javascript - JavaScript 函数报告的平均值不准确

Question

我有一个 average() 方法来计算两个值之间的平均值。平均值相差了小数点的“毛发”。

const measurements = [ 
  { timestamp: '2015-09-01T16:00:00.000Z',
    temperature: 27.1,
    dewPoint: 16.9 
  },
  { timestamp: '2015-09-01T16:10:00.000Z',
    temperature: 27.3,
    dewPoint: 0 
  },
  { timestamp: '2015-09-01T16:20:00.000Z',
    temperature: 27.5,
    dewPoint: 17.1 
   },
   { timestamp: '2015-09-01T16:30:00.000Z',
    temperature: 27.4,
    dewPoint: 17.3 
   },
   { timestamp: '2015-09-01T16:40:00.000Z',
    temperature: 27.2,
    dewPoint: 0 
   },
   { timestamp: '2015-09-01T17:00:00.000Z',
    temperature: 28.1,
    dewPoint: 18.3 
   } 
] 

为了简洁起见，我不会在这里分享 60 行代码:

假设:

• 我有一个 query() 方法来获取给定时间戳的范围。在我的示例中from 2015-09-01T16:00:00.000Z 至 2015-09-01T17:00:00.000Z

• 另一种方法循环遍历对象并输出min和ma​​x具有指定指标的值。我的指标是 dewPoint，上述数组的 min 和 ma​​x 值为 16.9 和 18.3分别是强>。

• 最后(见下文)一种获取最小值和最大值值之间平均值的方法

因此这是上面解释的结果:

// POSTMAN result

[
    {
        "metric": "dewPoint",
        "stat": "min",
        "value": 16.9
    },
    {
        "metric": "dewPoint",
        "stat": "max",
        "value": 18.3
    },
    {
        "metric": "dewPoint",
        "stat": "average",
        "value": 17.4
    }
]

我想获得最大值和最小值之间的平均值。即 16.9 和 18.3 应该是 17.6，但是，我得到的是 17.4

这是存在实际错误的一种方法。

function averageMetric(measurements, metric) {

  // => metric dewPoint
  // => measurements = the data array in example

  let value = 0
  let measurementsWithMetric = 0
  measurements.forEach(measurement => {
    if (measurement[metric]) {
      value += measurement[metric]
      measurementsWithMetric++
    }
  })
  //=> value = 69.6
  //=> measurementsWithMetric = 4
  const average = value / measurementsWithMetric // Is this the issue?
  // average =  17.4

  return isNaN(average) ? null : Math.round(average * 100) / 100
}

您能否帮我理解这里的问题，并建议与上述方法等效的 ES6 解决方案？

Answer 1

您的函数计算算术平均值，而不是中间范围。这是一个计算函数:

const measurements = [ 
  {temperature: 27.1, dewPoint: 16.9},
  {temperature: 27.3, dewPoint: 0}, // Isn't this the minimum value though?
  {temperature: 27.5, dewPoint: 17.1},
  {temperature: 27.4, dewPoint: 17.3},
  {temperature: 27.2, dewPoint: 0},
  {temperature: 28.1, dewPoint: 18.3}, 
  {temperature: 28.2} 
];
function averageMetric(meas, metr) {
  const valid = meas.filter(e => e[metr]);
  const min = Math.min(...valid.map(e => e[metr]));
  const max = Math.max(...valid.map(e => e[metr]));
  return (min + max) / 2;
}

console.log(averageMetric(measurements, 'dewPoint'));