r.push(t);
    }
    return r;
}, []);

There is a problem here, findIndex is not supported by at least two browsers, so if it is not supported, you have to write one yourself

Array.prototype.findIndex = Array.prototype.findIndex || function(predicate) {
    for (let i = 0; i < this.length; i++) {
        if (predicate(this[i])) {
            return i;
        }
    }
    return -1;
};

Other solutions

Classic solution, use Map

Because id is not a string, ES6’s Map class is used. When the amount of data is large, using a lookup table can significantly improve efficiency compared to linear search in a list.

const result = arr
    .reduce((m, t) => {
        const { map, list } = m;
        var index = map.get(t.id);
        if (index >= 0) {
            list[index] = t;
        } else {
            map.set(t.id, list.length);
            list.push(t);
        }
        return m;
    }, {
        map: new Map(),
        list: []
    })
    .list;

In fact, you can also use objects instead of maps, at least in this use case there will be no problem. Because there are no es6 features, we simply use es5 syntax. The code structure and logic are the same as the above paragraph

var result = arr
    .reduce(function(m, t) {
        var index = m.map[t.id];
        if (index >= 0) {
            m.list[index] = t;
        } else {
            m.map[t.id] = m.list.length;
            m.list.push(t);
        }
        return m;
    }, {
        map: {},
        list: []
    })
    .list;

Weird solution, using integer id

Because it is an integer id, you can directly put it in the array according to this id. If the same ID is encountered, it will be replaced directly. If the ids are not consecutive, you need to filter out the empty elements at the end

var result = arr
    .reduce(function(r, t) {
        r[t.id] = t;
        return r;
    }, [])
    .filter(function(t) { return t; });

There is another problem with this solution. It cannot maintain the order of the elements of the original array. Then someone will definitely think that the solution using Map can also reduce the code into similar code without making it so complicated. Of course, it may also lose the original order

const map = arr
    .reduce((m, t) => {
        m.set(t.id, t);
        return m;
    }, new Map());

const result = [...map.values()];

Note: All the above codes have been actually run and passed, and the running environment is Node v8.1.2

var result = arr.filter(function(val, index) {
    /**
     * 使用arr.slice(index + 1)獲取從當(dāng)前索引下一個元素到數(shù)組最后一個元素組成的數(shù)組
     * 使用findIndex在當(dāng)前項(xiàng)的后面選項(xiàng)中查找是否有和當(dāng)前項(xiàng)id值相同的選項(xiàng)
     */
    var index = arr.slice(index + 1).findIndex(function(item) {
         return item.id === val.id;
    });
    // 如果為-1，則說明后面沒有同名id了，所以這一項(xiàng)可以返回
    return index === -1;
});
console.log(result);

Using arrow functions simplifies as follows:

var result = arr.filter((val, index) => arr.slice(index + 1).findIndex(item => item.id === val.id) === -1);
console.log(result);

There are already many answers here, but there is no mention of Array’s built-in function reduceRight. In fact, the questioner’s requirement is to retain the last digit of the same ID, which is very convenient to implement with reduceRight.

arr.reduceRight((r,v)=>{
    if(!r[0].has(v.id)) r[0].add(v.id) && r[1].unshift(v)
    return r
},[new Set,[]])[1]

reduceRight starts looping from the end of your original array. The initial value here is an array. r[0] is used to store the Set of ids, and r[1] is used to store the result array. If there is no id in the Set, then add this id to Set and place this item at the head of the resulting array.

In the end, the topic owner’s needs were easily achieved, and the order was guaranteed.

function uniq(arr) {
    var idArr = [],arr2 = []
    for (var i = 0, len = arr.length; i < len; i++) {
        if (arr[i].id in idArr) {
            arr2.pop()
            arr2.push(arr[i])
        } else {
            idArr.push(arr[i].id)
            arr2.push(arr[i])
        }
    }
    return arr2
}

Personally tested and effective

arr = [ { id: 1, content: 'a' },
        { id: 2, content: 'b' },
        { id: 2, content: 'c' },
        { id: 3, content: 'd' },
        { id: 3, content: 'e' },
        { id: 3, content: 'f' },
        { id: 3, content: 'g' },
        { id: 4, content: 'h' } ]
        
tmp = []
for(k in arr){tmp[arr[k]['id']] = arr[k]['content']}

arr = []
for(k in tmp){arr.push({'id':+k, 'content':tmp[k]})}

console.log(arr)
[ { id: 1, content: 'a' },
  { id: 2, content: 'c' },
  { id: 3, content: 'g' },
  { id: 4, content: 'h' } ]

let newArr = [],

    result = [],
    status = false，
    lastResult = []
for (let i = arr.length - 1; i >= 0; i--) {
    newArr.push(arr[i])
}
for (let i = 0, len = newArr.length; i < len; i++) {
    if (result.length == 0) {
        result.push(newArr[0])
    }
    for (let j = 0; j < result.length; j++) {
        if (newArr[i].id == result[j].id) {
            console.log(newArr[i])
            status = true
        }
    }
    if (!status) {

        result.push(newArr[i])
    }
    status = false

}
for (let i = result.length - 1; i >= 0; i--) {
    lastResult.push(result[i])
}
console.log(lastResult) //為去掉相同的id 然后保留各個id的最后一項(xiàng)

var arr = [ ... ]; // 這個為給定的數(shù)組
var obj = {}; // 定義一個對象存儲

arr.forEach(function(v) {
    obj[v.id] = v;
    // 不管如何，直接將數(shù)組賦值給obj中下標(biāo)為v.id的項(xiàng)，這樣對應(yīng)的v.id的值到最后必然是同ID最后一項(xiàng)
});

// 下面考慮兼容性給出新語法和常規(guī)語法
// 新語法，Object.values部分瀏覽器版本不支持
var result = Object.values(obj);
// 常規(guī)語法，用Array.map實(shí)現(xiàn)Object.values的效果
var result = Object.keys(obj).map(function(id) {
    return obj[id];
});

Please refer to it

Array.from(arr.reduce((map, el) => map.set(el.id, el), new Map()).values())

You can refer to what I wrote on my blog. I wrote 8 methods on my blog. http://alfierichou.top/2017/0...