Javascript 根据一个条件，如何从处理过的数组中移除一项并将其移动到另一个数组

const tmpSelected = [
  { "Metric": "AAA", "Weight": 10, "Value": "xxx" },
  { "Metric": "BBB", "Weight": 20, "Value": "xx1" },
  { "Metric": "CCC", "Weight": 30, "Value": "xx2" },
];
const tmpNonSelected = [
  { "Metric": "DDD", "Weight": 5, "Value": "yy" },
  { "Metric": "EEE", "Weight": 15, "Value": "zz" },
  { "Metric": "DDD", "Weight": 5, "Value": "aa" },
  { "Metric": "FFF", "Weight": 25, "Value": "cc" },
  { "Metric": "DDD", "Weight": 5, "Value": "bb" },
];

const moveItems = (metric, weight) => {
  tmpSelected
    .push(
      ...reject(tmpNonSelected, item => {

        return item.Metric === metric && item.Weight === weight;
      }),
    );
};
moveItems('DDD', 5);

console.log({ tmpSelected, tmpNonSelected });

.as-console-wrapper { min-height: 100%!important; top: 0; }

<script>
function reject(target, condition) {
  const result = [];

  let idx = target.length;
  const copy = [...target];

  // - Processing the `target` array from RIGHT to LEFT
  //   keeps the `idx` always in sync with both related
  //   array items, the one of the mutated and also the
  //   one of the unmutated `copy` of the processed array
  //   reference.
  // - Thus, `condition` always gets passed the unmutated
  //   shallow `copy`.

  while (idx) {
    if (
      // - take a *sparse array* into account.
      target.hasOwnProperty(--idx) &&

      // - keep processing the unmutated array.
      condition(copy[idx], idx, copy)
    ) {
      // - keep filling the `result` array with each
      //   *rejected* item FROM its LEFT side while
      //   mutating the `target` array.
      result.unshift(target.splice(idx, 1)[0]);
    }
  }
  // - returns an array of rejected items, but not the
  //   processed and mutated `target` array reference.
  return result;
}
</script>