Question

我正在尝试计算数组的mode，但是我想排除0

这是我的代码：

const数据= [0，0，0，4，4，2，3，2，0];

function mode(numbers) {
  var modes = [],
    count = [],
    i,
    number,
    maxIndex = 0;

  for (i = 0; i < numbers.length; i += 1) {
    number = numbers[i];
    count[number] = (count[number] || 0) + 1;
    if (count[number] > maxIndex) {
      maxIndex = count[number];
    }
  }

  for (i in count)
    if (count.hasOwnProperty(i)) {
      if (count[i] === maxIndex) {
        modes.push(Number(i));
      }
    }

  return modes;
}

mode(datas); // output : [0] and I want [4] [2]

感谢您的时间。

Answer 1

您可以简单地filter清除零：

datas = [0, 0, 0, 4, 4, 2, 3, 2, 0];

function mode(numbers) {
  // we don't want to consider zeros
  // so filter them out
  numbers = numbers.filter(function(n) { return n !== 0 }); 

  var modes = [],
    count = [],
    i, number, maxIndex = 0;

  for (i = 0; i < numbers.length; i += 1) {
    number = numbers[i];
    count[number] = (count[number] || 0) + 1;
    if (count[number] > maxIndex) {
      maxIndex = count[number];
    }
  }

  for (i in count)
    if (count.hasOwnProperty(i)) {
      if (count[i] === maxIndex) {
        modes.push(Number(i));
      }
    }

  return modes;
}

console.log(mode(datas)) // output : [4] [2]

如果您使用的是ES6，则可以使用箭头函数语法：

numbers = numbers.filter(n => n !== 0);

Answer 2

我只是想分享一种无mode / for循环的forEach计算方法。

相关文档：

const counted = [0, 0, 0, 4, 4, 2, 3, 2, 0]
.filter(element => element)
.reduce(
  (accumulator, currentValue) => 
    ({
      ...accumulator,
      [currentValue]: (accumulator[currentValue] || 0) + 1
    }),
  {}
);

const maxCount = Math.max(...Object.values(counted));

const mode = Object.keys(counted).filter(key => counted[key] === maxCount);

console.log(mode);

我如何计算数组的模式并忽略数据

2 个答案: