Question

我以为我已经完成了我的功能，但事实证明它不能按我预期的那样工作。应该计算一个数字数组的标准偏差值。

function average (arr) {
  let result = 0
  for (let counter = 0; counter < arr.length; counter++) {
    result += arr[counter]
  }
  const averageValue = result / arr.length
  return averageValue
}

function standardDeviation (arr) {
  if (arr.length === 1) {
    return 0
  } else {
  const getAverage = arr.map(x => average(arr) - x)
  const squaredArr = getAverage.map(i => i * i)
  let sum = squaredArr.reduce(function (i, j) {
    return i + j
  })
  const dividedArr = sum / (arr.length - 1)
  return Math.sqrt(dividedArr)
 }
}

如果我传递数组[4、2、6、1、3、7、5、3]，则我的函数应返回1.8998，但返回2.031 .....为什么？以及如何解决该问题？

Answer 1

不需要average函数，可以通过以下几行轻松更改它：

  // instead of
  const getAverage = arr.map(x => average(arr) - x)
  // calculate average 
  let arrSum = arr.reduce(function (i, j) {
    return i + j
  });
  let getAverage = arr.map(x => arrSum/arr.length - x);

  ...

  // also this was changed
  const dividedArr = sum / (arr.length - 1)
  // with this 
  let dividedArr = sumsQuaredArr / arr.length;

您可以检查以下示例。控制台中提供了部分输出，可轻松检查功能的每个步骤。

编辑：您的示例数组已添加用于测试。

const arr = [30,28,45,33,34,10,13,19];
const arr2 = [4, 2, 6, 1, 3, 7, 5, 3];
function standardDeviation (arr) {
  if (arr.length === 1) {
    return 0;
  } 
  console.log("InitialArray ",arr);
  let arrSum = arr.reduce(function (i, j) {
    return i + j
  });
  console.log("Average ",arrSum/arr.length);
  let getAverage = arr.map(x => arrSum/arr.length - x);
  console.log("getAverage ",getAverage);
  let squaredArr = getAverage.map(i => i * i);
  console.log("squaredArr ",squaredArr );
  let sumsQuaredArr = squaredArr.reduce(function (i, j) {
    return i + j
  });
  console.log("sumsQuaredArr ",sumsQuaredArr );
  let dividedArr = sumsQuaredArr / arr.length;
  console.log("dividedArr ",dividedArr );
  return Math.sqrt(dividedArr);
}
console.log("standardDeviation ",standardDeviation (arr));
console.log("standardDeviation ",standardDeviation (arr2));

找到标准偏差函数

1 个答案: