javascript - 来自对象数组的递归数组父匹配

Question

我想创建一个函数，在递归数组循环中获取“父级”，并生成那些“已用”父级的输出数组。

解释起来有点费劲，但是看例子:

const regions = [{
  name: 'Europe',
  subRegions: [{
    name: 'BeNeLux',
    territories: [{
      code: 'NL',
      name: 'Netherlands'
    }, {
      code: 'DE',
      name: 'Germany'
    }, {
      code: 'LU',
      name: 'Luxembourg'
    }]
  }],
  territories: [{
    code: 'UK',
    name: 'United Kingdom'
  }, {
    code: 'AL',
    name: 'Albania'
  }, {
    code: 'ZW',
    name: 'Switzerland'
  }]
}, {
  name: 'Africa',
  territories: [{
    code: 'GH',
    name: 'Ghana'
  }]
}]

const selectedTerritories = ['NL', 'UK', 'GH']

此时我需要一个函数，通过代码搜索领土的所有 TOP 区域，因此输出将如下所示:

const activeRegions = ['Europe', 'Africa']

需要注意的是，Europe (BeNeLux) 中有一个子区域，递归到那个点，它不应该返回 BeNeLux 作为事件区域，但 Europe 而不是。

这是我尝试过的方法，但是有重复的子区域名称并且它忽略了“父”递归发现要求:

const getRegionsLabelFromTerritoryList = (activeTerritories, regions, activeRegions = []) => {
  regions.forEach((region) => {
    if (region.territories) {
      region.territories.forEach(t => {
        if (activeTerritories.includes(t.code)) {
          activeRegions.push(region)
        }
      })
    }

    if (region.subRegions) {
      getRegionsLabelFromTerritoryList(region.subRegions, activeRegions)
    }
  })

  return activeRegions
}

Answer 1

假设 Africa 对象应该在顶层，根据它们的 subRegions 是否满足 filter 顶层对象递归测试(检查 selectedTerritories 是否包含被迭代对象的代码，或者是否有任何 subRegions 子项通过测试的测试):

const regions = [{
  name: 'Europe',
  subRegions: [{
    name: 'BeNeLux',
    territories: [{
      code: 'NL',
      name: 'Netherlands'
    }, {
      code: 'DE',
      name: 'Germany'
    }, {
      code: 'LU',
      name: 'Luxembourg'
    }]
  }],
  territories: [{
    code: 'UK',
    name: 'United Kingdom'
  }, {
    code: 'AL',
    name: 'Albania'
  }, {
    code: 'ZW',
    name: 'Switzerland'
  }]
}, {
  name: 'Africa',
  territories: [{
    code: 'GH',
    name: 'Ghana'
  }]
}];

const selectedTerritories = ['NL', 'UK', 'GH'];

const regionPasses = ({ subRegions, territories }) => (
  territories.some(({ code }) => selectedTerritories.includes(code))
  || (subRegions && subRegions.some(regionPasses))
);
    
const topSelected = regions
  .filter(regionPasses)
  .map(({ name }) => name);
console.log(topSelected);

为了降低计算复杂度，您可以先将 selectedTerritories 转换为 Set(将 O(n) 操作转换为 O(1) 操作):

const regions = [{
  name: 'Europe',
  subRegions: [{
    name: 'BeNeLux',
    territories: [{
      code: 'NL',
      name: 'Netherlands'
    }, {
      code: 'DE',
      name: 'Germany'
    }, {
      code: 'LU',
      name: 'Luxembourg'
    }]
  }],
  territories: [{
    code: 'UK',
    name: 'United Kingdom'
  }, {
    code: 'AL',
    name: 'Albania'
  }, {
    code: 'ZW',
    name: 'Switzerland'
  }]
}, {
  name: 'Africa',
  territories: [{
    code: 'GH',
    name: 'Ghana'
  }]
}];

const selectedTerritories = new Set(['NL', 'UK', 'GH']);

const regionPasses = ({ subRegions, territories }) => (
  territories.some(({ code }) => selectedTerritories.has(code))
  || (subRegions && subRegions.some(regionPasses))
);
    
const topSelected = regions
  .filter(regionPasses)
  .map(({ name }) => name);
console.log(topSelected);

您可以使用单个外部循环而不是两个(.reduce 或类似的东西而不是 .filter 后跟 .map 来获得相同的结果>)，但我认为这更清楚。