blocks|key|1313237|text|需要将字符串更改为+format，如下所示|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1313238|new+Date(Month+Date+Year)|code-block|syntax|javascript|1313239|两个阵列的更新正则表达式|1313240|https://regex101.com/r/h1cm1z/2/|1313241|基于第一数组排序索引的更新的排序第二数组|style|BOLD|1313242|​|1313243|var+arr1+=++['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var+arr2+=['55','2','3','0','21','132'];

function+datesort(arr){
+return+arr.concat().sort((a,b)=>{
+a+=+a.replace(/(\d%2B)(.*)/g,'+1+$1');+//+Month+1+YEAR
+b+=+b.replace(/(\d%2B)(.*)/g,'+1+$1');+//+Month+1+YEAR
++return+new+Date(a)+-+new+Date(b)
})
}

var+after_arr1+=new+datesort(arr1);
var+after_arr2+=+arr1.reduce(function(a,b,c){
+++++var+ind+=+after_arr1.indexOf(b);
+++++a[ind]+=+arr2[c]
+++++return+a
},[]);


console.log(after_arr1.join(','));
console.log(after_arr2.join(','))|1313244|1313245|entityMap|0|LINK|mutability|MUTABLE|url|https://www.w3schools.com/js/js_dates.asp|1^0|9|7|0|0|0|0|0|W|1|0|B|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|19|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]]|C|$]]|$1|M|3|N|5|6|7|1E|8|@$A|1F|B|1G|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|6|7|1H|8|@]|9|@]|C|$]]|$1|S|3|T|5|F|7|1I|8|@]|9|@]|C|$G|H]]|$1|U|3|R|5|6|7|1J|8|@]|9|@]|C|$]]|$1|V|3|-4|5|6|7|1K|8|@]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|Y|Z|10|C|$11|L]]]]

<p>You need change the string to <a href="https://www.w3schools.com/js/js_dates.asp" rel="nofollow noreferrer"><code>new Date(dateString)</code> format</a> like </p>

<pre><code>new Date(Month Date Year)
</code></pre>

<p><strong>Updated regex Pattern for Both Array</strong> </p>

<p><a href="https://regex101.com/r/h1cm1z/2/" rel="nofollow noreferrer">https://regex101.com/r/h1cm1z/2/</a></p>

<p><strong>Updated</strong> Sort Second array based on first array sorting index</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var arr1 =  ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
var arr2 =['55','2','3','0','21','132'];

function datesort(arr){
 return arr.concat().sort((a,b)=&gt;{
 a = a.replace(/(\d+)(.*)/g,' 1 $1'); // Month 1 YEAR
 b = b.replace(/(\d+)(.*)/g,' 1 $1'); // Month 1 YEAR
  return new Date(a) - new Date(b)
})
}

var after_arr1 =new datesort(arr1);
var after_arr2 = arr1.reduce(function(a,b,c){
     var ind = after_arr1.indexOf(b);
     a[ind] = arr2[c]
     return a
},[]);


console.log(after_arr1.join(','));
console.log(after_arr2.join(','))</code></pre>
</div>
</div>
</p>


blocks|key|244262|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|244263|const+input+=+['Mar19',+'Apr18',+'Jun18',+'Jul18',+'May18',+'Jan19'];

//const+output+:+['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can+be+done+easily+by+using+momentjs,+darte-fns,+but+here+i+will+do+it+natively

const+t+=+{
++Jan:+1,
++Feb:+2,
++Mar:+3,
++Apr:+4,
++May:+5,
++Jun:+6,
++Jul:+7,
++Aug:+8,
++Sep:+9,
++Oct:+10,
++Nov:+11,
++Dec:+12
}
const+giveConcatString+=+(a,+t)+=>+{
++const+monthPart+=+a.substr(0,+3)
++const+yearPart+=+a.substr(3)
++return+`${yearPart}${t[monthPart]}`
}
const+sortedArray+=+input.sort((a,+b)+=>+{
++const+concatString+=+giveConcatString(a,+t)
++const+concatStringB+=+giveConcatString(b,+t)
++return+concatString+<=+concatStringB+?+-1+:+1

})
console.log(sortedArray)|code-block|syntax|javascript|244264|244265|这可能会帮助你解决这个问题。本地人做的。|244266|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|M|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const input = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];

//const output : ['Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
//Can be done easily by using momentjs, darte-fns, but here i will do it natively

const t = {
  Jan: 1,
  Feb: 2,
  Mar: 3,
  Apr: 4,
  May: 5,
  Jun: 6,
  Jul: 7,
  Aug: 8,
  Sep: 9,
  Oct: 10,
  Nov: 11,
  Dec: 12
}
const giveConcatString = (a, t) =&gt; {
  const monthPart = a.substr(0, 3)
  const yearPart = a.substr(3)
  return `${yearPart}${t[monthPart]}`
}
const sortedArray = input.sort((a, b) =&gt; {
  const concatString = giveConcatString(a, t)
  const concatStringB = giveConcatString(b, t)
  return concatString &lt;= concatStringB ? -1 : 1

})
console.log(sortedArray)</code></pre>
</div>
</div>
</p>

<p>This may help you solve this problem. Did it natively.</p>


blocks|key|1313273|text|我使用lodash进行分类，并使用substring方法将日期划分为月和日。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1313274|​|1313275|const+data+=+['Mar20','Mar21',+'Mar01',+'Mar19','Apr18','Jun18','Jul18','May18','Jan19']

const+monthMap+=+["Jan",+"Feb",+"Mar",+"Apr",+"May",+"Jun",+"Jul",+"Aug",+"Sep",+"Oct",+"Nov",+"Dec"];

const+sorted+=+_.sortBy(data,+date+=>+monthMap.indexOf(date.substring(0,3)),+date+=>+date.substring(3,5))

console.log(sorted)|code-block|syntax|javascript|1313276|<script+src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>|1313277|1313278|entityMap^0|H|9|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|V|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|O|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|Q|$]]

<p>I used lodash for sorting and <code>substring</code> method to divide date into month and day parts</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const data = ['Mar20','Mar21', 'Mar01', 'Mar19','Apr18','Jun18','Jul18','May18','Jan19']

const monthMap = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"];

const sorted = _.sortBy(data, date =&gt; monthMap.indexOf(date.substring(0,3)), date =&gt; date.substring(3,5))

console.log(sorted)</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>


blocks|key|1313317|text|您可以将日期作为可排序字符串获取并对其进行排序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1313318|要获得由一个签名数组排序的多个数组，可以使用map进行排序，在映射中对索引数组进行排序，指示最终的排序，然后使用此排序重新分配所有数组。|1313319|getD函数通过获取用于排序的array0索引来返回格式化的字符串。在函数的内部，字符串被分解为月份和年份部分，并被它的ISO+8601表示所取代。替换函数回调接受匹配的项，返回一个具有格式化年份的数组和一个具有月份名称和相关月份号的对象月份。然后，将这个数组连接起来并返回。|offset|length|style|CODE|1313320|排序是通过与String#localeCompare进行比较进行的。|1313321|​|1313322|var+array1+=+['Mar19',+'Apr18',+'Jun18',+'Jul18',+'May18',+'Jan19'],
++++array2+=+['Mar19_value',+'Apr18_value',+'Jun18_value',+'Jul18_value',+'May18_value',+'Jan19_value'],
++++array3+=+['55',+'2',+'3',+'0',+'21',+'132'],
++++indices+=+Object
++++++++.keys(array1)
++++++++.sort(function+(a,+b)+{
++++++++++++function+getD(i)+{
++++++++++++++++var+months+=+{+Jan:+'01',+Feb:+'02',+Mar:+'03',+Apr:+'04',+May:+'05',+Jun:+'06',+Jul:+'07',+Aug:+'08',+Sep:+'09',+Oct:+'10',+Nov:+'11',+Dec:+'12'+},
++++++++++++++++++++s+=+array1[i];
++++++++++++++++return+s.replace(/%5E(...)(.%2B)$/,+(_,+m,+y)+=>+[y.padStart(4,+'0'),+months[m]].join('-'));
++++++++++++}
++++++++++++return+getD(a).localeCompare(getD(b));
++++++++}),
++++result+=+[array1,+array3].map(a+=>+indices.map(i+=>+a[i]));
++
result.forEach(a+=>+console.log(...a));|code-block|syntax|javascript|1313323|1313324|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/ISO_8601|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/localeCompare^0|0|0|0|4|F|6|1O|8|0|0|6|K|6|K|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|15|8|@$F|16|G|17|H|I]|$F|18|G|19|H|I]]|9|@$F|1A|G|1B|1|1C]]|A|$]]|$1|J|3|K|5|6|7|1D|8|@$F|1E|G|1F|H|I]]|9|@$F|1G|G|1H|1|1I]]|A|$]]|$1|L|3|M|5|6|7|1J|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1K|8|@]|9|@]|A|$Q|R]]|$1|S|3|M|5|6|7|1L|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|U|$V|$5|W|X|Y|A|$Z|10]]|11|$5|W|X|Y|A|$Z|12]]]]

<p>You could get the date as a sortable string and sort it. </p>

<p>For getting more than one array sorted by one signature array, you could take sorting with map, where you sort an array of indices, indicating the final sorting and then reassign all arrays with this sorting.</p>

<p>The <code>getD</code> function returns a formatted string by taking an index of <code>array0</code> for sorting. Inside of the function the string is destructed into month and year parts, and replaced by its <a href="https://en.wikipedia.org/wiki/ISO_8601" rel="noreferrer"><strong>ISO 8601</strong></a> representation. The replacement function callback takes the matched items, returns an array with a formatted year and the month of an object with the month names and the related month numbers. Then, this array is joined and returned.</p>

<p>Sorting takes place with a comparison with <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/localeCompare" rel="noreferrer"><code>String#localeCompare</code></a>.</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var array1 = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'],
    array2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'],
    array3 = ['55', '2', '3', '0', '21', '132'],
    indices = Object
        .keys(array1)
        .sort(function (a, b) {
            function getD(i) {
                var months = { Jan: '01', Feb: '02', Mar: '03', Apr: '04', May: '05', Jun: '06', Jul: '07', Aug: '08', Sep: '09', Oct: '10', Nov: '11', Dec: '12' },
                    s = array1[i];
                return s.replace(/^(...)(.+)$/, (_, m, y) =&gt; [y.padStart(4, '0'), months[m]].join('-'));
            }
            return getD(a).localeCompare(getD(b));
        }),
    result = [array1, array3].map(a =&gt; indices.map(i =&gt; a[i]));
  
result.forEach(a =&gt; console.log(...a));</code></pre>
</div>
</div>
</p>


blocks|key|1313331|text|要正确地订购，你需要知道月份的顺序。这不是按字母顺序排列的，所以您可以使用一个按月份顺序排列的数组，然后查找它们。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1313332|monthOrder+=“一月”、“二月”、“三月”、“四月”、“五月”、“六月”、“七月”、“八月”、“九月”、“十月”、“十一月”、“十二月”|1313333|已经正确排序，可以与.indexOf()一起使用以获得一个月的职位。|1313334|​|1313335|const+myArr+=++['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const+myArr2+=+['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
+++,'Jan19_value'];

const+monthOrder+=+['Jan',+'Feb',+'Mar',+'Apr',+'May',+'Jun',+'Jul',+'Aug',+'Sep',+'Oct',+'Nov',+'Dec']

let+sortYearMonth+=+(a,+b)+=>+{
++++let+monthA+=+monthOrder.indexOf(a.slice(0,3))
	let+yearA+=+a.slice(3,6)
	let+monthB+=+monthOrder.indexOf(b.slice(0,3))
	let+yearB+=+b.slice(3,6)
	return+(`${yearA}-${monthA}`+<+`${yearB}-${monthB}`)+?+-1+:+(`${yearA}-${monthA}`+>+`${yearB}-${monthB}`)+?+1+:+0
}

let+sortedMonths+=+myArr.sort(sortYearMonth)
let+sortedMonths2+=+myArr2.sort(sortYearMonth)

console.log(sortedMonths+)
console.log(sortedMonths2+)|code-block|syntax|javascript|1313336|1313337|更新:相同位置的值|offset|length|style|BOLD|1313338|更新后的版本将两个数组链接在一起，然后对第一个数组进行排序，同时保持相对位置与第二个数组的相对位置。|1313339|思想:将两个数组与一个临时对象连接起来，然后使用Object.entries提取键/值对。然后根据对的第一个值(即array1的值)对数组进行排序。然后，它以正确的顺序返回键/值对，您可以使用.map()将这些值再次提取到两个数组中。|CODE|1313340|我使用基于字符串的示例和下面的实际值示例添加了一个run|1313341|const+myArr+=+['Mar19',+'Apr18',+'Jun18',+'Jul18',+'May18',+'Jan19'];
const+myArr2+=+['Mar19_value',+'Apr18_value',+'Jun18_value',+'Jul18_value',+'May18_value',+'Jan19_value'];

const+myArr3+=+['55','2','3','0','21','132'];

const+monthOrder+=+['Jan',+'Feb',+'Mar',+'Apr',+'May',+'Jun',+'Jul',+'Aug',+'Sep',+'Oct',+'Nov',+'Dec']

let+sortYearMonth+=+(a,+b)+=>+{
++let+monthA+=+monthOrder.indexOf(a.slice(0,+3))
++let+yearA+=+a.slice(3,+6)
++let+monthB+=+monthOrder.indexOf(b.slice(0,+3))
++let+yearB+=+b.slice(3,+6)
++return+(`${yearA}-${monthA}`+<+`${yearB}-${monthB}`)+?+-1+:+(`${yearA}-${monthA}`+>+`${yearB}-${monthB}`)+?+1+:+0
}

function+sortByFirst(myArr,+myArr2)+{
++let+keyValue+=+myArr.reduce((links,+item,+i)+=>+{
++++links[item]+=+myArr2[i];
++++return+links
++},+{})
++let+entries+=+Object.entries(keyValue)
++return+entries.sort((a,+b)+=>+sortYearMonth(a[0],+b[0]))
}

let+sortedEntries+=+sortByFirst(myArr,+myArr2)
let+sortedMonths+=+sortedEntries.map(i+=>+i[0])
let+sortedValues+=+sortedEntries.map(i+=>+i[1])

let+sortedEntries2+=+sortByFirst(myArr,+myArr3)
let+sortedMonths2+=+sortedEntries2.map(i+=>+i[0])
let+sortedValues2+=+sortedEntries2.map(i+=>+i[1])

console.log(sortedMonths)
console.log(sortedValues)

console.log(sortedMonths2)
console.log(sortedValues2)|1313342|1313343|entityMap^0|0|0|0|0|0|0|0|9|0|0|O|E|2O|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|17|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|18|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|19|8|@]|9|@]|A|$K|L]]|$1|M|3|G|5|6|7|1A|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1B|8|@$P|1C|Q|1D|R|S]]|9|@]|A|$]]|$1|T|3|U|5|6|7|1E|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1F|8|@$P|1G|Q|1H|R|X]|$P|1I|Q|1J|R|X]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1K|8|@]|9|@]|A|$]]|$1|10|3|11|5|J|7|1L|8|@]|9|@]|A|$K|L]]|$1|12|3|G|5|6|7|1M|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|14|$]]

<p>To order properly you need to know the order of the months. This is not alphabetical so you can use an array with the order of the months then look them up.</p>

<p>const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']</p>

<p>Is already properly sorted and can be used in conjunction with .indexOf() to get the position of a month.</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const myArr =  ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const myArr2 = ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
   ,'Jan19_value'];

const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']

let sortYearMonth = (a, b) =&gt; {
    let monthA = monthOrder.indexOf(a.slice(0,3))
	let yearA = a.slice(3,6)
	let monthB = monthOrder.indexOf(b.slice(0,3))
	let yearB = b.slice(3,6)
	return (`${yearA}-${monthA}` &lt; `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` &gt; `${yearB}-${monthB}`) ? 1 : 0
}

let sortedMonths = myArr.sort(sortYearMonth)
let sortedMonths2 = myArr2.sort(sortYearMonth)

console.log(sortedMonths )
console.log(sortedMonths2 )</code></pre>
</div>
</div>
</p>

<p><strong>Update: values in same position</strong></p>

<p>Updated version links the two arrays together then sorts the first while keeping the relative position to the second.</p>

<p>Idea: Link two arrays with a temporary Object then extract the key/value pair using <code>Object.entries</code>. Then sorting the array based on the first value of the pair, which is the value of array1. Then it returns the key/value pair in the right order you can extract the values into two arrays again by using <code>.map()</code> </p>

<p>I added a run with the string based example and the real values examples below
<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const myArr = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
const myArr2 = ['Mar19_value', 'Apr18_value', 'Jun18_value', 'Jul18_value', 'May18_value', 'Jan19_value'];

const myArr3 = ['55','2','3','0','21','132'];

const monthOrder = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']

let sortYearMonth = (a, b) =&gt; {
  let monthA = monthOrder.indexOf(a.slice(0, 3))
  let yearA = a.slice(3, 6)
  let monthB = monthOrder.indexOf(b.slice(0, 3))
  let yearB = b.slice(3, 6)
  return (`${yearA}-${monthA}` &lt; `${yearB}-${monthB}`) ? -1 : (`${yearA}-${monthA}` &gt; `${yearB}-${monthB}`) ? 1 : 0
}

function sortByFirst(myArr, myArr2) {
  let keyValue = myArr.reduce((links, item, i) =&gt; {
    links[item] = myArr2[i];
    return links
  }, {})
  let entries = Object.entries(keyValue)
  return entries.sort((a, b) =&gt; sortYearMonth(a[0], b[0]))
}

let sortedEntries = sortByFirst(myArr, myArr2)
let sortedMonths = sortedEntries.map(i =&gt; i[0])
let sortedValues = sortedEntries.map(i =&gt; i[1])

let sortedEntries2 = sortByFirst(myArr, myArr3)
let sortedMonths2 = sortedEntries2.map(i =&gt; i[0])
let sortedValues2 = sortedEntries2.map(i =&gt; i[1])

console.log(sortedMonths)
console.log(sortedValues)

console.log(sortedMonths2)
console.log(sortedValues2)</code></pre>
</div>
</div>
</p>


blocks|key|2501501|text|let+mth+=+['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];

mth.sort((itemA,+itemB)=>{
+++let+mthAstr+=+itemA.slice(0,3)%2B"+01+"%2BitemA.slice(3,5);
+++let+mthA+=+new+Date(mthAstr);
+++let+mthBstr+=+itemB.slice(0,3)%2B"+01+"%2BitemB.slice(3,5);
+++let+mthB+=+new+Date(mthBstr);
+++if+(mthA+<+mthB)
++++++return+-1;
+++if+(mthA+>+mthB)
++++++return+1;
+++return+0;
});|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2501502|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>let mth = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];

mth.sort((itemA, itemB)=&gt;{
   let mthAstr = itemA.slice(0,3)+" 01 "+itemA.slice(3,5);
   let mthA = new Date(mthAstr);
   let mthBstr = itemB.slice(0,3)+" 01 "+itemB.slice(3,5);
   let mthB = new Date(mthBstr);
   if (mthA &lt; mthB)
      return -1;
   if (mthA &gt; mthB)
      return 1;
   return 0;
});
</code></pre>


blocks|key|2501512|text|你可以试试这个原始片段。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2501513|​|2501514|var+MY+=+['Mar19',+'Apr18',+'Jun18',+'Jul18',+'May18',+'Jan19'];
var+s+=+"JanFebMarAprMayJunJulAugSepOctNovDec";
//+converting+to+raw+date+format
for+(i+in+MY)+{
++num+=+MY[i].match(/\d%2B/g);
++letr+=+MY[i].match(/[a-zA-Z]%2B/g);
++MY[i]+=+(s.indexOf(letr)+/+3+%2B+1)+%2B+'-'+%2B+'20'+%2B+num;
}
//+sorting+logic
var+sorted+=+MY.sort(function(a,+b)+{
++a+=+a.split("-");
++b+=+b.split("-")
++return+new+Date(a[1],+a[0],+1)+-+new+Date(b[1],+b[0],+1)
});
//+converting+back+to+original+array+after+sorting
res+=+[];
for+(i+in+sorted)+{
++var+a+=+sorted[i].split('-');
++res[i]+=+(s.substr((parseInt(a[0])+-+1)+*+3,+3))+%2B+'-'+%2B+a[1].substr(2,+2);
}

console.log(res);|code-block|syntax|javascript|2501515|2501516|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>You can try this raw snippet.</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var MY = ['Mar19', 'Apr18', 'Jun18', 'Jul18', 'May18', 'Jan19'];
var s = "JanFebMarAprMayJunJulAugSepOctNovDec";
// converting to raw date format
for (i in MY) {
  num = MY[i].match(/\d+/g);
  letr = MY[i].match(/[a-zA-Z]+/g);
  MY[i] = (s.indexOf(letr) / 3 + 1) + '-' + '20' + num;
}
// sorting logic
var sorted = MY.sort(function(a, b) {
  a = a.split("-");
  b = b.split("-")
  return new Date(a[1], a[0], 1) - new Date(b[1], b[0], 1)
});
// converting back to original array after sorting
res = [];
for (i in sorted) {
  var a = sorted[i].split('-');
  res[i] = (s.substr((parseInt(a[0]) - 1) * 3, 3)) + '-' + a[1].substr(2, 2);
}

console.log(res);</code></pre>
</div>
</div>
</p>


blocks|key|242916|text|你可以试试这样的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|242917|var+Array+=++['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
Array.sort(function(a,+b)+{
++++a+=+[a.slice(0,3),+'+20',+a.slice(3)].join('');
++++b+=+[b.slice(0,3),+'+20',+b.slice(3)].join('')
++++return+new+Date()+-+new+Date(b);
});

console.log(Array);|code-block|syntax|javascript|242918|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<h2>You can try something like that :</h2>

<pre><code>var Array =  ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
Array.sort(function(a, b) {
    a = [a.slice(0,3), ' 20', a.slice(3)].join('');
    b = [b.slice(0,3), ' 20', b.slice(3)].join('')
    return new Date() - new Date(b);
});

console.log(Array);
</code></pre>


blocks|key|2501545|text|我认为你的数据结构是不幸的。使用共享索引比组合它们的数据结构(如对象数组[{label:+'Mar19',+value:+55},+...]+)困难得多。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2501546|如果您的数据来自您无法控制的上游解决方案，您仍然可以在您自己的工作中管理它，在使用之前进行转换。(如果你真的要把它转换成传递给别人的话)|2501547|将两个数组组合在一起的函数的一个常见名称是zip+--想想它的作用就像拉链一样。这个版本使用一个函数来说明如何组合成对的元素。(在其他地方，这样的函数可能称为zipWith。)|2501548|这里，sortArraysByDate调用zip传递一个函数，该函数将'Mar19'和55转换为{label:+'Mar19,+value:+55,+month:+'Mar',+year:+19}，使用dateFields从该标签中提取月份和年份，然后使用直截了当的dateSort对它们进行排序。|2501549|​|2501550|const+months+=+{Jan:+1,+Feb:+2,+Mar:+3,+Apr:+4,++May:+5,++Jun:+6,
++++++++++++++++Jul:+7,+Aug:+8,+Sep:+9,+Oct:+10,+Nov:+11,+Dec:+12}
++++++++++++++++
const+dateFields+=+(d)+=>+({
++month:+d.slice(0,+3),
++year:+Number(d.slice(3))
})

const+dateSort+=+({month:+m1,+year:+y1},+{month:+m2,+year:+y2})+=>
++(y1+-+y2)+%7C%7C+(months[m1]+-+months[m2])

const+zip+=+(fn,+a1,+a2)+=>+a1.map((a,+i)+=>+fn(a,+a2[i]))

const+sortArraysByDate+=+(a1,+a2)+=>
++zip((label,+value)+=>+({label,+value,+...dateFields(label)}),+a1,+a2)
++++.sort(dateSort)
++++.map(({label,+value})+=>+({label,+value}))


const+Array1+=+['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const+Array2+=+++['55','2','3','0','21','132'];+//real+values

const+result+=+sortArraysByDate(Array1,+Array2)

console.log(result)|code-block|syntax|javascript|2501551|2501552|map调用在sortArraysByDate中很可能是不必要的。没有它，生成的数据包括额外的year和month字段；这可能不是一个问题。|2501553|如果确实需要以原始格式更新的数组，只需map结果：|2501554|const+newArray1+=+result.map(o+=>+o.label)
const+newArray2+=+result.map(o+=>+o.value)|2501555|但我劝你不要这么做除非是绝对必要的。这个结构真的很有用。成对数组的情况要少得多。|2501556|此外，如果需要组合两个以上的数组，则可以编写稍微复杂一些的zip版本。|2501557|const+zip+=+(fn,+...as)+=>+as[0].map((_,+i)+=>+fn(...as.map(a+=>+a[i])))|2501558|这将在n参数和n数组上获得一个函数，并生成一个新数组，其中包含对每个数组中的连续项分别调用该函数的结果。|2501559|entityMap|0|LINK|mutability|MUTABLE|url|https://ramdajs.com/docs/#zipWith^0|10|Y|0|0|L|3|27|7|27|7|0|0|3|G|L|3|Z|7|17|2|1C|1E|2T|A|3P|8|0|0|0|0|0|3|6|G|1A|4|1F|5|0|J|3|0|0|0|T|3|0|0|3|1|7|1|0^^$0|@$1|2|3|4|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1I|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1J|8|@$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]]|D|@$9|1O|A|1P|1|1Q]]|E|$]]|$1|J|3|K|5|6|7|1R|8|@$9|1S|A|1T|B|C]|$9|1U|A|1V|B|C]|$9|1W|A|1X|B|C]|$9|1Y|A|1Z|B|C]|$9|20|A|21|B|C]|$9|22|A|23|B|C]|$9|24|A|25|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|26|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|27|8|@]|D|@]|E|$Q|R]]|$1|S|3|M|5|6|7|28|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|29|8|@$9|2A|A|2B|B|C]|$9|2C|A|2D|B|C]|$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|2I|8|@$9|2J|A|2K|B|C]]|D|@]|E|$]]|$1|X|3|Y|5|P|7|2L|8|@]|D|@]|E|$Q|R]]|$1|Z|3|10|5|6|7|2M|8|@]|D|@]|E|$]]|$1|11|3|12|5|6|7|2N|8|@$9|2O|A|2P|B|C]]|D|@]|E|$]]|$1|13|3|14|5|P|7|2Q|8|@]|D|@]|E|$Q|R]]|$1|15|3|16|5|6|7|2R|8|@$9|2S|A|2T|B|C]|$9|2U|A|2V|B|C]]|D|@]|E|$]]|$1|17|3|-4|5|6|7|2W|8|@]|D|@]|E|$]]]|18|$19|$5|1A|1B|1C|E|$1D|1E]]]]

<p>I would suggest that your data structure is unfortunate.  Working with shared indices makes it much more difficult than a data structure that combines them, such as an array of objects, <code>[{label: 'Mar19', value: 55}, ...]</code>.</p>

<p>If your data is coming from an upstream solution you cannot control, you can still manage this in your own work, converting before you use it.  (And if you really have to converting back to pass to others.)</p>

<p>A common name for a function combining two arrays is <code>zip</code> -- think of it acting like a zipper.  This version uses one that takes a function to say how the paired elements should be combined.  (Elsewhere such a function might be called <a href="https://ramdajs.com/docs/#zipWith" rel="nofollow noreferrer"><code>zipWith</code></a>.)</p>

<p>Here <code>sortArraysByDate</code> calls <code>zip</code> passing a function that turns <code>'Mar19'</code> and <code>55</code> into <code>{label: 'Mar19, value: 55, month: 'Mar', year: 19}</code> using <code>dateFields</code> to extract the month and year from that label, then sorts these using the straightforward  <code>dateSort</code></p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const months = {Jan: 1, Feb: 2, Mar: 3, Apr: 4,  May: 5,  Jun: 6,
                Jul: 7, Aug: 8, Sep: 9, Oct: 10, Nov: 11, Dec: 12}
                
const dateFields = (d) =&gt; ({
  month: d.slice(0, 3),
  year: Number(d.slice(3))
})

const dateSort = ({month: m1, year: y1}, {month: m2, year: y2}) =&gt;
  (y1 - y2) || (months[m1] - months[m2])

const zip = (fn, a1, a2) =&gt; a1.map((a, i) =&gt; fn(a, a2[i]))

const sortArraysByDate = (a1, a2) =&gt;
  zip((label, value) =&gt; ({label, value, ...dateFields(label)}), a1, a2)
    .sort(dateSort)
    .map(({label, value}) =&gt; ({label, value}))


const Array1 = ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'];
const Array2 =   ['55','2','3','0','21','132']; //real values

const result = sortArraysByDate(Array1, Array2)

console.log(result)</code></pre>
</div>
</div>
</p>

<p>The <code>map</code> call in <code>sortArraysByDate</code> is quite possibly not necessary.  Without it, the resulting data includes extra <code>year</code> and <code>month</code> fields; that might not be an issue.</p>

<p>If you really need those updated arrays in the original format, you can just <code>map</code> the result:</p>

<pre><code>const newArray1 = result.map(o =&gt; o.label)
const newArray2 = result.map(o =&gt; o.value)
</code></pre>

<p>But I would urge you not to do this unless it's absolutely necessary.  This structure is really useful.  The paired arrays are much less so.</p>

<p>Also, if you need to combine more than two arrays, you could write a slightly more sophisticated version of <code>zip</code>:</p>

<pre><code>const zip = (fn, ...as) =&gt; as[0].map((_, i) =&gt; fn(...as.map(a =&gt; a[i])))
</code></pre>

<p>This takes a function on <code>n</code> arguments, and <code>n</code> arrays, and yields a new array containing the result of calling that function respectively on the successive items in each of the arrays.</p>


<p>I need help with sorting an array by month and year to display on chart respectively.</p>

<pre><code>Array1: ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'....];
Desired Output : [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
</code></pre>

<p>I also have another array with values for each month according to the array above respectively</p>

<pre><code>Array2: ['Mar19_value','Apr18_value','Jun18_value','Jul18_value','May18_value'
       ,'Jan19_value'....];

Array2: ['55','2','3','0','21','132'....]; //real values
</code></pre>

<p>When the <code>monthyear</code> array sorts, I want the data in this array to move to a new position according to the <code>monthyear</code> position. Like this:</p>

<pre><code>Desired Array1: [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];
Desired Array2: [....'Apr18_value','May18_value','Jun18_value','Jul18_value','Jan19_value','Mar19_value'];
</code></pre>

<p>So I can pick the data later like this:</p>

<pre><code>var label = array[4];
var value = array2[4];
</code></pre>

<p>How can this be accomplished?</p>


Sort array by month and year

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我需要帮助排序的数组按月和年分别显示在图表上。Array1: ['Mar19','Apr18','Jun18','Jul18','May18','Jan19'....];Desired Output : [....'Apr18','May18','Jun18','Jul18','Jan19','Mar19'];我也有另...

问按月和年排序数组
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