通过与另一个数组的值比较对数组进行排序 [英] Sort an array by comparison with values of another array

问题描述

我有两个数组:

const tags = ['one', 'two', 'three'];
let posts = [
  { tags: ['four', 'five'] },
  { tags: ['one', 'six'] },
  { tags: ['seven'] },
  { tags: ['nine', 'two'] },
];

我需要以这种方式对posts数组进行排序:从tags数组中至少带有一个标签的元素必须位于数组的开头.其余元素(没有匹配标签)的顺序并不重要.

I need to sort posts array in this way: elements with at least one tag from tags array must be at the beginning of the array. The order of the remaining elements (without matching tags) is not important.

预期结果:

posts = [
  { tags: ['one', 'six'] },
  { tags: ['nine', 'two'] },
  { tags: ['four', 'five'] },
  { tags: ['seven'] },
];

推荐答案

您可以使用

You could check if any of the tags of each object exists in tags array using some and includes. Then subtract the value for 2 objects being compared.

const tags = ['one', 'two', 'three'];
let posts = [
  { tags: ['four', 'five'] },
  { tags: ['one', 'six'] },
  { tags: ['seven'] },
  { tags: ['nine', 'two'] },
];

posts.sort((a, b) => 
  tags.some(t => b.tags.includes(t)) - tags.some(t => a.tags.includes(t))
)

console.log(posts)

如果a具有匹配的标签，而b没有，则compareFunction返回-1(false - true)，并且a相对于b优先.

If a has a matching tag and b doesn't, then the compareFunction returns -1 (false - true) and a is prioritized relative to b.

在相反的情况下，它返回1

For the reverse situation, it returns 1

如果a和b都具有匹配的标签或没有标签，则compareFunction将返回零.因此，它们彼此之间相对不动

If both of a and b have a matching tag or don't have a tag, compareFunction will return zero. So, they are unmoved relative to each other

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