在 JavaScript 中编写排序方法的最简洁方法，该方法对多个 object 属性进行排序

Question

This is a sort of code golf / code cleanliness and conciseness question. I have an object I want to sort in JavaScript, and I'm wondering what the shortest and cleanest way to do this.

Given objest like this:

items = [
 {name: "Pierre",  boosted: true,  rank: 1   },
 {name: "Burhan",  boosted: false, rank: null},
 {name: "Ellise",  boosted: false, rank: 1   },
 {name: "Glenn",   boosted: true,  rank: 2   },
 {name: "Zidane",  boosted: false, rank: null},
 {name: "Antonia", boosted: false, rank: 2   },
];

I need to sort with these requirements:

1. Boosted items first, then by name ascending within multiple boosted items.
2. Next put rank , ascending, where rank.= null. Doesn't matter how we sort if there's a tie in rank .
3. When rank = null, sort by name ascending.

So essentially three buckets to sort the items by, then within each bucket sort by name. Rank isn't normalized, it can be any valid number.

The resulting list above sorted should look like this:

sorted = [
 {name: "Glenn",   boosted: true,  rank: 2   },
 {name: "Pierre",  boosted: true,  rank: 1   },
 {name: "Ellise",  boosted: false, rank: 1   },
 {name: "Antonia", boosted: false, rank: 2   },
 {name: "Burhan",  boosted: false, rank: null},
 {name: "Zidane",  boosted: false, rank: null},
];

I wrote this long method:

items.sort((a, b) => {
    // Boosted
    if (a.boosted && b.boosted) {
        return a.name.localeCompare(b.name);
    } else if (a.boosted && !b.boosted) {
        return -1;
    } else if (!a.boosted && b.boosted) {
        return 1;
    }

    // Ranked
    if (a.rank != null && b.rank != null) {
        return a - b;
    } else if (a.rank != null && b.rank == null) {
        return -1;
    } else if (a.rank == null && b.rank != null) {
        return 1;
    }

    // Default by name
    return a.name.localeCompare(b.name);
});

It works, but I think there are much better ways to do this, and I'm curious what styles other people recommend. What's the shortest solution I could write? What's the cleanest? What's the fastest? (Since the sort method could get called many times during a sort operation)

Answer 1

You can use - on booleans and numbers. You can use || to chain multiple sort criteria, short-circuiting when the first doesn't return 0 .

In your case that might be

items.sort((a, b) =>
  b.boosted - a.boosted ||
  (a.rank == null) - (b.rank == null) ||
  a.rank - b.rank ||
  a.name.localeCompare(b.name)
);

if you were just to sort by differently priotised columns. However, you've got the special requirement that boosted items should be sorted differently than non-boosted items, so you indeed need a conditional statement thrown in:

items.sort((a, b) =>
  b.boosted - a.boosted ||
  (a.boosted
    ? a.name.localeCompare(b.name)
    : (a.rank == null) - (b.rank == null) ||
      (a.rank != null
        ? a.rank - b.rank
        : a.name.localeCompare(b.name)
      )
  )
);

Answer 2

Personally I do this way:

 const items = [ {name: "Pierre", boosted: true, rank: 1 }, {name: "Burhan", boosted: false, rank: null}, {name: "Ellise", boosted: false, rank: 1 }, {name: "Glenn", boosted: true, rank: 2 }, {name: "Zidane", boosted: false, rank: null}, {name: "Antonia", boosted: false, rank: 2 }, ]; const gKey=({name,boosted,rank})=>`${boosted?0:1}_${(rank?=null).('000'+rank):slice(-3).'999'}_${name}` items,sort((ab)=>gKey(a):localeCompare(gKey(b)) ) // proof. items.forEach(e=>console.log(JSON,stringify(e),' (key->',gKey(e),')'))

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