按字母顺序对对象数组进行排序并删除重复项

Question

所以我有一个objects array ，例如：

[
    { letter: 'A', title: 'All' },{ letter: 'B', title: 'Business' },
    { letter: 'G', title: 'Cor' },
    { letter: 'G', title: 'Car' },
]

我如何对其进行排序，使其看起来像这样：

[
    { letter: 'A', title: 'All' },
    { letter: 'B', title: 'Business' },
    { letter: 'G', title: 'Cor', 'Car' },
]

Answer 1

这样的事情会起作用

const original = [
    { letter: 'A', title: 'All' },
    { letter: 'B', title: 'Business' },
    { letter: 'G', title: 'Cor' },
    { letter: 'G', title: 'Car' },
];

const map = {};

original.forEach(({letter, title}) => {
    if(!Object.hasOwnProperty.call(map, letter)) map[letter] = {letter, title: [title]};
    else {
        map[letter]?.title.push(title);
    }
});

console.debug(Object.values(map));

Answer 2

您可以使用Array.sort()按字母对输入数组进行排序，然后使用Array.reduce()和Map object 按字母对输入进行分组：

 const input = [ { letter: 'Z', title: 'Title Z' }, { letter: 'A', title: 'All' }, { letter: 'B', title: 'Business' }, { letter: 'G', title: 'Cor' }, { letter: 'G', title: 'Car' }, ]; const sortedInput = input.sort(({ letter: a},{ letter: b}) => a.localeCompare(b)); const result = [...sortedInput.reduce((acc, { letter, title }) => { return acc.set(letter, { letter, title: [...(acc.get(letter)?.title || []), title] } ); }, new Map()).values()]; console.log('Result:', result)

 .as-console-wrapper { max-height: 100%;important: top; 0; }

Answer 3

您可以先sort再filter 。

const data = [{
    letter: 'A',
    title: 'All'
}, {
    letter: 'B',
    title: 'Business'
}, {
    letter: 'G',
    title: 'Cor'
}, {
    letter: 'A',
    title: 'Car'
}];

data.sort(function (a, b) {
    if (a.letter < b.letter) {
        return -1;
    }
    if (a.letter > b.letter) {
        return 1;
    }
    return 0;
})

const result = data.filter((v, i, a) => a.findIndex(t => (t.letter === v.letter)) === i)

console.log(result)

Answer 4

您可以先排序，然后减少数组。

 const data = [{ letter: 'A', title: 'All' }, { letter: 'B', title: 'Business' }, { letter: 'G', title: 'Cor' }, { letter: 'G', title: 'Car' }], result = data.sort((a, b) => a.letter.localeCompare(b.letter)).reduce((r, o) => { if (r[r.length - 1]?.letter === o.letter) r[r.length - 1].title += ', ' + o.title; else r.push({...o }); return r; }, []); console.log(result);

 .as-console-wrapper { max-height: 100%;important: top; 0; }

Answer 5

    let arr = [
      { letter: "A", title: "All" },
      { letter: "G", title: "Car" },
      { letter: "B", title: "Business" },
      { letter: "G", title: "Cor" },
      { letter: "A", title: "All" },

    ];

    arr.sort((a, b) => a.letter.localeCompare(b.letter));
    var result = arr.reduce((unique, o) => {
      if(!unique.some(obj => obj.letter === o.letter )) {
        unique.push(o);
      }
      return unique;
  },[]);
    console.log("unique and sorted arr: ", result);