Question

我应该以更简单的方式询问，而不是展示我一直想做的事情。假设我有这个数组：

function scr(array, n){

}
scr([187,187,187,187,187,62,62,43,43,43,43,43,5,5,5], 5)

我想得到一个数组，显示出现一定次数的数字。在这种情况下，我希望出现5次的数字，然后给出：

[187,43]

我该怎么办？

PS：对于复杂的上一个问题抱歉

Answer 1

您可以使用reduce对数组中的项目进行分组，为每个数字计算一个数字，然后按计数为5的数字对其进行过滤：

const input = [187,187,187,187,187,62,62,43,43,43,43,43,5,5,5];
const count = input.reduce((a, num) => {
  a[num] = (a[num] || 0) + 1;
  return a;
}, {});
const fives = Object.entries(count)
  .filter(([, occurrences]) => occurrences === 5)
  .map(([num]) => Number(num));
console.log(fives);

Answer 2

不是性能最高但可读性最强，它是减少

的好地方

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce

＆＃13;

const numbers = [187, 187, 187, 187, 187, 62, 62, 43, 43, 43, 43, 43, 5, 5, 5];

function equalFive(array, num) {
   return array.filter(i => i === num).length === 5;
}

function getNumbersThatAppearFiveTimes(array) {
  return array.reduce((prev, curr) => {
    if (!prev.includes(curr) && equalFive(array, curr)) {
      return [...prev, curr]
    }
    return prev;
  }, []);
}
console.log(getNumbersThatAppearFiveTimes(numbers))

＆＃13;

Answer 3

这是一个使用lodash的解决方案，它已经内置了许多这样的功能，因此您的代码库可以更清晰。

const inputArray = [187,187,187,187,187,62,62,43,43,43,43,43,5,5,5];
const inputValue = 5;

const count = _.countBy(inputArray); // count each instance of values
const equality = _.pickBy(count, x => x === inputValue); // filter values with 5 appearances
const output = Object.keys(equality).map(Number); // convert to desired format

console.log(output); // [43, 187]

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.js"></script>

Answer 4

＆＃13;

var input_array = [187, 187, 187, 187, 187, 62, 62, 43, 43, 43, 43, 43, 5, 5, 5];
//Sort input array
input_array.sort();

var output_array = [];
var count = 0;
var current = input_array[0];
for(var i = 0; i < input_array.length; i++){
  //Loop until change to next number
  //Check if total quantity equals 5
  if(input_array[i] != current){
    //save into output if equals 5
    if(count == 5){
      output_array.push(current);
    }
    //reset counter
    current = input_array[i];
    count = 1;
  }else{
    count++;
  }
}

console.log(output_array);

＆＃13;

Answer 5

使用Array＃reduce（）将计数映射到Map对象，然后使用Array＃filter（）返回结果

function scr(array, n) {
  const counts = array.reduce((a, c) => a.set(c, (a.get(c) || 0) + 1), new Map);
  return [...counts.keys()].filter(v => counts.get(v) >= n);
}


const input = [187, 187, 187, 187, 187, 62, 62, 43, 43, 43, 43, 43, 5, 5, 5];

console.log(scr(input, 5))

Answer 6

您可以使用计数器并使用单个循环过滤 nth 计数元素。

＆＃13;

function scr(array, n) {
    var count = Object.create(null);
    return array.filter(v => (count[v] = (count[v] || 0) + 1) === n);
}

console.log(scr([187, 187, 187, 187, 187, 62, 62, 43, 43, 43, 43, 43, 5, 5, 5], 5));

＆＃13;

如何获取在数组中出现一定次数的数字

6 个答案: