Learn Advanced Array Methods by Building a Statistics Calculator - Step 27

Telling me to:

Use the .forEach() method to loop through the array . In the callback, use the el parameter to access the counts object and increment the count for each number.

Your code so far

const getMode = (array) => {
  const counts = {};
  array.forEach(el => {

    counts[el] = counts[el] ? counts[el] + 1 : 1;
  });
}
<!-- file: index.html -->
<!DOCTYPE html>
<html lang="en">
  <head>
    <meta charset="utf-8" />
    <meta name="viewport" content="width=device-width, initial-scale=1" />
    <link rel="stylesheet" href="./styles.css" />
    <script src="./script.js"></script>
    <title>Statistics Calculator</title>
  </head>
  <body>
    <h1>Statistics Calculator</h1>
    <p>Enter a list of comma-separated numbers.</p>
    <form onsubmit="calculate(); return false;">
      <label for="numbers">Numbers:</label>
      <input type="text" name="numbers" id="numbers" />
      <button type="submit">Calculate</button>
    </form>
    <div class="results">
      <p>
        The <dfn>mean</dfn> of a list of numbers is the average, calculated by
        taking the sum of all numbers and dividing that by the count of numbers.
      </p>
      <p class="bold">Mean: <span id="mean"></span></p>
      <p>
        The <dfn>median</dfn> of a list of numbers is the number that appears in
        the middle of the list, when sorted from least to greatest.
      </p>
      <p class="bold">Median: <span id="median"></span></p>
      <p>
        The <dfn>mode</dfn> of a list of numbers is the number that appears most
        often in the list.
      </p>
      <p class="bold">Mode: <span id="mode"></span></p>
      <p>
        The <dfn>range</dfn> of a list of numbers is the difference between the
        largest and smallest numbers in the list.
      </p>
      <p class="bold">Range: <span id="range"></span></p>
      <p>
        The <dfn>variance</dfn> of a list of numbers measures how far the values
        are from the mean, on average.
      </p>
      <p class="bold">Variance: <span id="variance"></span></p>
      <p>
        The <dfn>standard deviation</dfn> of a list of numbers is the square
        root of the variance.
      </p>
      <p class="bold">
        Standard Deviation: <span id="standardDeviation"></span>
      </p>
    </div>
  </body>
</html>
/* file: styles.css */
body {
  margin: 0;
  background-color: rgb(27, 27, 50);
  text-align: center;
  color: #fff;
}

button {
  cursor: pointer;
  background-color: rgb(59, 59, 79);
  border: 3px solid white;
  color: white;
}

input {
  background-color: rgb(10, 10, 35);
  color: white;
  border: 1px solid rgb(59, 59, 79);
}

.bold {
  font-weight: bold;
}
/* file: script.js */
const getMean = (array) => array.reduce((acc, el) => acc + el, 0) / array.length;

const getMedian = (array) => {
  const sorted = array.sort((a, b) => a - b);
  const median =
    array.length % 2 === 0
      ? getMean([sorted[array.length / 2], sorted[array.length / 2 - 1]])
      : sorted[Math.floor(array.length / 2)];
  return median;
}


// User Editable Region

const getMode = (array) => {
  const counts = {};
  array.forEach(el => {

    counts[el] = counts[el] ? counts[el] + 1 : 1;
  });
}

// User Editable Region



const calculate = () => {
  const value = document.querySelector("#numbers").value;
  const array = value.split(/,\s*/g);
  const numbers = array.map(el => Number(el)).filter(el => !isNaN(el));
  
  const mean = getMean(numbers);
  const median = getMedian(numbers);

  document.querySelector("#mean").textContent = mean;
  document.querySelector("#median").textContent = median;
}

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/121.0.0.0 Safari/537.36

Challenge Information:

Learn Advanced Array Methods by Building a Statistics Calculator - Step 27

the tests want you to use the OR operator to check that count[el] exist

This topic was automatically closed 182 days after the last reply. New replies are no longer allowed.