JS，字典列表到列表字典，基于键

Question

我有一个字典列表，其中有一些属性，比如一个 url 和一些关于 url 的信息：[{    url:"https://example1.com/a"    something:"ABC"},{    url:"https://example1.com/b"    something:"DEF"},{    url:"https://example2.com/c"    something:"GHI"},{    url:"https://example2.com/d"    something:"JKL"}]现在我想把它分成一个列表字典，根据 url 分组。对于上述，我的目标数据结构是这样的：{    "example1.com" : [{        url:"https://example1.com/a"        something:"ABC"    },{        url:"https://example1.com/b"        something:"DEF"    }],    "example2.com" : [{        url:"https://example2.com/c"        something:"GHI"    },{        url:"https://example2.com/d"        something:"JKL"    }]}在 python 中，这可以使用 itertools 包和一些列表理解技巧来实现，但我需要在 javascript/nodejs 中完成。有人可以引导我朝着正确的方向在 javascript 中执行此操作吗？

Answer 1

您可以reduce在数组对象上使用该方法。

let data = [{

    url:"https://example1.com/a",

    something:"ABC"

},{

    url:"https://example1.com/b",

    something:"DEF"

},{

    url:"https://example2.com/c",

    something:"GHI"

},{

    url:"https://example2.com/d",

    something:"JKL"

}];

let ret = data.reduce((acc, cur) => {

  let host = cur['url'].substring(8, 20); // hardcoded please use your own 

  if (acc[host])

    acc[host].push(cur);

  else

    acc[host] = [cur];

  return acc;

}, {})

console.log(ret);

Answer 2

const dataFromQuestion = [{

    url:"https://example1.com/a",

    something:"ABC"

},{

    url:"https://example1.com/b",

    something:"DEF"

},{

    url:"https://example2.com/c",

    something:"GHI"

},{

    url:"https://example2.com/d",

    something:"JKL"

}];

function listOfDictionaryToDictionaryOfList(input, keyMapper) {

  const result = {};

  for (const entry of input) {

    const key = keyMapper(entry);

    if (!Object.prototype.hasOwnProperty.call(result, key)) {

      result[key] = [];

    }

    result[key].push(entry);

  }

  return result;

}

function getHost(data) {

  const url = new URL(data.url);

  return url.host;

}

console.log(listOfDictionaryToDictionaryOfList(dataFromQuestion, getHost)); 

Answer 3

data.reduce((groups, item) => {

    let host = new URL(item.url).hostname;

    (groups[host] || (groups[host] = [])).push(item);

    return groups;

}, {});

单行（虽然很神秘）

data.reduce((g, i, _1, _2, h = new URL(i.url).hostname) => ((g[h] || (g[h] =[])).push(i), g), {});