在Javascript中，我试图采取数字值的初始数组，并计算其中的元素。理想情况下，结果将是两个新数组，第一个数组指定每个唯一元素，第二个数组包含每个元素出现的次数。但是，我愿意听取关于输出格式的建议。
例如，如果初始数组是:
5, 5, 5, 2, 2, 2, 2, 2, 9, 4
然后将创建两个新数组。第一个将包含每个唯一元素的名称:
5, 2, 9, 4
第二个将包含该元素在初始数组中出现的次数:
3, 5, 1, 1
因为数字5在初始数组中出现了三次，数字2出现了五次，9和4都出现了一次。
我一直在寻找解决方案，但似乎没有一个可行，而且我自己尝试过的每件事最后都出奇地复杂。任何帮助都将不胜感激!
谢谢:)
当前回答
function countOcurrences(arr){
return arr.reduce((aggregator, value, index, array) => {
if(!aggregator[value]){
return aggregator = {...aggregator, [value]: 1};
}else{
return aggregator = {...aggregator, [value]:++aggregator[value]};
}
}, {})
}
其他回答
function countOcurrences(arr){
return arr.reduce((aggregator, value, index, array) => {
if(!aggregator[value]){
return aggregator = {...aggregator, [value]: 1};
}else{
return aggregator = {...aggregator, [value]:++aggregator[value]};
}
}, {})
}
你可以使用一个对象来保存结果:
Const arr = [5,5,5,2,2,2,2,2,2,2,2,9,4]; Const counts = {}; (const num of arr) { Counts [num] = Counts [num] ?计数[num] + 1: 1; ｝ console.log(重要); Console.log(计数[5]，计数[2]，计数[9]，计数[4]);
所以，现在你的counts对象可以告诉你一个特定数字的计数是多少:
console.log(counts[5]); // logs '3'
如果您想获取成员数组，只需使用keys()函数即可
keys(counts); // returns ["5", "2", "9", "4"]
使用减法和波浪号(~)操作符的较短版本。
Const data = [2,2,2,2,2,2,4,5,5,5,9]; 函数频率(nums) { num返回。Reduce ((acc, curr) => { Acc [curr] = -~ Acc [curr]; 返回acc; }, {}); ｝ console.log(频率(数据));
试试这个:
Array.prototype.getItemCount = function(item) {
var counts = {};
for(var i = 0; i< this.length; i++) {
var num = this[i];
counts[num] = counts[num] ? counts[num]+1 : 1;
}
return counts[item] || 0;
}
下面是一种计算对象数组中出现次数的方法。它还将第一个数组的内容放在一个新数组中，以便对值进行排序，这样原始数组中的顺序就不会被打乱。然后使用递归函数遍历每个元素并计算数组中每个对象的quantity属性。
var big_array = [
{ name: "Pineapples", quantity: 3 },
{ name: "Pineapples", quantity: 1 },
{ name: "Bananas", quantity: 1 },
{ name: "Limes", quantity: 1 },
{ name: "Bananas", quantity: 1 },
{ name: "Pineapples", quantity: 2 },
{ name: "Pineapples", quantity: 1 },
{ name: "Bananas", quantity: 1 },
{ name: "Bananas", quantity: 1 },
{ name: "Bananas", quantity: 5 },
{ name: "Coconuts", quantity: 1 },
{ name: "Lemons", quantity: 2 },
{ name: "Oranges", quantity: 1 },
{ name: "Lemons", quantity: 1 },
{ name: "Limes", quantity: 1 },
{ name: "Grapefruit", quantity: 1 },
{ name: "Coconuts", quantity: 5 },
{ name: "Oranges", quantity: 6 }
];
function countThem() {
var names_array = [];
for (var i = 0; i < big_array.length; i++) {
names_array.push( Object.assign({}, big_array[i]) );
}
function outerHolder(item_array) {
if (item_array.length > 0) {
var occurrences = [];
var counter = 0;
var bgarlen = item_array.length;
item_array.sort(function(a, b) { return (a.name > b.name) ? 1 : ((b.name > a.name) ? -1 : 0); });
function recursiveCounter() {
occurrences.push(item_array[0]);
item_array.splice(0, 1);
var last_occurrence_element = occurrences.length - 1;
var last_occurrence_entry = occurrences[last_occurrence_element].name;
var occur_counter = 0;
var quantity_counter = 0;
for (var i = 0; i < occurrences.length; i++) {
if (occurrences[i].name === last_occurrence_entry) {
occur_counter = occur_counter + 1;
if (occur_counter === 1) {
quantity_counter = occurrences[i].quantity;
} else {
quantity_counter = quantity_counter + occurrences[i].quantity;
}
}
}
if (occur_counter > 1) {
var current_match = occurrences.length - 2;
occurrences[current_match].quantity = quantity_counter;
occurrences.splice(last_occurrence_element, 1);
}
counter = counter + 1;
if (counter < bgarlen) {
recursiveCounter();
}
}
recursiveCounter();
return occurrences;
}
}
alert(JSON.stringify(outerHolder(names_array)));
}
