目前，我有一个这样的数组：
var uniqueCount = Array();
经过几个步骤，我的数组看起来像这样：
uniqueCount = [a,b,c,d,d,e,a,b,c,f,g,h,h,h,e,a];
如何计算阵列中有多少a，b，c？我希望得到一个结果：
a = 3
b = 1
c = 2
d = 2
等
答案 0 :(得分：222)
var counts = {};
your_array.forEach(function(x) { counts[x] = (counts[x] || 0)+1; });
答案 1 :(得分：45)
这样的事情：
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);
如果您不希望在旧版浏览器中使用它，请使用简单的for循环而不是forEach。
答案 2 :(得分：25)
我偶然发现了这个（非常古老的）问题。有趣的是，最明显和最优雅的解决方案（imho）缺失：Array.prototype.reduce(...)。自2011年左右（IE）或更早（所有其他）以来，所有主流浏览器都支持此功能：
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}
答案 3 :(得分：15)
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');
}
}
答案 4 :(得分：8)
基于减少数组函数的单行
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));
答案 5 :(得分：6)
没有人对此似乎使用内置的Map()，这往往是我与Array.prototype.reduce()结合使用的目的：
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polyfill Map()。
答案 6 :(得分：4)
我认为这是如何计算数组中具有相同值的事件的最简单方法。
var a = [true, false, false, false];
a.filter(function(value){
return value === false;
}).length
答案 7 :(得分：4)
你可以在不使用任何for / while循环的情况下解决它。
function myCounter(inputWords) {
return inputWords.reduce( (countWords, word) => {
countWords[word] = ++countWords[word] || 1;
return countWords;
}, {});
}
希望它可以帮到你！
答案 8 :(得分：3)
您可以拥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]);
}
答案 9 :(得分：3)
你可以这样做：
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;
}
}
现在你有一张包含所有字符数的地图
答案 10 :(得分：2)
const obj = {};
const uniqueCount = [ 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a' ];
for (let i of uniqueCount) obj[i] ? obj[i]++ : (obj[i] = 1);
console.log(obj);
答案 11 :(得分：2)
简单更好，一个变量，一个函数：）
const counts = arr.reduce((acc, value) => ({
...acc,
[value]: (acc[value] || 0) + 1
}), {});
答案 12 :(得分：1)
包含字母的数组中的重复：
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：
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>");
}
答案 13 :(得分：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);
答案 14 :(得分：1)
import pickle, zlib
def store_obj(file_, obj):
compressed = zlib.compress(pickle.dumps(obj, protocol=-1), level=9)
file_.write(len(compressed).to_bytes(4, "little"))
file_.write(compressed)
def get_obj(file_):
obj_size = int.from_bytes(file_.read(4), "little")
if obj_size == 0:
return None
data = zlib.decompress(self.file_.read(obj_size))
return pickle.loads(data)
答案 15 :(得分：0)
好的答案组合：
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]);
}
}
希望它有用。
答案 16 :(得分：0)
通过使用array.map，我们可以减少循环，请参阅jsfiddle
eruptions = [3.6, 1.8, 3.333, 2.283, 4.533, 2.883, 4.7, 3.6, 1.95, 4.35, 1.833, 3.917, 4.2, 1.75, 4.7, 2.167, 1.75, 4.8, 1.6, 4.25, 1.8, 1.75, 3.45, 3.067, 4.533, 3.6, 1.967, 4.083, 3.85, 4.433, 4.3, 4.467, 3.367, 4.033, 3.833, 2.017, 1.867, 4.833, 1.833, 4.783, 4.35, 1.883, 4.567, 1.75, 4.533, 3.317, 3.833, 2.1, 4.633, 2.0, 4.8, 4.716, 1.833, 4.833, 1.733, 4.883, 3.717, 1.667, 4.567, 4.317, 2.233, 4.5, 1.75, 4.8, 1.817, 4.4, 4.167, 4.7, 2.067, 4.7, 4.033, 1.967, 4.5, 4.0, 1.983, 5.067, 2.017, 4.567, 3.883, 3.6, 4.133, 4.333, 4.1, 2.633, 4.067, 4.933, 3.95, 4.517, 2.167, 4.0, 2.2, 4.333, 1.867, 4.817, 1.833, 4.3, 4.667, 3.75, 1.867, 4.9, 2.483, 4.367, 2.1, 4.5, 4.05, 1.867, 4.7, 1.783, 4.85, 3.683, 4.733, 2.3, 4.9, 4.417, 1.7, 4.633, 2.317, 4.6, 1.817, 4.417, 2.617, 4.067, 4.25, 1.967, 4.6, 3.767, 1.917, 4.5, 2.267, 4.65, 1.867, 4.167, 2.8, 4.333, 1.833, 4.383, 1.883, 4.933, 2.033, 3.733, 4.233, 2.233, 4.533, 4.817, 4.333, 1.983, 4.633, 2.017, 5.1, 1.8, 5.033, 4.0, 2.4, 4.6, 3.567, 4.0, 4.5, 4.083, 1.8, 3.967, 2.2, 4.15, 2.0, 3.833, 3.5, 4.583, 2.367, 5.0, 1.933, 4.617, 1.917, 2.083, 4.583, 3.333, 4.167, 4.333, 4.5, 2.417, 4.0, 4.167, 1.883, 4.583, 4.25, 3.767, 2.033, 4.433, 4.083, 1.833, 4.417, 2.183, 4.8, 1.833, 4.8, 4.1, 3.966, 4.233, 3.5, 4.366, 2.25, 4.667, 2.1, 4.35, 4.133, 1.867, 4.6, 1.783, 4.367, 3.85, 1.933, 4.5, 2.383, 4.7, 1.867, 3.833, 3.417, 4.233, 2.4, 4.8, 2.0, 4.15, 1.867, 4.267, 1.75, 4.483, 4.0, 4.117, 4.083, 4.267, 3.917, 4.55, 4.083, 2.417, 4.183, 2.217, 4.45, 1.883, 1.85, 4.283, 3.95, 2.333, 4.15, 2.35, 4.933, 2.9, 4.583, 3.833, 2.083, 4.367, 2.133, 4.35, 2.2, 4.45, 3.567, 4.5, 4.15, 3.817, 3.917, 4.45, 2.0, 4.283, 4.767, 4.533, 1.85, 4.25, 1.983, 2.25, 4.75, 4.117, 2.15, 4.417, 1.817, 4.467]
waiting = [79, 54, 74, 62, 85, 55, 88, 85, 51, 85, 54, 84, 78, 47, 83, 52, 62, 84, 52, 79, 51, 47, 78, 69, 74, 83, 55, 76, 78, 79, 73, 77, 66, 80, 74, 52, 48, 80, 59, 90, 80, 58, 84, 58, 73, 83, 64, 53, 82, 59, 75, 90, 54, 80, 54, 83, 71, 64, 77, 81, 59, 84, 48, 82, 60, 92, 78, 78, 65, 73, 82, 56, 79, 71, 62, 76, 60, 78, 76, 83, 75, 82, 70, 65, 73, 88, 76, 80, 48, 86, 60, 90, 50, 78, 63, 72, 84, 75, 51, 82, 62, 88, 49, 83, 81, 47, 84, 52, 86, 81, 75, 59, 89, 79, 59, 81, 50, 85, 59, 87, 53, 69, 77, 56, 88, 81, 45, 82, 55, 90, 45, 83, 56, 89, 46, 82, 51, 86, 53, 79, 81, 60, 82, 77, 76, 59, 80, 49, 96, 53, 77, 77, 65, 81, 71, 70, 81, 93, 53, 89, 45, 86, 58, 78, 66, 76, 63, 88, 52, 93, 49, 57, 77, 68, 81, 81, 73, 50, 85, 74, 55, 77, 83, 83, 51, 78, 84, 46, 83, 55, 81, 57, 76, 84, 77, 81, 87, 77, 51, 78, 60, 82, 91, 53, 78, 46, 77, 84, 49, 83, 71, 80, 49, 75, 64, 76, 53, 94, 55, 76, 50, 82, 54, 75, 78, 79, 78, 78, 70, 79, 70, 54, 86, 50, 90, 54, 54, 77, 79, 64, 75, 47, 86, 63, 85, 82, 57, 82, 67, 74, 54, 83, 73, 73, 88, 80, 71, 83, 56, 79, 78, 84, 58, 83, 43, 60, 75, 81, 46, 90, 46, 74]
df = pd.DataFrame({'eruptions': eruptions, 'waiting': waiting})
s = sorted([str(round(i * 10, 1)) for i in df['eruptions']])
s = (
pd.Series(s)
.groupby([str(int(float(i) // 2 * 2)) for i in s])
.apply(lambda group: "".join((x[-1] for x in group)))
.reset_index()
)
print("\n".join(s['index'] +' | ' + s[0]))
# Output
# 16 | 070355555588
# 18 | 000022233333335577777777888822335777888
# 20 | 00002223378800035778
# 22 | 0002335578023578
# 24 | 00228
# 26 | 23
# 28 | 080
# 30 | 7
# 32 | 2337
# 34 | 250077
# 36 | 0000823577
# 38 | 2333335582225577
# 40 | 0000003357788888002233555577778
# 42 | 03335555778800233333555577778
# 44 | 02222335557780000000023333357778888
# 46 | 0000233357700000023578
# 48 | 00000022335800333
# 50 | 0370
测试：
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;
}
答案 17 :(得分：0)
示例如何返回字符串中最常用的字符。
function maxChar(str) {
const charMap = {};
let maxCharacter = '';
let maxNumber = 0;
for (let item of str) {
charMap[item] = charMap[item] + 1 || 1;
}
for (let char in charMap) {
if (charMap[char] > maxNumber) {
maxNumber = charMap[char];
maxCharacter = char;
}
}
return maxCharacter;
}
console.log(maxChar('abcccccccd'))答案 18 :(得分：0)
计算字符串中提供的字母
function countTheElements(){
var str = "ssdefrcgfrdesxfdrgs";
var arr = [];
var count = 0;
for(var i=0;i<str.length;i++){
arr.push(str[i]);
}
arr.sort();
for(var i=0;i<arr.length;i++){
if(arr[i] == arr[i-1]){
count++;
}else{
count = 1;
}
if(arr[i] != arr[i+1]){
console.log(arr[i] +": "+(count));
}
}
}
countTheElements()
答案 19 :(得分：0)
let arr=[1,2,3,3,4,5,5,6,7,7]
let obj={}
for(var i=0;i<arr.length;i++){
obj[arr[i]]=obj[arr[i]]!=null ?obj[arr[i]]+1:1 //stores duplicate in an obj
}
console.log(obj)
//returns object {1:1,:1,3:2,.....}
答案 20 :(得分：0)
用法：
wrap.common.getUniqueDataCount(, columnName);
代码：
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;
}
代码段
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;
}
答案 21 :(得分：0)
// 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]);
答案 22 :(得分：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");
}
}
答案 23 :(得分：0)
声明一个对象arr来保存唯一的键集。通过使用map遍历数组来填充arr。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。
给出testArray：
var testArray = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
解决方案：
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
答案 24 :(得分：0)
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);
答案 25 :(得分：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;
}
答案 26 :(得分：0)
最快的方法：
Сomputational complexity是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']);
答案 27 :(得分：0)
// 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));
答案 28 :(得分：0)
例如创建一个文件demo.js，并在节点为demo.js的控制台中运行该文件，您将发现矩阵形式的元素。
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如下结果：
[ 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 ] ]
答案 29 :(得分：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
*/
答案 30 :(得分：-1)
在JavaScript中使用数组化简方法很简单：
var arr = ['a','d','r','a','a','f','d'];
console.log(arr.reduce((json,val)=>({...json,[val]:(json[val]|0)+1}),{}))
//{ a:3,d:2,r:1,f:1 }