Determine the mean of each group of n elements within a given array

Question

Determine the mean of each group of n elements within a given array

I am looking for assistance with a coding challenge involving an array of numbers. The goal is to calculate the average of every 3 elements in the array and store these averages in a new array.

Below is the code I currently have:


var total = 0;
//Array containing the numbers
for(i=0; i < arr.length; i++) {
    total += arr[i];
}

var avg = total / arr.length;
avgArray.push(Math.round(avg));

However, this code only gives me the average of all the elements in the array. I need help modifying it so that it calculates the average of every set of 3 elements instead.

For example, the avgArray should display: - Number 1: Average of first 3 elements - Number 2: Average of the second 3 elements - Number 3: Average of the third 3 elements ... Can you please provide guidance on how to achieve this?

javascript arrays

Answer 1

Answer №1

When approaching this problem from a functional perspective, rather than procedural, one possible solution is:

function calculateAverage(arr, n) {
  if (!arr || !n){
    return false;
  }

  let groups = [];

  while (arr.length) {
    groups.push(arr.splice(0, n));
  }

  return groups.map(
    group =>   group.reduce(
      (a, b) => a + b
    ) / group.length
  );
}

console.log(
  calculateAverage([1, 2, 3, 4, 5, 6], 3)
); // [2, 5]

If you wish to round the average values, you can modify the console.log() statement as follows:

console.log(
  calculateAverage([1, 2, 4, 5, 6, 7], 3).map(Math.round)
);

Alternatively, the function itself can be adjusted to return rounded averages right away:

function calculateAverage(arr, n) {
  if (!arr || !n) {
    return false;
  }
  let groups = [];
  
  while (arr.length) {
    groups.push(arr.splice(0, n));
  }
  
  return groups.map(
    group => Math.round(group.reduce(
      (a, b) => a + b
    ) / group.length)
  );
}

console.log(
  calculateAverage([1, 2, 4, 5, 6, 7], 3)
);

For additional information and references regarding JavaScript functions used in these solutions, refer to the following:

Answer 2

When approaching this problem from a functional perspective, rather than procedural, one possible solution is:

function calculateAverage(arr, n) {
  if (!arr || !n){
    return false;
  }

  let groups = [];

  while (arr.length) {
    groups.push(arr.splice(0, n));
  }

  return groups.map(
    group =>   group.reduce(
      (a, b) => a + b
    ) / group.length
  );
}

console.log(
  calculateAverage([1, 2, 3, 4, 5, 6], 3)
); // [2, 5]

If you wish to round the average values, you can modify the console.log() statement as follows:

console.log(
  calculateAverage([1, 2, 4, 5, 6, 7], 3).map(Math.round)
);

Alternatively, the function itself can be adjusted to return rounded averages right away:

function calculateAverage(arr, n) {
  if (!arr || !n) {
    return false;
  }
  let groups = [];
  
  while (arr.length) {
    groups.push(arr.splice(0, n));
  }
  
  return groups.map(
    group => Math.round(group.reduce(
      (a, b) => a + b
    ) / group.length)
  );
}

console.log(
  calculateAverage([1, 2, 4, 5, 6, 7], 3)
);

For additional information and references regarding JavaScript functions used in these solutions, refer to the following:

Answer 3

Answer №2

One way to approach this problem is by creating a function that takes an array and batch size as parameters. The function then calculates the total sum of each batch based on the given size, and divides it by the batch size to determine the average for each group.

function computeGroupAverage(arr, n) {
  var result = [];
  
  for (var i = 0; i < arr.length;) {
    var sum = 0;
    
    for(var j = 0; j< n; j++){
      // Check if value is a number. If not, default to 0
      sum += +arr[i++] || 0
    }
    
    result.push(sum/n);
  }
  
  return result;
}

var arr = [1, 3, 2, 1, 5, 6, 7, 89,"test", 2, 3, 6, 8];

console.log(computeGroupAverage(arr, 3));
console.log(computeGroupAverage(arr, 2));

Answer 4

One way to approach this problem is by creating a function that takes an array and batch size as parameters. The function then calculates the total sum of each batch based on the given size, and divides it by the batch size to determine the average for each group.

function computeGroupAverage(arr, n) {
  var result = [];
  
  for (var i = 0; i < arr.length;) {
    var sum = 0;
    
    for(var j = 0; j< n; j++){
      // Check if value is a number. If not, default to 0
      sum += +arr[i++] || 0
    }
    
    result.push(sum/n);
  }
  
  return result;
}

var arr = [1, 3, 2, 1, 5, 6, 7, 89,"test", 2, 3, 6, 8];

console.log(computeGroupAverage(arr, 3));
console.log(computeGroupAverage(arr, 2));

Answer 5

Answer №3

give this a shot

var numbers = [10, 20, 30, 40, 50, 60];
var averages = [];
total = 0;
for (var j = 0; j < numbers.length; j++) {
  total = total + numbers[j];
  if ((j + 1) % 3 == 0) {
    averages.push(total / 3);
    total = 0;
  }
}
console.log(averages);

Answer 6

give this a shot

var numbers = [10, 20, 30, 40, 50, 60];
var averages = [];
total = 0;
for (var j = 0; j < numbers.length; j++) {
  total = total + numbers[j];
  if ((j + 1) % 3 == 0) {
    averages.push(total / 3);
    total = 0;
  }
}
console.log(averages);

Answer 7

Answer №4

To keep track of the count, simply create a variable called 'counter' and increase its value with every loop iteration.

var counter;

for ...
conuter++;
if(counter % 3 === 0)
  // perform certain actions

Answer 8

To keep track of the count, simply create a variable called 'counter' and increase its value with every loop iteration.

var counter;

for ...
conuter++;
if(counter % 3 === 0)
  // perform certain actions

Answer 9

Answer №5

To achieve this, you can utilize the functions reduce() and map().

var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var count = 0

var average = numbers.reduce(function(result, element, index) {
  if(index % 3 === 0) count++
  result[count] = (result[count] || 0) + element / 3
  if(index === numbers.length - 1) result = Object.keys(result).map(e => result[e])
  return result;
}, {})

console.log(average)

Answer 10

To achieve this, you can utilize the functions reduce() and map().

var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var count = 0

var average = numbers.reduce(function(result, element, index) {
  if(index % 3 === 0) count++
  result[count] = (result[count] || 0) + element / 3
  if(index === numbers.length - 1) result = Object.keys(result).map(e => result[e])
  return result;
}, {})

console.log(average)

Answer 11

Answer №6

Instead of providing a code snippet, let's delve into the rationale behind the code. One approach they took was to only calculate and add the average value after every third iteration. This was achieved by utilizing a counter variable and checking if it is divisible by 3 with the modulo operator:

if (counter % 3 == 0)
  {calculate and add the average}

By evaluating whether the counter is divisible by 3 using the modulo operation, any action can be triggered specifically when the counter is a multiple of 3.

Answer 12

Instead of providing a code snippet, let's delve into the rationale behind the code. One approach they took was to only calculate and add the average value after every third iteration. This was achieved by utilizing a counter variable and checking if it is divisible by 3 with the modulo operator:

if (counter % 3 == 0)
  {calculate and add the average}

By evaluating whether the counter is divisible by 3 using the modulo operation, any action can be triggered specifically when the counter is a multiple of 3.

Answer 13

Answer №7

To calculate the average of every third element in an array, first add up the values and then divide by the count.

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
    nth = 3,
    avg = array.reduce(function (r, a, i) {
        var ii = Math.floor(i / nth);
        r[ii] = ((r[ii] || 0) + a) / (i % nth !== nth - 1 || nth);
        return r;
    }, []);

console.log(avg);

Answer 14

To calculate the average of every third element in an array, first add up the values and then divide by the count.

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
    nth = 3,
    avg = array.reduce(function (r, a, i) {
        var ii = Math.floor(i / nth);
        r[ii] = ((r[ii] || 0) + a) / (i % nth !== nth - 1 || nth);
        return r;
    }, []);

console.log(avg);

Answer 15

Answer №8

Take a look at this unique tail call recursive method for some added variety:

let         numbers = Array(20).fill().map(_=> ~~(Math.random()*20)),
groupAverage = (arr, size, index=0, result=[]) => arr.length-index > 0 ? groupAverage(arr, size, index+size, result.concat(arr.slice(index,index+size).reduce((prev,current,_,source) => prev+current/source.length,0))) : result;
console.log(JSON.stringify(numbers));
console.log(JSON.stringify(groupAverage(numbers,3)));

Answer 16

Take a look at this unique tail call recursive method for some added variety:

let         numbers = Array(20).fill().map(_=> ~~(Math.random()*20)),
groupAverage = (arr, size, index=0, result=[]) => arr.length-index > 0 ? groupAverage(arr, size, index+size, result.concat(arr.slice(index,index+size).reduce((prev,current,_,source) => prev+current/source.length,0))) : result;
console.log(JSON.stringify(numbers));
console.log(JSON.stringify(groupAverage(numbers,3)));

Answer 17

Answer №9

Have you thought about using a for loop, but this time incrementing by 3 each iteration and storing the average in the avgArray list?

Here's an example:

var avgArray = [];
for(i=0; i < arr.length; i+=3) {
    total = arr[i] + (arr[i+1] || 0) + (arr[i+2] || 0);
    avg = total / 3;
    avgArray.push(Math.round(avg));
}

Answer 18

Have you thought about using a for loop, but this time incrementing by 3 each iteration and storing the average in the avgArray list?

Here's an example:

var avgArray = [];
for(i=0; i < arr.length; i+=3) {
    total = arr[i] + (arr[i+1] || 0) + (arr[i+2] || 0);
    avg = total / 3;
    avgArray.push(Math.round(avg));
}

Determine the mean of each group of n elements within a given array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

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