Javascript 匹配数组中的3个或更多相同元素并将其添加到列表中

我试图在数组中找到3个或更多匹配项，但它只匹配前3个，而对数组的其余部分不匹配。如果有人能帮忙，那就太好了：

var网格=[2,2,2,5,5,3,3,3]； 检查结果； 函数检查结果{ var list_matches=[]；//存储找到的所有匹配项 var listcurrent=[]；//存储当前值 var maxitems=3； var last=-1；//最后一个单元格 forlet j=0；j=maxitems{ list_matches.pushlistcurrent； } //重置为空 last=-1； listcurrent=[]； } } console.loglist_匹配； console.logCols:+grid.length； } 预期结果：从[2,2,2,5,5,5,3,3,3]

0:222

1:555

2:3333

电流输出为：

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0:222就这样

你可以这样做：

var grid = [ 1, 1, 2, 3, 4, 5 ];
var hashMap = {};


for( var i = 0; i < grid.length; i++ ) {


  if( hashMap.hasOwnProperty( grid[i] ) ) {
    hashMap[ grid[i] ]++;
  } else {
    hashMap[ grid[i] ] = 1;
  }

}

---

您可以这样做：

var grid = [ 1, 1, 2, 3, 4, 5 ];
var hashMap = {};


for( var i = 0; i < grid.length; i++ ) {


  if( hashMap.hasOwnProperty( grid[i] ) ) {
    hashMap[ grid[i] ]++;
  } else {
    hashMap[ grid[i] ] = 1;
  }

}

---

您可以使用一个临时数组来收集相同的值，并在长度达到所需的最小长度时推送此数组

函数getMorearray，最小值{ var结果=[]， 临时雇员 array.forEachv，i，a=>{ 如果v！==a[i-1]返回温度=[v]； 温度pushv； 如果temp.length==min result.pushtemp； }; 返回结果； }

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loggetMore[2,2,2,5,5,5,3,3,3,3]，3 您可以使用一个临时数组来收集相同的值，并在长度达到所需的最小长度时推送此数组

函数getMorearray，最小值{ var结果=[]， 临时雇员 array.forEachv，i，a=>{ 如果v！==a[i-1]返回温度=[v]； 温度pushv； 如果temp.length==min result.pushtemp； }; 返回结果； } loggetMore[2,2,2,5,5,5,3,3,3,3]，3 它会帮助你的

//toLowerCase for get unique on words, not on character. if i ignore this, it will return two words=> developers. and developers
//first Split and join for remove '.' character form text and finally last split is for convert string to an Array of words that splited by Space character


let uniqWords = Array.from(new Set(text));
//using Set to get unique words and convert it to Array to do more on Array.

let count = {};
// declare varriable for store counts of words.

uniqWords.map(item => {
    count[item] = text.filter(elem => {
        //create property with the name of words and filter common words to an Array 
        return elem == item
    }).length
    //get Array length for get words repeated count.
})

console.table(count)
//log Result into Console 

它会帮助你的

//toLowerCase for get unique on words, not on character. if i ignore this, it will return two words=> developers. and developers
//first Split and join for remove '.' character form text and finally last split is for convert string to an Array of words that splited by Space character


let uniqWords = Array.from(new Set(text));
//using Set to get unique words and convert it to Array to do more on Array.

let count = {};
// declare varriable for store counts of words.

uniqWords.map(item => {
    count[item] = text.filter(elem => {
        //create property with the name of words and filter common words to an Array 
        return elem == item
    }).length
    //get Array length for get words repeated count.
})

console.table(count)
//log Result into Console 

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另一个使用Array.prototype[reduce/map/filter]的解决方案

常量someArray=[2,2,2,5,5,5,3,3,3,9]； console.logaggregatesomeArray； 函数聚合器{ 返回arr //检索唯一值 .reduceacc，val=>！acc.includeval和acc.concatval | | acc，[] //使用唯一值将arr值映射到字符串 //如果匹配数>=3 .mapval=>{ 常量过滤=arr.filterv=>v==val； 返回filtered.length>2？filtered.join:false } //过滤非虚假值 .filterval=>val；

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} 另一个使用Array.prototype[reduce/map/filter]的解决方案

常量someArray=[2,2,2,5,5,5,3,3,3,9]； console.logaggregatesomeArray； 函数聚合器{ 返回arr //检索唯一值 .reduceacc，val=>！acc.includeval和acc.concatval | | acc，[] //使用唯一值将arr值映射到字符串 //如果匹配数>=3 .mapval=>{ 常量过滤=arr.filterv=>v==val； 返回filtered.length>2？filtered.join:false } //过滤非虚假值 .filterval=>val； }