（javascript）如何對具有季節和年份的字串陣列進行排序？

季節需要按以下順序排列：春季、夏季、秋季、冬季，每個季節都有 2025 年和 2026 年。

所有 2025 年都需要彼此相處，所有 2026 年都需要彼此相處（2025 年和 2026 年只是示例，年份可以是任何東西：1945、3005、7980 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']

我知道我可能必須比較這些年份，但由于它們是字串，我很難只比較數字。

這是我想到的：

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)

uj5u.com熱心網友回復：

我將字串分成年份和季節，比較年份并比較同一年份的季節。我為訂單使用了一系列季節。

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);

我創建了一個淺拷貝

[...s]

保持原始陣列不變。

相同的邏輯，大型陣列的性能更好

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);

uj5u.com熱心網友回復：

該解決方案基本上建立在您的建議之上。

首先，將值拆分為具有season和的物件陣列year。year然后按季節的索引進行排序。然后將這些值重新組合在一起。

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))

uj5u.com熱心網友回復：

season.split(/([0-9] )/)使用String#split()之后你幾乎完成了

• 剛剛創建seasonObj了所有季節所需的排序順序，并結合了Array#map()、解構分配和Array#sort()

代碼：

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)

標籤：javascript 数组 排序

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