javascript - 过滤JavaScript中具有相同嵌套属性的对象数组

Question

[{"code":"96","value":"RemoveTest","parentLabelCode":1,"color":null,"level":1,"children":[{"code":"97","value":"R1","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":1,"labelTranslations":null},{"code":"98","value":"R2","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":true,"systemLabel":true,"removeOnArchive":true,"displayNumber":2,"labelTranslations":null},{"code":"99","value":"R3","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":true,"systemLabel":true,"removeOnArchive":true,"displayNumber":2,"labelTranslations":null}],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":0,"labelTranslations":null}]

我想要这个输出:基于退休:假，退休:真过滤，不应该显示并且有多个对象并且有深层嵌套的子项，我正在递归执行但没有得到预期的输出

 var list = label_lists.filter(function f(item) {
            return item.retired==false ||
            item.children && (item.children = item.children.filter(f)).length
          })

[{"code":"96","value":"RemoveTest","parentLabelCode":1,"color":null,"level":1,"children":[{"code":"97","value":"R1","parentLabelCode":96,"color":null,"level":2,"children":[],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":1,"labelTranslations":null}],"visible":true,"retired":false,"systemLabel":true,"removeOnArchive":true,"displayNumber":0,"labelTranslations":null}]

Answer 1

这是一个递归解决方案，它也复制而不是修改源对象:

function notRetired(arr) {
    return arr.filter(item => !item.retired)
        .map(item => Object.assign({}, item, {
            children: notRetired(item.children)
        }));
}

这是一个(有点老套的)变体，适用于循环结构:

function notRetired(arr, cache = null) {
    return arr.filter(item => !item.retired)
        .map((item, ret) => (cache || (cache = new Map())).get(item) ||
            (cache.set(item, ret = {}) && Object.assign(ret, item, {
                children: notRetired(item.children, cache)
            })));
}