如何计算JavaScript中数组中的重复值


97

目前,我得到了这样的数组:

var uniqueCount = Array();

几个步骤后,我的数组如下所示:

uniqueCount = [a,b,c,d,d,e,a,b,c,f,g,h,h,h,e,a];

我如何计算数组中有多少a,b,c?我想要一个类似的结果:

a = 3
b = 1
c = 2
d = 2

等等



@Nirk我想,musical_coder的意思是像{},而不是函数式编程的那样map
马特·鲍尔

Answers:


28

function count() {
    array_elements = ["a", "b", "c", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];

    array_elements.sort();

    var current = null;
    var cnt = 0;
    for (var i = 0; i < array_elements.length; i++) {
        if (array_elements[i] != current) {
            if (cnt > 0) {
                document.write(current + ' comes --> ' + cnt + ' times<br>');
            }
            current = array_elements[i];
            cnt = 1;
        } else {
            cnt++;
        }
    }
    if (cnt > 0) {
        document.write(current + ' comes --> ' + cnt + ' times');
    }

}

count();

演示小提琴

您也可以使用高阶函数进行操作。 看到这个答案


1
循环后不需要多余的if语句...只需使用for (var i = 0; i <= array_elements.length; i++) {<=代替<
EmmaGamma 2015年

嗨@Vinay,也许您可​​以在这里帮我吗?stackoverflow.com/questions/57819850/…–
SMPLYJR

318
var counts = {};
your_array.forEach(function(x) { counts[x] = (counts[x] || 0)+1; });

9
这绝对是最简单的答案
Josh Beam

3
(counts [x] || 0)+1这怎么给count?
jsduniya

5
@SidBhalke:表达式counts[x] || 0返回值counts[x]如果设置了,否则返回0。然后只需添加一个,然后再次在对象中进行设置即可完成计数。
Constantinius

1
@SheetJS,如果您想知道为什么投票失败-是我;我在移动设备上浏览时,没有实际注意到点击了按钮。一旦发现,现在就来不及了。对此表示歉意,答案确实很好。如果您要编辑它,我很乐意撤消。
Todor Minakov '19

4
另外还有reducevar counts = your_array.reduce((map, val) => {map[val] = (map[val] || 0)+1; return map}, {} );
Alberto89

70

像这样:

uniqueCount = ["a","b","c","d","d","e","a","b","c","f","g","h","h","h","e","a"];
var count = {};
uniqueCount.forEach(function(i) { count[i] = (count[i]||0) + 1;});
console.log(count);

如果您不希望在较旧的浏览器中出现这种情况,请使用简单的for循环而不是forEach。


4
@web_dev他创建了一个名为count的关联数组对象,该数组对象将为数组中的每个唯一元素都有一个键值对,其中键是唯一元素值,而值是count。他遍历数组,并为每个值增加值或创建键值对(不存在的键的值求值为undefined,因此||或运算符取零,然后加1)
robisrob

@neelmeg也许为“ forEach”编写所有参数有助于更好地理解(“ i”是每个数组值,而不是índex):uniqueCount.forEach(function(value, index) { count[value] = (count[value] || 0) + 1; });
Pedro Ferreira

37

我偶然发现了这个(非常古老的)问题。有趣的是,缺少了最明显,最优雅的解决方案(imho):Array.prototype.reduce(...)。自2011年(IE)或更早版本(所有其他版本)开始,所有主流浏览器均支持此功能:

var arr = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = arr.reduce(function(prev, cur) {
  prev[cur] = (prev[cur] || 0) + 1;
  return prev;
}, {});

// map is an associative array mapping the elements to their frequency:
document.write(JSON.stringify(map));
// prints {"a": 3, "b": 2, "c": 2, "d": 2, "e": 2, "f": 1, "g": 1, "h": 3}


10

基于减少数组功能的单行

const uniqueCount =  ["a", "b", "c", "d", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];
const distribution = uniqueCount.reduce((acum,cur) => Object.assign(acum,{[cur]: (acum[cur] | 0)+1}),{});
console.log(JSON.stringify(distribution,null,2));


我刚刚意识到@ isnot2bad(stackoverflow.com/a/32886673/621058)与我的差不多。我只是碰巧使用了



5

// Initial array
let array = ['a', 'b', 'c', 'd', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a'];

// Unique array without duplicates ['a', 'b', ... , 'h']
let unique = [...new Set(array)];

// This array counts duplicates [['a', 3], ['b', 2], ... , ['h', 3]] 
let duplicates = unique.map(value => [value, array.filter(str => str === value).length]);

5

似乎没有人响应使用Map()内置功能,这往往是我的首选Array.prototype.reduce()

const data = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
const result = data.reduce((a, c) => a.set(c, (a.get(c) || 0) + 1), new Map());
console.log(...result);

注意,如果要在较旧的浏览器中使用它,则必须填充Map()


您能否深入解释一下它是如何工作的?(特别是set / get部分)。我试图将reducer分解为一个函数,但是我得到的“ get”不是一个响应函数。
Antoine Nedelec

好的getset功能来自Map对象。但是初始累加器不是Map对象,那么为什么reducer的简化版本需要一个?
Antoine Nedelec

@AntoineNedelec初始值一个新Map对象;参见reduce的第二个参数。Map.prototype.set返回地图对象,并Map.prototype.get返回undefined或提供给它的任何键的值。这使我们可以获取每个字母的当前计数(0如果未定义),然后将其递增一,然后将该字母的计数设置为新计数,这将返回映射并成为新的累加器值。
分发

4

您可以有一个包含计数的对象。遍历列表并增加每个元素的计数:

var counts = {};

uniqueCount.forEach(function(element) {
  counts[element] = (counts[element] || 0) + 1;
});

for (var element in counts) {
  console.log(element + ' = ' + counts[element]);
} 

你为什么要设置这个条件 counts[element] || 0
AskMen

4

您可以解决它,而无需使用forEach的任何for / while循环。

function myCounter(inputWords) {        
    return inputWords.reduce( (countWords, word) => {
        countWords[word] = ++countWords[word] || 1;
        return countWords;
    }, {});
}

希望对您有帮助!


4

// new example.
var str= [20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5];

function findOdd(para) {
  var count = {};
  para.forEach(function(para) {
  count[para] = (count[para] || 0) + 1;
  });
  return count;
}

console.log(findOdd(str));


3

您可以执行以下操作:

uniqueCount = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = new Object();

for(var i = 0; i < uniqueCount.length; i++) {
 if(map[uniqueCount[i]] != null) {
    map[uniqueCount[i]] += 1;
} else {
    map[uniqueCount[i]] = 1;
    }
}

现在您有了一张包含所有字符的地图


1
var uniqueCount = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
// here we will collect only unique items from the array
var uniqueChars = [];

// iterate through each item of uniqueCount
for (i of uniqueCount) {
// if this is an item that was not earlier in uniqueCount, 
// put it into the uniqueChars array
  if (uniqueChars.indexOf(i) == -1) {
    uniqueChars.push(i);
  } 
}
// after iterating through all uniqueCount take each item in uniqueChars
// and compare it with each item in uniqueCount. If this uniqueChars item 
// corresponds to an item in uniqueCount, increase letterAccumulator by one.
for (x of uniqueChars) {
  let letterAccumulator = 0;
  for (i of uniqueCount) {
    if (i == x) {letterAccumulator++;}
  }
  console.log(`${x} = ${letterAccumulator}`);
}

感谢您对其进行更新,这对那些开始的人有很大帮助。
定期的乔

1

包含字母的数组中的重复项:

var arr = ["a", "b", "a", "z", "e", "a", "b", "f", "d", "f"],
  sortedArr = [],
  count = 1;

sortedArr = arr.sort();

for (var i = 0; i < sortedArr.length; i = i + count) {
  count = 1;
  for (var j = i + 1; j < sortedArr.length; j++) {
    if (sortedArr[i] === sortedArr[j])
      count++;
  }
  document.write(sortedArr[i] + " = " + count + "<br>");
}

包含数字的数组中的重复项:

var arr = [2, 1, 3, 2, 8, 9, 1, 3, 1, 1, 1, 2, 24, 25, 67, 10, 54, 2, 1, 9, 8, 1],
  sortedArr = [],
  count = 1;
sortedArr = arr.sort(function(a, b) {
  return a - b
});
for (var i = 0; i < sortedArr.length; i = i + count) {
  count = 1;
  for (var j = i + 1; j < sortedArr.length; j++) {
    if (sortedArr[i] === sortedArr[j])
      count++;
  }
  document.write(sortedArr[i] + " = " + count + "<br>");
}


1

var testArray = ['a','b','c','d','d','e','a','b','c','f','g','h ','h','h','e','a'];

var newArr = [];
testArray.forEach((item) => {
    newArr[item] = testArray.filter((el) => {
            return el === item;
    }).length;
})
console.log(newArr);

1
uniqueCount = ["a","b","a","c","b","a","d","b","c","f","g","h","h","h","e","a"];
var count = {};
uniqueCount.forEach((i) => { count[i] = ++count[i]|| 1});
console.log(count);

1

简化的sheet.js软件

var counts = {};
var aarr=['a','b','a'];
aarr.forEach(x=>counts[x]=(counts[x] || 0)+1 );
console.log(counts)


0

好答案的组合:

var count = {};
var arr = ['a', 'b', 'c', 'd', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a'];
var iterator = function (element) {
    count[element] = (count[element] || 0) + 1;
}

if (arr.forEach) {
    arr.forEach(function (element) {
        iterator(element);
    });
} else {
    for (var i = 0; i < arr.length; i++) {
        iterator(arr[i]);
    }
}  

希望对您有所帮助。


0
public class CalculateCount {
public static void main(String[] args) {
    int a[] = {1,2,1,1,5,4,3,2,2,1,4,4,5,3,4,5,4};
    Arrays.sort(a);
    int count=1;
    int i;
    for(i=0;i<a.length-1;i++){
        if(a[i]!=a[i+1]){
            System.out.println("The Number "+a[i]+" appears "+count+" times");
            count=1;                
        }
        else{
            count++;
        }
    }
    System.out.println("The Number "+a[i]+" appears "+count+" times");

}   

}


您能否围绕此添加一些上下文?

0

通过使用array.map我们可以减少循环,请参见jsfiddle

function Check(){
    var arr = Array.prototype.slice.call(arguments);
    var result = [];
    for(i=0; i< arr.length; i++){
        var duplicate = 0;
        var val = arr[i];
        arr.map(function(x){
            if(val === x) duplicate++;
        })
        result.push(duplicate>= 2);
    }
    return result;
}

去测试:

var test = new Check(1,2,1,4,1);
console.log(test);

0

var string = ['a','a','b','c','c','c','c','c','a','a','a'];

function stringCompress(string){

var obj = {},str = "";
string.forEach(function(i) { 
  obj[i] = (obj[i]||0) + 1;
});

for(var key in obj){
  str += (key+obj[key]);
}
  console.log(obj);
  console.log(str);
}stringCompress(string)

/*
Always open to improvement ,please share 
*/


0

例如demo.js,创建一个文件并在带有node的控制台中运行它demo.js,您将以矩阵形式出现元素。

var multipleDuplicateArr = Array(10).fill(0).map(()=>{return Math.floor(Math.random() * Math.floor(9))});
console.log(multipleDuplicateArr);

var resultArr = Array(Array('KEYS','OCCURRENCE'));

for (var i = 0; i < multipleDuplicateArr.length; i++) {
  var flag = true;
  for (var j = 0; j < resultArr.length; j++) {
     if(resultArr[j][0] == multipleDuplicateArr[i]){
       resultArr[j][1] = resultArr[j][1] + 1;
       flag = false;
      }
  }
  if(flag){
    resultArr.push(Array(multipleDuplicateArr[i],1));
  }
}

console.log(resultArr);

您将在控制台中获得以下结果:

[ 1, 4, 5, 2, 6, 8, 7, 5, 0, 5 ] . // multipleDuplicateArr
[ [ 'KEYS', 'OCCURENCE' ],        // resultArr
  [ 1, 1 ],
  [ 4, 1 ],
  [ 5, 3 ],
  [ 2, 1 ],
  [ 6, 1 ],
  [ 8, 1 ],
  [ 7, 1 ],
  [ 0, 1 ] ]

0

最快的方法:

计算复杂度为O(n)。

function howMuchIsRepeated_es5(arr) {
	const count = {};
	for (let i = 0; i < arr.length; i++) {
		const val = arr[i];
		if (val in count) {
			count[val] = count[val] + 1;
		} else {
			count[val] = 1;
		}
	}

	for (let key in count) {
		console.log("Value " + key + " is repeated " + count[key] + " times");
	}
}

howMuchIsRepeated_es5(['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a']);

最短的代码:

使用ES6。

function howMuchIsRepeated_es6(arr) {
	// count is [ [valX, count], [valY, count], [valZ, count]... ];
	const count = [...new Set(arr)].map(val => [val, arr.join("").split(val).length - 1]);

	for (let i = 0; i < count.length; i++) {
		console.log(`Value ${count[i][0]} is repeated ${count[i][1]} times`);
	}
}

howMuchIsRepeated_es6(['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a']);


0
var arr = ['a','d','r','a','a','f','d'];  

//call function and pass your array, function will return an object with array values as keys and their count as the key values.
duplicatesArr(arr);

function duplicatesArr(arr){
    var obj = {}
    for(var i = 0; i < arr.length; i++){
        obj[arr[i]] = [];
        for(var x = 0; x < arr.length; x++){
            (arr[i] == arr[x]) ? obj[arr[i]].push(x) : '';
        }
        obj[arr[i]] = obj[arr[i]].length;
    }

    console.log(obj);
    return obj;
}

0

声明一个对象arr以将唯一集作为键。arr通过使用map遍历数组来填充。如果先前未找到密钥,则添加密钥并分配零值。在每次迭代中,增加密钥的值。

给定testArray:

var testArray = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];

解:

var arr = {};
testArray.map(x=>{ if(typeof(arr[x])=="undefined") arr[x]=0; arr[x]++;});

JSON.stringify(arr) 将输出

{"a":3,"b":2,"c":2,"d":2,"e":2,"f":1,"g":1,"h":3}

Object.keys(arr) 将返回 ["a","b","c","d","e","f","g","h"]

查找任何项目的出现,例如b arr['b']将输出2


请不要只发布代码作为答案,还请提供解释,说明代码的作用以及如何解决问题。附有解释的答案通常质量较高,并且更有可能吸引投票。
Mark Rotteveel

0

用法:

wrap.common.getUniqueDataCount(, columnName);

码:

function getUniqueDataCount(objArr, propName) {
        var data = [];
        objArr.forEach(function (d, index) {
            if (d[propName]) {
                data.push(d[propName]);
            }
        });

        var uniqueList = [...new Set(data)];

        var dataSet = {};
        for (var i=0; i < uniqueList.length; i++) {
            dataSet[uniqueList[i]] = data.filter(x => x == uniqueList[i]).length;
        }
        
        return dataSet;
    }

片段

var data= [
          {a:'you',b:'b',c:'c',d:'c'},
          {a: 'you', b: 'b', c: 'c', d:'c'},
          {a: 'them', b: 'b', c: 'c', d:'c'},
          {a: 'them', b: 'b', c: 'c', d:'c'},
          {a: 'okay', b: 'b', c: 'c', d:'c'},
          {a: 'okay', b: 'b', c: 'c', d:'c'},
          ];
          
  console.log(getUniqueDataCount(data, 'a'));       
  
  function getUniqueDataCount(objArr, propName) {
        var data = [];
        objArr.forEach(function (d, index) {
            if (d[propName]) {
                data.push(d[propName]);
            }
        });

        var uniqueList = [...new Set(data)];

        var dataSet = {};
        for (var i=0; i < uniqueList.length; i++) {
            dataSet[uniqueList[i]] = data.filter(x => x == uniqueList[i]).length;
        }

        return dataSet;
    }


-1

在javascript中,使用数组reduce方法很简单:

const arr = ['a','d','r','a','a','f','d'];
const result =  arr.reduce((json,val)=>({...json, [val]:(json[val] | 0) + 1}),{});
console.log(result)
//{ a:3,d:2,r:1,f:1 }

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