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javascript - 如何通过多个相同的属性过滤对象数组

问题描述

如何按照问题中的说明过滤此数组

描述

请注意，entry1和entry4共享和的相同值。property: 'subject'property: 'field'

问题

我正在寻找一种高效且干净的方法来过滤此数组并获取与这些ies共享两个 values的条目。property

更新：

返回值

我不是试图转换数据，而是分析它。因此分析的返回值应如下所示：

[['entry1', 'entry4'],...]

并且通过这个分析列表，我可以轻松地将我triples = [...]的列表转换为三元组列表，在其中删除一个条目（无关紧要，可以是“entry1”或“entry4”），并更新另一个

[
  { subject: "entry1", property: "subject", value: "sport" },
  { subject: "entry1", property: "field", value: "category" },
  { subject: "entry1", property: "content", value: "football" },
  { subject: "entry1", property: "content", value: "basketball" },
]

附言

  1. 我不是在寻找这样的解决方案：

    array.filter(({property, value})=> property === 'sport' && value === 'category')

我不知道“运动”或“类别”。这些是动态值。

  1. 我的实际数据要大得多，并且每个条目包含更多的属性类型。它也没有我在这里展示的那么好。我确实简化了它，所以请记住性能。

代码片段：

const triples = [
  { subject: "entry1", property: "subject", value: "sport" },
  { subject: "entry1", property: "field", value: "category" },
  { subject: "entry1", property: "content", value: "football" },
  
  { subject: "entry4", property: "subject", value: "sport" },
  { subject: "entry4", property: "field", value: "category" },
  { subject: "entry4", property: "content", value: "basketball" },
  
  { subject: "entry2", property: "subject", value: "music" },
  { subject: "entry2", property: "field", value: "category" },
  { subject: "entry2", property: "content", value: "notes" },
  
  { subject: "entry3", property: "subject", value: "painting" },
  { subject: "entry3", property: "field", value: "category" },
  { subject: "entry3", property: "content", value: "drawings" }
];

标签： javascriptarraysperformancefilteringlodash

解决方案

我必须说输入数据结构不是最优的，并且使用“主题”作为真实对象属性和值将 property使其更加混乱。我将第一个概念（真正的 subject）称为“条目”，因为样本值为“entry1”、“entry2”、...。

这是一种提取["entry1", "entry4"]样本数据的方法：

  1. 通过将数据输入到对象中进行分组，其中“属性”和“值”被转换为键/值对，因此您将得到如下内容：

    {
    entry1: { subject: "sport", field: "category", content: "football" },
    entry4: { subject: "sport", field: "category", content: "basketball" },
    entry2: { subject: "music", field: "category", content: "notes" },
    entry3: { subject: "painting", field: "category", content: "drawings" }
}

    这将更容易使用。下面的代码实际上会创建一个Map而不是一个普通的对象，但它是相同的原理。

  2. 为这些对象定义一个新group属性，其中值由主题和字段组成，字符串化为 JSON。例如，上述结果的第一个对象将扩展为：

    group: '["sport","category"]'

  3. 创建一个条目映射，由它们的组值键控。所以这将给出这个结果：

    {
    '["sport","category"]': ["entry1","entry4"],
    '["music","category"]': ["entry2"],
    '["painting","category"]': ["entry3"]
}

  4. 现在，仅列出值（子数组）以及仅列出具有多个条目值的值是一个简单的步骤。

这是实现：

const triples = [{subject: "entry1", property: "subject", value: "sport"},{subject: "entry1", property: "field", value: "category"},{subject: "entry1", property: "content", value: "football"},{subject: "entry4", property: "subject", value: "sport"},{subject: "entry4", property: "field", value: "category"},{subject: "entry4", property: "content", value: "basketball"},{subject: "entry2", property: "subject", value: "music"},{subject: "entry2", property: "field", value: "category"},{subject: "entry2", property: "content", value: "notes"},{subject: "entry3", property: "subject", value: "painting"},{subject: "entry3", property: "field", value: "category"},{subject: "entry3", property: "content", value: "drawings"},];

// 1. Group the data by subject into objects where "property" and "value" are translated into key/value pairs:
const entries = new Map(triples.map(o => [o.subject, { entry: o.subject }]));
triples.forEach(o => entries.get(o.subject)[o.property] = o.value);
// 2. Define a group value for these objects (composed of subject and field)
entries.forEach(o => o.group = JSON.stringify([o.subject, o.field]));
// 3. Create Map of entries, keyed by their group value
const groups = new Map(Array.from(entries.values(), o => [o.group, []]));
entries.forEach(o => groups.get(o.group).push(o.entry));
// 4. Keep only the subarrays that have more than one value
const result = [...groups.values()].filter(group => group.length > 1);
console.log(result);

请注意，输出是一个嵌套数组，因为理论上可能有更多组合条目，例如[ ["entry1", "entry4"], ["entry123", "entry521", "entry951"] ]

以上可以修改/扩展以获得最终的过滤结果。在第三步中，您仍将收集对象（不仅仅是条目值），然后将过滤后的结果映射回原始格式：

const triples = [{subject: "entry1", property: "subject", value: "sport"},{subject: "entry1", property: "field", value: "category"},{subject: "entry1", property: "content", value: "football"},{subject: "entry4", property: "subject", value: "sport"},{subject: "entry4", property: "field", value: "category"},{subject: "entry4", property: "content", value: "basketball"},{subject: "entry2", property: "subject", value: "music"},{subject: "entry2", property: "field", value: "category"},{subject: "entry2", property: "content", value: "notes"},{subject: "entry3", property: "subject", value: "painting"},{subject: "entry3", property: "field", value: "category"},{subject: "entry3", property: "content", value: "drawings"},];

// 1. Group the data by subject into objects where "property" and "value" are translated into key/value pairs:
const entries = new Map(triples.map(o => [o.subject, { entry: o.subject }]));
triples.forEach(o => entries.get(o.subject)[o.property] = o.value);
// 2. Define a group value for these objects (composed of subject and field)
entries.forEach(o => o.group = JSON.stringify([o.subject, o.field]));
// 3. Create Map of objects(*), keyed by their group value
const groups = new Map(Array.from(entries.values(), o => [o.group, []]));
entries.forEach(o => groups.get(o.group).push(o));
// 4. Keep only the subarrays that have more than one value
const result = [...groups.values()].filter(group => group.length > 1)
// 5. ...and convert it back to the original format:
    .flatMap(group => [
        { subject: group[0].entry, property: "subject", value: group[0].subject },
        { subject: group[0].entry, property: "field", value: group[0].field },
        ...group.map(o => ({ subject: group[0].entry, property: "content", value: o.content }))
    ]);

console.log(result);

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