按对象属性从数组中删除对象

问题描述

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？

How do I remove an object from the array by matching object property?

请仅使用原生JavaScript。

我在使用拼接时遇到问题，因为每次删除都会缩短长度。
在orignal索引上使用克隆和拼接仍然会让你减少长度的问题。

I am having trouble using splice because length diminishes with each deletion.Using clone and splicing on orignal index still leaves you with the problem of diminishing length.

推荐答案

我假设你使用过 splice 这样的事情？

I assume you used splice something like this?

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

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

你需要做的就是减少错误我下一次，然后（并向后循环也是一个选项）：

All you need to do to fix the bug is decrement i for the next time around, then (and looping backwards is also an option):

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，您可以在阵列上编写要保持的数组元素：

To avoid linear-time deletions, you can write array elements you want to keep over the array:

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;

并且为了避免在现代运行时中进行线性时间查找，您可以使用哈希集：

and to avoid linear-time lookups in a modern runtime, you can use a hash set:

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;

可以包含在一个很好的函数中：

which can be wrapped up in a nice function:

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);

如果您不需要这样做，那就是数组＃过滤器：

If you don’t need to do it in place, that’s Array#filter:

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

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