Question

我有一个这样的数组数组：

let arr = [
      ["do you ever", "have you ever", "worried"],
      ["another", ""],
      ["about"],
      ["the notion"],
      ["that"],
      ["did you ever"],
      [""]
];

我需要从嵌套数组和作为结果变为空的嵌套数组中删除空字符串。例如，

[ ["do you ever", "have you ever", "worried"],
  ["another"],
  ["about"],
  ["the notion"],
  ["that"],
  ["did you ever"]
];

Answer 1

您可以按照以下方式尝试使用Array.prototype.map()和Array.prototype.filter()：

let arr = [
      ["do you ever", "have you ever", "worried"],
      ["another", ""],
      ["about"],
      ["the notion"],
      ["that"],
      ["did you ever"],
      [""]
];
arr = arr.map(i => i.filter(i => i.trim())).filter(i => i.length);
console.log(arr);

Answer 2

这是另一种实现方式：

let arr = [
      ["do you ever", ,"","have you ever", "worried"],
      ["another", ""],
      ["about"],
      ["the notion"],
      ["that"],
      ["did you ever"],
      [""]
];
arr = arr.map(a => a.filter(e => e !== "")).filter(e=> e.length !== 0)
console.log(arr)

Answer 3

据我了解，您所有的数组（父数组和子数组）都可以减小大小。单独使用Array#filter是不可能的，并且Array#map不会减小数组的大小。

您可以为此使用Array#reduce或Array#flatMap。

在这种情况下，我会选择后者。它允许您取消嵌套在迭代过程中产生的子数组。这意味着您可以使用Array#flatMap来增长数组：

[1, 2].flatMap(n => [n, n]);
//=> [1, 1, 2, 2]

这意味着您也可以清空数组：

[1, 2].flatMap(n => []);
//=> []

两个示例都无法用Array#map完成。

因此，我们既可以用来变换数组的元素，也可以用来变换数组本身（即大小变大或变小）：

const xs = [
      ["do you ever", "have you ever", "worried"],
      ["another", ""],
      ["about"],
      ["the notion"],
      ["that"],
      ["did you ever"],
      [""]
];

console.log(

  xs.flatMap(ys => {
    const zs = ys.filter(y => y !== "");
    return zs.length > 0 ? [zs] : [];
  })

)

Answer 4

通过地图和过滤器的组合，您可以实现；

我假设arr仅包含数组，并且该数组包含字符串。否则，应该进行类型检查。

let arr = [
  ["do you ever", "have you ever", "worried"],
  ["another", ""],
  ["about"],
  ["the notion"],
  ["that"],
  ["did you ever"],
  [""]
];

arr.map(items => items.filter(_item => _item.length > 0)).filter(i => i.length > 0);

从嵌套数组中过滤出空字符串

4 个答案: