我假设你使用了像这样的splice
函数?
for (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
}
}
修复错误的方法就是在下一次循环中将 i
减少,然后(倒序循环也是一个选项):
for (var i = 0; i < arrayOfObjects.length; i++) {
var obj = arrayOfObjects[i];
if (listToDelete.indexOf(obj.id) !== -1) {
arrayOfObjects.splice(i, 1);
<b>i--;</b>
}
}
为了避免线性时间的删除操作,您可以在数组上写入您想要保留的数组元素:
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,您可以使用哈希集合:
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;
可以用一个好的函数来封装:
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);
如果你不需要原地操作,那就使用Array#filter
:
const toDelete = new Set(['abc', 'efg']);
const newArray = arrayOfObjects.filter(obj => !toDelete.has(obj.id));