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。
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);
}
}
修复该错误所需要做的就是i
在下一次减少,然后(也可以向后循环):
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));
您可以通过其属性之一删除某项,而无需使用任何第三方库,例如:
var removeIndex = array.map(item => item.id)
.indexOf("abc");
~removeIndex && array.splice(removeIndex, 1);
使用lodash /下划线:
如果要修改现有数组本身,则必须使用splice。这是使用underscore / lodash的findWhere的更好/更易理解的方法:
var items= [{id:'abc',name:'oh'}, // delete me
{id:'efg',name:'em'},
{id:'hij',name:'ge'}];
items.splice(_.indexOf(items, _.findWhere(items, { id : "abc"})), 1);
使用ES5或更高版本
(不带下划线/下划线)
从ES5开始,我们findIndex
对数组有了方法,因此无需lodash /下划线就很容易
items.splice(items.findIndex(function(i){
return i.id === "abc";
}), 1);
(几乎所有现代浏览器都支持ES5)
findIndex适用于现代浏览器:
var myArr = [{id:'a'},{id:'myid'},{id:'c'}];
var index = myArr.findIndex(function(o){
return o.id === 'myid';
})
if (index !== -1) myArr.splice(index, 1);
通过给定数组中的ID删除对象;
const hero = [{'id' : 1, 'name' : 'hero1'}, {'id': 2, 'name' : 'hero2'}];
//remove hero1
const updatedHero = hero.filter(item => item.id !== 1);
如果您只想将其从现有阵列中删除而不创建一个新阵列,请尝试:
var items = [{Id: 1},{Id: 2},{Id: 3}];
items.splice(_.indexOf(items, _.find(items, function (item) { return item.Id === 2; })), 1);
通过递减i
来反向循环,避免出现此问题:
for (var i = arrayOfObjects.length - 1; i >= 0; i--) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
}
}
或使用filter
:
var newArray = arrayOfObjects.filter(function(obj) {
return listToDelete.indexOf(obj.id) === -1;
});
使用Set和ES6过滤器进行检查。
let result = arrayOfObjects.filter( el => (-1 == listToDelete.indexOf(el.id)) );
console.log(result);
这是JsFiddle:https ://jsfiddle.net/jsq0a0p1/1/
请只使用本机JavaScript。
作为使用ECMAScript 5的另一种更“实用”的解决方案,您可以使用:
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
arrayOfObjects.reduceRight(function(acc, obj, idx) {
if (listToDelete.indexOf(obj.id) > -1)
arrayOfObjects.splice(idx,1);
}, 0); // initial value set to avoid issues with the first item and
// when the array is empty.
console.log(arrayOfObjects);
[ { id: 'hij', name: 'ge' } ]
根据ECMA-262中'Array.prototype.reduceRight'的定义:
reduceRight不会直接改变在其上调用它的对象,但是可以通过对callbackfn的调用来改变该对象。
因此,这是的有效用法reduceRight
。
var arrayOfObjects = [{id:'abc',name:'oh'}, // delete me
{id:'efg',name:'em'}, // delete me
{id:'hij',name:'ge'}] // all that should remain
根据您的答案将是这样。当您单击某个特定对象时,将参数中的索引发送给Delete me功能。这个简单的代码将像魅力一样工作。
function deleteme(i){
if (i > -1) {
arrayOfObjects.splice(i, 1);
}
}
与filter&indexOf
withLodash = _.filter(arrayOfObjects, (obj) => (listToDelete.indexOf(obj.id) === -1));
withoutLodash = arrayOfObjects.filter(obj => listToDelete.indexOf(obj.id) === -1);
带过滤器和包含
withLodash = _.filter(arrayOfObjects, (obj) => (!listToDelete.includes(obj.id)))
withoutLodash = arrayOfObjects.filter(obj => !listToDelete.includes(obj.id));
如果您喜欢简短的自我描述参数,或者您不想使用splice
直接筛选器,或者您只是像我这样的SQL人员:
function removeFromArrayOfHash(p_array_of_hash, p_key, p_value_to_remove){
return p_array_of_hash.filter((l_cur_row) => {return l_cur_row[p_key] != p_value_to_remove});
}
以及示例用法:
l_test_arr =
[
{
post_id: 1,
post_content: "Hey I am the first hash with id 1"
},
{
post_id: 2,
post_content: "This is item 2"
},
{
post_id: 1,
post_content: "And I am the second hash with id 1"
},
{
post_id: 3,
post_content: "This is item 3"
},
];
l_test_arr = removeFromArrayOfHash(l_test_arr, "post_id", 2); // gives both of the post_id 1 hashes and the post_id 3
l_test_arr = removeFromArrayOfHash(l_test_arr, "post_id", 1); // gives only post_id 3 (since 1 was removed in previous line)
您可以使用filter
。如果条件为true,则此方法始终返回元素。因此,如果要按ID删除,则必须保留所有与给定ID不匹配的元素。这是一个例子:
arrayOfObjects = arrayOfObjects.filter(obj => obj.id!= idToRemove)
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