Question

我有一张这样的地图：

我想从地图值内的数组中删除{frequency:"WN"}。

我正在尝试以这种方式将其删除，但价值并没有被删除：

const map2 = new Map();
map2.set(0, [{frequency:"DN"},{frequency:"WN"}, {frequency:"KK"}]);
map2.set(1, [{frequency:"DN"},{frequency:"WN"}, {frequency:"KK"}]);

const strNanFrq = ['DN', 'WN', 'MN', 'QN', 'YN'];
for (tsRequest of map2.values()) {
  tsRequest.filter((req, index) => {
    if (strNanFrq.includes(req['frequency'])) {
      tsRequest.splice(index, 1);
    }
  });
}

console.log(Array.from(map2.values())); // still has all the values in map

Answer 1

您可以循环访问Map和filter的每个值，以删除frequency中包含的strNanFrq的对象。为strNanFrq创建一个Set，并检查集合has是否是当前对象的频率。

const map = new Map([
  [0, [{ frequency:"DN" },{ frequency:"WN" }, { frequency:"KK" } ]],
  [1, [{ frequency:"YN" },{ frequency:"AA" }, { frequency:"BB" } ]],
]);

const strNanFrq = new Set(['DN', 'WN', 'MN', 'QN', 'YN']);

for (const [key, value] of map)
  map.set(key, value.filter(v => !strNanFrq.has(v.frequency)))

console.log(map.get(0))
console.log(map.get(1))

Answer 2

再次使用前一个数组的过滤后的数组设置映射值。如果地图中还有更多数组，请使用forEach遍历地图键值对，并检查该值是否实际上是一个数组，并使用filter对其进行重置：

var map2 = new Map();

map2.set(0, [{frequency:"DN"},{frequency:"WN"}, {frequency:"KK"}]);
map2.set(1, [{frequency:"DN"},{frequency:"WN"}, {frequency:"KK"}]);


map2.forEach((val, key) => {  
  if (Array.isArray(val)) {
      map2.set(key, val.filter(el => el.frequency !== 'WN'));
  }
});

console.log(map2.values());

如何从地图中删除特定值

2 个答案: