新闻元数据多维度统计及结果按标签排序技术咨询

需求背景 #

我有一款新闻推送应用，会从其他站点拉取JSON格式的数据，仅需汇总各类元数据用于后续性能评估，不关注文章实际内容。

源数据 #

let articles = [
    {id: "0",title: "article 1",tags: ["110"],source: "Company A", countries:["USA","CAN"]},
    {id: "1",title: "article 2",tags: ["120","160","100"],source: "Company B", countries:["USA","CHN"]},
    {id: "2",title: "article 3",tags: ["130","150"],source: "Company C", countries:["USA"]},
    {id: "3",title: "article 4",tags: ["140","120","110","130"],source: "Company A", countries:["JPN","IRE"]},
    {id: "4",title: "article 5",tags: ["110","130"],source: "Company C", countries:["SWE"]},
  ]

期望输出 #

按来源统计结果 #

let output = [
    {source: "Company A",tags:[{code:"110",count:2},{code:"140",count:1},{code:"120",count:1},{code:"130",count:1}],countries:[{code:"USA",count:1},{code:"CAN",count:1},{code:"JPN",count:1},{code:"IRE",count:1}]},
    {source: "Company B",tags:[{code:"120",count:1},{code:"160",count:1},{code:"100",count:1}],countries:[{code:"USA",count:1},{code:"CHN",count:1}]},
    {source: "Company C",tags:[{code:"110",count:2},{code:"130",count:2},{code:"150",count:1}],countries:[{code:"USA",count:1},{code:"SWE",count:1}]}
  ]

按标签统计结果（需按标签数字顺序排序） #

let output2 = [
    {tag:"110",sources:[{source:"Company A",count:2},{source:"Company C",count:1}],countries:[{country:"USA",count:1},{country:"CAN",count:1},{country:"SWE",count:1}]},
    {tag:"120",sources:[{source:"Company A",count:1},{source:"Company B",count:1}],countries:[{country:"USA",count:1},{country:"CHN",count:1},{country:"JPN",count:1},{country:"IRE",count:1}]},
    {tag:"130",sources:[{source:"Company A",count:1},{source:"Company C",count:1}],countries:[{country:"USA",count:1},{country:"SWE",count:1}]},
    {tag:"140",sources:[{source:"Company A",count:1}],countries:[{country:"JPN",count:1},{country:"IRE",count:1}]},
    {tag:"150",sources:[{source:"Company C",count:1}],countries:[{country:"USA",count:1}]},
    {tag:"160",sources:[{source:"Company B",count:1}],countries:[{country:"USA",count:1},{country:"CHN",count:1}]},
    {tag:"100",sources:[{source:"Company B",count:1}],countries:[{country:"USA",count:1},{country:"CHN",count:1}]}
]

初始思路与顾虑 #

最初考虑用三个Set分别存储标签、国家和来源，再基于这些集合构建对象并迭代统计，但担心扩展性——文章数量可达100-5000条，单篇文章标签1-10+个、国家1-50+个。相关代码：

tags = new Set();
countries = new Set();
sources = new Set();

articles.forEach(item =>{
   item.tags.forEach(element =>{
      tags.add(element);
    })
})

articles.forEach(item =>{
    item.countries.forEach(element =>{
        countries.add(element);
    })
})

articles.forEach(item =>{
    sources.add(item.source);
})

现有Map实现与问题 #

目前已经用Map实现了统计逻辑，但不知道在哪里添加排序逻辑，让按标签统计的结果按标签数字顺序排列。现有代码：

按标签统计代码 #

//-----By Tag------//

articles.forEach((a) => {
    a.tags.forEach((tag) => {
        if (!tags.has(tag)) tags.set(tag, { sources: new Map(), countries: new Map() });
    
        tags.get(tag).sources.set(a.source, (tags.get(tag).sources.get(a.source) ?? 0) + 1);
    
        a.countries.forEach((c) => {
            tags.get(tag).countries.set(c, (tags.get(tag).countries.get(c) ?? 0) + 1);
        });
    });
});

const result = Array.from(tags)
    .map(([tag, { sources, countries }]) => ({
        tag,
        sources: Array.from(sources).map(([source, count]) => ({ source, count })),
        countries: Array.from(countries).map(([country, count]) => ({ country, count })),
    }));

console.log(result);

按来源统计代码 #

//-----By Source------//

articles.forEach((a) => {
    if (!sources.has(a.source)) sources.set(a.source, { tags: new Map(), countries: new Map() });
    
        a.tags.forEach((b) => {
            sources.get(a.source).tags.set(b, (sources.get(a.source).tags.get(b) ?? 0) + 1);
        });

        a.countries.forEach((c) => {
            sources.get(a.source).countries.set(c, (sources.get(a.source).countries.get(c) ?? 0) + 1);
    });
});

const result2 = Array.from(sources)
    .map(([source, { tags, countries }]) => ({
        source,
        tags: Array.from(tags).map(([tags, count]) => ({ tags, count })),
        countries: Array.from(countries).map(([country, count]) => ({ country, count })),
    }));

console.log(result2);

解决方案 #

要实现按标签数字顺序排序，只需在将Map转换为数组后添加sort方法，把标签字符串转为数字进行比较；同时修正现有代码中与期望输出不一致的字段名：

修改后的按标签统计代码 #

//-----By Tag------//

const tags = new Map(); // 声明变量避免全局污染

articles.forEach((a) => {
    a.tags.forEach((tag) => {
        if (!tags.has(tag)) tags.set(tag, { sources: new Map(), countries: new Map() });
    
        tags.get(tag).sources.set(a.source, (tags.get(tag).sources.get(a.source) ?? 0) + 1);
    
        a.countries.forEach((c) => {
            tags.get(tag).countries.set(c, (tags.get(tag).countries.get(c) ?? 0) + 1);
        });
    });
});

const result = Array.from(tags)
    // 新增排序逻辑：将标签转为数字比较大小
    .sort(([tagA], [tagB]) => Number(tagA) - Number(tagB))
    .map(([tag, { sources, countries }]) => ({
        tag,
        sources: Array.from(sources).map(([source, count]) => ({ source, count })),
        countries: Array.from(countries).map(([country, count]) => ({ country, count })),
    }));

console.log(result);

修改后的按来源统计代码（匹配期望输出格式+可选排序） #

//-----By Source------//

const sources = new Map(); // 声明变量避免全局污染

articles.forEach((a) => {
    if (!sources.has(a.source)) sources.set(a.source, { tags: new Map(), countries: new Map() });
    
    a.tags.forEach((tag) => {
        sources.get(a.source).tags.set(tag, (sources.get(a.source).tags.get(tag) ?? 0) + 1);
    });

    a.countries.forEach((country) => {
        sources.get(a.source).countries.set(country, (sources.get(a.source).countries.get(country) ?? 0) + 1);
    });
});

const result2 = Array.from(sources)
    // 可选：按来源名称排序
    .sort(([sourceA], [sourceB]) => sourceA.localeCompare(sourceB))
    .map(([source, { tags, countries }]) => ({
        source,
        // 修正字段名为code，同时按标签数字排序
        tags: Array.from(tags)
            .sort(([tagA], [tagB]) => Number(tagA) - Number(tagB))
            .map(([tag, count]) => ({ code: tag, count })),
        // 修正字段名为code
        countries: Array.from(countries).map(([country, count]) => ({ code: country, count }))
    }));

console.log(result2);

内容的提问来源于stack exchange，提问作者Eric Smith