小编典典

通过对象属性从数组中删除对象

javascript

var listToDelete = ['abc', 'efg'];

var arrayOfObjects = [{id:'abc',name:'oh'}, // delete me
                      {id:'efg',name:'em'}, // delete me
                      {id:'hij',name:'ge'}] // all that should remain

如何通过匹配对象属性从数组中删除对象?

请只使用本机JavaScript。

我在使用接头时遇到麻烦,因为每次删除的长度都会减少。使用克隆并在原始索引上进行拼接仍然会给您带来长度减少的问题。


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2020-04-25

共1个答案

小编典典

我以为你用过splice这样的东西?

for (var i = 0; i < arrayOfObjects.length; i++) {
    var obj = arrayOfObjects[i];

    if (listToDelete.indexOf(obj.id) !== -1) {
        arrayOfObjects.splice(i, 1);
    }
}

修复bug所需要做的就是i在下一次减少,然后(也可以选择向后循环):

for (var i = 0; i < arrayOfObjects.length; i++) {
    var obj = arrayOfObjects[i];

    if (listToDelete.indexOf(obj.id) !== -1) {
        arrayOfObjects.splice(i, 1);
        **i--;**
    }
}

为了避免线性时间删除,可以编写要 保留 在数组上的数组元素:

var end = 0;

for (var i = 0; i < arrayOfObjects.length; i++) {
    var obj = arrayOfObjects[i];

    if (listToDelete.indexOf(obj.id) === -1) {
        arrayOfObjects[end++] = obj;
    }
}

arrayOfObjects.length = end;

为了避免在现代运行时中进行线性时间查找,可以使用哈希集:

const setToDelete = new Set(listToDelete);
let end = 0;

for (let i = 0; i < arrayOfObjects.length; i++) {
    const obj = arrayOfObjects[i];

    if (setToDelete.has(obj.id)) {
        arrayOfObjects[end++] = obj;
    }
}

arrayOfObjects.length = end;

可以包装成一个不错的功能:

const filterInPlace = (array, predicate) => {

    let end = 0;



    for (let i = 0; i < array.length; i++) {

        const obj = array[i];



        if (predicate(obj)) {

            array[end++] = obj;

        }

    }



    array.length = end;

};



const toDelete = new Set(['abc', 'efg']);



const arrayOfObjects = [{id: 'abc', name: 'oh'},

                        {id: 'efg', name: 'em'},

                        {id: 'hij', name: 'ge'}];



filterInPlace(arrayOfObjects, obj => !toDelete.has(obj.id));

console.log(arrayOfObjects);

如果您不需要就位,那就是Array#filter

const toDelete = new Set(['abc', 'efg']);
const newArray = arrayOfObjects.filter(obj => !toDelete.has(obj.id));
2020-04-25