如何获得包含字符串和数字的数组中的平均数

Question

我正在编写一个函数来通过数组包含数字和字符串的测试。

这里是测试用例average-numner.test.js:

var average = require("./get-average");

test("Average", function() {
  var numbers = [4, "-", 8, 11, "hello", "57", 0, 2];
  var expected = 5;

  var output = average(numbers);

  expect(output).toEqual(expected);
});

这是我写的函数average-number.js：

// the input is an array of numbers and strings
// return the average of all the numbers
// be sure to exclude the strings
function averageNumbers(str) {
  let sum = 0;
  let newArr = [];
  var filtered = str.filter(function(item) {
    return parseInt(item);
  });
  return filtered;
}

console.log(averageNumbers([4, "-", 8, 11, "hello", "57", 0, 2]));

//    module.exports = averageNumbers;

我得到了只有数字的新数组，但我的filtered 数组中仍然有字符串"57"。我怎样才能摆脱它并找到总和以获得平均数？

Answer 1

function averageNumbers(arr) {
  var filter = arr.filter(item => typeof item === 'number');
  var sum = filter.reduce((acc, n) => acc + n, 0);
  return sum / filter.length;
}

console.log(averageNumbers([4, "-", 8, 11, "hello", "57", 0, 2]));

//    module.exports = averageNumbers;

这给了你预期的5

Answer 2

映射和过滤

注意我把变量名 str 改成了 arr 因为它不是字符串。

我测试类型以及数字是否是有限的并添加了其余的计算

代码现在返回预期的5

// the input is an array of numbers and strings
// return the average of all the numbers
// be sure to exclude the strings
function averageNumbers(arr) {
  const filtered = arr.filter(item => typeof item === 'number' && isFinite(item))
  const sum = filtered.reduce((a,b) => a+b)
  return +parseFloat(sum/filtered.length).toPrecision(12); // this can be ugly
}

console.log(averageNumbers([4, "-", 8, 11, "hello", "57", 0, 2]));
console.log(averageNumbers([3, "-", 4, 11, "hello", "57", 0, 2, 3]));

只是一个reduce返回一个对象来查看子部分

// the input is an array of numbers and strings
// return the average of all the numbers
// be sure to exclude the strings
const averageNumbers = arr => {
  const averageObject = arr.reduce((acc,item,i) => {
    if (typeof item === 'number' && isFinite(item)) {
      acc.sum+=item;
      acc.nofItems++;
    }  
    if (i===arr.length-1) acc.average = acc.sum/acc.nofItems;
    return acc;
  },{sum:0,nofItems:0, average:0})
  return averageObject;
}  

const ave1 = averageNumbers([4, "-", 8, 11, "hello", "57", 0, 2]);
console.log(`Average of ${ave1.nofItems} numeric items was ${ave1.average}. The sum was ${ave1.sum}`) 
const ave2 = averageNumbers([3, "-", 4, 11, "hello", "57", 0, 2, 3]);
console.log(`Average of ${ave2.nofItems} numeric items was ${ave2.average}. The sum was ${ave2.sum}`) 

Answer 3

filter 只检查给定元素的条件是否为真，如果是则保留它。但它保留原始值，而不是谓词函数的输出。如果您只想对数字求和，这应该可以：

var filtered = str.filter(function (item) {
    return parseInt(item);
  }).map(item => parseInt(item));

但是，如果一个数字被解析为0，它将被认为是虚假的，过滤器会丢弃它。计算总和没关系，因为0 没有区别，但如果你想保留它，你可以这样做：

const filtered = str.map(item => parseInt(item))
    .filter(item => item !== NaN);

这将尝试解析所有值，并仅保留成功解析的值，包括现有整数。

Answer 4

您可以使用reduce 和内部回调使用typeof 来检查元素是否为非数字。累加器对象由总计、计数和平均值三个键组成。

function averageNumbers(c) {
  return c.reduce(function(a, b) {
    "number" === typeof b && (a.total += b, a.count += 1, a.average = a.total / a.count);
    return a;
  }, {
    total: 0,
    count: 0,
    average: 0
  }).average;
};

console.log(averageNumbers([4, '-', 8, 11, 'hello', '57', 0, 2]));