Question

这里的问题很简单。无法弄清。
我正在尝试从primaryWork和secondaryWork获取结果，并将这些结果分配给变量myWorkList。请让我知道我在这里做错了。

谢谢

  let myWorkList
  let primaryWork = this.list.filter(r => r.worker === null)
  let secondaryWork = this.list.filter(r => r.worker === this.currentWorker.id)
  if (this.list) {
    if (this.superuser && this.currentWorker) myWorkList = primaryWork && secondaryWork
  }
  return myWorkList

Answer 1

听起来primaryWork和secondaryWork都是数组。您可能正在寻找.concat() method：

  let myWorkList
  let primaryWork = this.list.filter(r => r.worker === null)
  let secondaryWork = this.list.filter(r => r.worker === this.currentWorker.id)
  if (this.list) {
    if (this.superuser && this.currentWorker) myWorkList = primaryWork.concat(secondaryWork)
  }
  return myWorkList

或者，修复代码中的一些潜在错误：

  // whoever is using the return value from this function expects an array, so if this.list is undefined (or if this.superuser is false) we should return an empty array instead of undefined
  let myWorkList = []
  // if this.list is undefined, this.list.filter will fail - so we do it inside the conditional block
  if (this.list) {
    let primaryWork = [];
    let secondaryWork = [];

    // if this.superuser or this.currentWorker are false, we don't need to waste CPU cycles computing this.list.filter()
    if (this.superuser)
      // I made the assumption (correct me if I'm wrong) that if r.worker is null, the work belongs to the superuser
      primaryWork = this.list.filter(r => r.worker === null)

    // if this.currentWorker is undefined, this.currentWorker.id will fail -- so we perform this filter inside yet another conditional block
    if (this.currentWorker)
      secondaryWork = this.list.filter(r => r.worker === this.currentWorker.id)

    myWorkList = primaryWork.concat(secondaryWork)
  }
  return myWorkList

最后，您可以将所有内容都放入一个单独的filter中，并且只对列表进行一次而不是两次的迭代，就像这样：

  return (
    // Check that this.list is defined before filtering
    this.list ?
      this.list.filter(r =>
        // superuser case
        (this.superuser && r.worker === null)
        || // or
        // non-superuser case
        (this.currentWorker && r.worker === this.currentWorker.id)
      )
    // Return an empty array if this.list was undefined
    : []
  );

请注意，在此最终版本中，我们不会实例化myWorkList，primaryWork，或 secondaryWork。如果我们可以直接返回所需的最终值，则无需在内存中分配空数组以稍后进行垃圾回收。最终表单的运行速度应快2-3倍：

因为我们将this.list数组迭代一次而不是两次，所以速度快了两倍
快一点 ，因为我们避免了不必要的内存分配

在我的计算机上，初步基准测试将其速度提高了约2.4：

var list = [{worker: null}, {worker: null}, {worker: 1}, {worker: 2}, {worker: 2}, {worker: 3}, {worker: 4}, {worker: null}, {worker: 2}]

var d0 = new Date(); for (var i = 0; i < 500000; i++) { var primary = list.filter(r => r.worker === null); var secondary = list.filter(r => r.worker === 2); primary.concat(secondary); } console.log(new Date() - d0);
// 659

var d0 = new Date(); for (var i = 0; i < 500000; i++) { list.filter(r => r.worker === null || r.worker === 2); } console.log(new Date() - d0);
// 272

如何在第三个变量中分配2个单独变量的结果？

1 个答案: