(javascript)如何对具有季节和年份的字符串数组进行排序？

Question

季节应该是这样的:春天，夏天，秋天，冬天，每一个季节都有2025年和2026年。

2025年的一切都需要彼此在一起，2026年的一切都需要在一起(2025年和2026年只是一个例子:1945年、3005、7980等等)。

例如：

const seasonArr = ['Spring2026',' Spring2025','Summer2026','Summer2025','Fall2025','Fall2026','Winter2026','Winter2025']

let sortedArr = []

const someFunction = () => {
   ...
}

someFunction(seasonArr) // output: sortedArr = ['Spring2025', 'Summer2025', 'Fall2025', 'Winter2025', 'Spring2026', 'Summer2026', 'Fall2026', 'Winter2026']

我知道我可能需要比较这些年，但因为它们是字符串，所以我很难仅仅比较数字。

这是我想过的：

const seasonArr = ['Spring2026',' Spring2025','Summer2026','Summer2025','Fall2025','Fall2026','Winter2026','Winter2025']

let sortedArr = []

const someFunction = (seasonArr) => {
   for (const season of seasonArr) {
     let year = season.split(/([0-9]+)/)
     // unsure where to go from here
   }
}

someFunction(seasonArr)

Answer 1

我把琴弦分成年份和季节，比较一年，比较同一年份的季节。我用了一系列的季节来订购。

​

const seasonArr = ['Spring2026','Spring2025','Summer2026','Summer2025','Fall2025','Fall2026','Winter2026','Winter2025'];

const seasons = ['Spring', 'Summer', 'Fall', 'Winter'];

const regexp = /(.+)(\d{4})/;
const someFunction = (s) => {
   return [...s].sort((lhs, rhs) => {
     const [seasonL, yearL] = regexp.exec(lhs).slice(1);
     const [seasonR, yearR] = regexp.exec(rhs).slice(1);
     return +yearL - +yearR || seasons.indexOf(seasonL) - seasons.indexOf(seasonR);
   });
}

let sortedArr = someFunction(seasonArr);
console.log(sortedArr);

​

我创建一个浅拷贝

[...s]

若要保持原始数组不变，请执行以下操作。

具有更好的大数组性能的相同逻辑

​

const seasonArr = ['Spring2026','Spring2025','Summer2026','Summer2025','Fall2025','Fall2026','Winter2026','Winter2025'];

const seasons = ['Spring', 'Summer', 'Fall', 'Winter'];

const regexp = /(.+)(\d{4})/;
const someFunction = (s) => {
   return s
     .map(el => {
       const [season, year] = regexp.exec(el).slice(1);
       return [season, year, seasons.indexOf(season[0])];
     })
     .sort((lhs, rhs) => {
       return +lhs[1] - +rhs[1] || lhs[2] - rhs[2];
     })
     .map(el => el[0] + el[1]);
}

let sortedArr = someFunction(seasonArr);
console.log(sortedArr);

​

Answer 2

这个解决方案基本上建立在你建议的基础上。

首先，将值拆分为具有season和year的对象数组。然后根据year和季节指数进行排序。然后把这些值重新组合起来。

const seasonArr = ['Spring2026',' Spring2025','Summer2026','Summer2025','Fall2025','Fall2026','Winter2026','Winter2025']

const SEASONS = ["Spring", "Summer", "Fall", "Winter"]

function comparator(a, b) {
  if (a.year == b.year) {
    const aSeasonIndex = SEASONS.indexOf(a.season)
    const bSeasonIndex = SEASONS.indexOf(b.season)
    return aSeasonIndex - bSeasonIndex;
  }
  return a.year - b.year;
}

function seasonYearToObject(obj) {
  const matches = obj.match(/([^\d]*)(\d+)/)
  if (matches) {
    return {season: matches[1], year: matches[2]}
  }
}

function objectToSeasonYear(obj) {
  return `${obj.season}${obj.year}`
}
  
function sortByYearAndSeason(arr) {
  return arr
    .map(entry => seasonYearToObject(entry))
    .sort(comparator)
    .map(objectToSeasonYear);
}

console.log (sortByYearAndSeason(seasonArr))

Answer 3

在使用season.split(/([0-9]+)/)和String#split()之后，您几乎完成了

• 只需为所有季节创建具有所需排序顺序的seasonObj，并将Array#map()、解构分配和Array#sort()组合起来

代码：

​

const seasonArr = ['Spring2026', 'Spring2025', 'Summer2026', 'Summer2025', 'Fall2025', 'Fall2026', 'Winter2026', 'Winter2025']
const seasonObj = { Spring: 0, Summer: 1, Fall: 2, Winter: 3 }

const result = seasonArr
  .map((season) => season.split(/([0-9]+)/))
  .sort(([aSeason, aYear], [bSeason, bYear]) =>
      +aYear - bYear || seasonObj[aSeason] - seasonObj[bSeason])
  .map(([season, year]) => `${season}${year}`)

console.log(result)

​