Most efficient method for organizing a nested array (comparing output versus sorting and using math.max)

Question

Most efficient method for organizing a nested array (comparing output versus sorting and using math.max)

I have always wondered if it is possible to retrieve a returned value from a nested function or code block.

For instance, consider the scenario where you want to sort values in an array from largest to smallest using the .sort() method and then extract the largest value from each nested array and add it to a new array:


function largestOfFour(arr) {
  for (var i = 0; i < arr.length; i++) {
    console.log(arr[i]);
    arr[i].sort(function(a,b) {
      return b - a; // attempt to access this organized array here
    });
    var newArr = [];
    newArr.push(arr[i][0]);
  }
}

largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);

javascript arrays sorting return-value

Answer 1

Answer №1

Sorting the array is not necessary in this case. One could simply utilize the Math.max method.

function findLargestNumbers(arr) {
  return arr.map(function (el) {
    return Math.max.apply(null, el);
  });
} // [ 5, 27, 39, 1001 ]

Check out the DEMO here

Answer 2

Sorting the array is not necessary in this case. One could simply utilize the Math.max method.

function findLargestNumbers(arr) {
  return arr.map(function (el) {
    return Math.max.apply(null, el);
  });
} // [ 5, 27, 39, 1001 ]

Check out the DEMO here

Answer 3

Answer №2

Is this the correct way?

function findLargestNumbers(arrays) {
  var finalResults = [];
  for (var index = 0; index < arrays.length; index++) {
    console.log(arrays[index]);
    arrays[index].sort(function(a, b) {
      return b - a; // aim is to access this newly sorted array
    });
    finalResults.push(arrays[index][0]);
  }
  return finalResults;
}

var finalResult = findLargestNumbers([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);

Answer 4

Is this the correct way?

function findLargestNumbers(arrays) {
  var finalResults = [];
  for (var index = 0; index < arrays.length; index++) {
    console.log(arrays[index]);
    arrays[index].sort(function(a, b) {
      return b - a; // aim is to access this newly sorted array
    });
    finalResults.push(arrays[index][0]);
  }
  return finalResults;
}

var finalResult = findLargestNumbers([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);

Answer 5

Answer №3

Is it feasible to retrieve a returned value from a function/code block nested within another code block?

The only way to obtain the return value is by calling the function. Since you are not directly invoking the .sort callback, but rather it is executed within .sort, retrieving its return value is not possible.

I assume your inquiry stems from encountering unexpected behavior in your code. Upon reviewing your script, it seems evident that although you are sorting the array, you are not utilizing it post-sorting. As mentioned earlier in my comments, your method of using .sort and accessing the largest value is correct.

To address this issue, I suggest declaring newArr outside the loop and returning it from the function:

function largestOfFour(arr) {
  var newArr = [];
  for (var i = 0; i < arr.length; i++) {
    arr[i].sort(function(a, b) {
      return b - a;
    });
    newArr.push(arr[i][0]);
  }
  return newArr;
}

console.log(largestOfFour([
  [4, 5, 1, 3],
  [13, 27, 18, 26],
  [32, 35, 37, 39],
  [1000, 1001, 857, 1]
]));

Alternatively, there is a more concise approach that avoids mutating the input array:

function largestOfFour(arr) {
  return arr.map(function(innerArray) {
    return innerArray.reduce(function(a, b) {
      return a > b ? a : b;
    });
  });
}

Answer 6

Is it feasible to retrieve a returned value from a function/code block nested within another code block?

The only way to obtain the return value is by calling the function. Since you are not directly invoking the .sort callback, but rather it is executed within .sort, retrieving its return value is not possible.

I assume your inquiry stems from encountering unexpected behavior in your code. Upon reviewing your script, it seems evident that although you are sorting the array, you are not utilizing it post-sorting. As mentioned earlier in my comments, your method of using .sort and accessing the largest value is correct.

To address this issue, I suggest declaring newArr outside the loop and returning it from the function:

function largestOfFour(arr) {
  var newArr = [];
  for (var i = 0; i < arr.length; i++) {
    arr[i].sort(function(a, b) {
      return b - a;
    });
    newArr.push(arr[i][0]);
  }
  return newArr;
}

console.log(largestOfFour([
  [4, 5, 1, 3],
  [13, 27, 18, 26],
  [32, 35, 37, 39],
  [1000, 1001, 857, 1]
]));

Alternatively, there is a more concise approach that avoids mutating the input array:

function largestOfFour(arr) {
  return arr.map(function(innerArray) {
    return innerArray.reduce(function(a, b) {
      return a > b ? a : b;
    });
  });
}

Answer 7

Answer №4

Arranging values in an array from largest to smallest using the .sort() method, then extracting the largest value from nested arrays and placing it into a new array:

Consider utilizing a combination of while loop, .push(), .sort(), .reverse(), and largestOfFour function which can provide two arrays. The first array contains the largest number from each sub-array, while the second array comprises sorted sub-arrays from largest to smallest.

function largestOfFour(arr) {
  var res = [], i = arr.length - 1;
  while(i > -1) {
    res.push(Math.max.apply(Math, arr[i].sort(function(a, b) {
      return Number(b) - Number(a)})
      )); 
  --i;
  };     
  return [res, arr.reverse()]
}

var input = [
    [4, 5, 1, 3],
    [13, 27, 18, 26],
    [32, 35, 37, 39],
    [1000, 1001, 857, 1]
  ];

function largestOfFour(arr) {
  var res = [], i = arr.length - 1;
  while(i > -1) {
    res.push(Math.max.apply(Math, arr[i].sort(function(a, b) {
      return Number(b) - Number(a)})
      )); 
  --i;
  };     
  return [res, arr.reverse()]
}

document.getElementsByTagName("pre")[0]
.textContent = JSON.stringify(largestOfFour(input), null, 2);

<pre></pre>

Answer 8

Arranging values in an array from largest to smallest using the .sort() method, then extracting the largest value from nested arrays and placing it into a new array:

Consider utilizing a combination of while loop, .push(), .sort(), .reverse(), and largestOfFour function which can provide two arrays. The first array contains the largest number from each sub-array, while the second array comprises sorted sub-arrays from largest to smallest.

function largestOfFour(arr) {
  var res = [], i = arr.length - 1;
  while(i > -1) {
    res.push(Math.max.apply(Math, arr[i].sort(function(a, b) {
      return Number(b) - Number(a)})
      )); 
  --i;
  };     
  return [res, arr.reverse()]
}

var input = [
    [4, 5, 1, 3],
    [13, 27, 18, 26],
    [32, 35, 37, 39],
    [1000, 1001, 857, 1]
  ];

function largestOfFour(arr) {
  var res = [], i = arr.length - 1;
  while(i > -1) {
    res.push(Math.max.apply(Math, arr[i].sort(function(a, b) {
      return Number(b) - Number(a)})
      )); 
  --i;
  };     
  return [res, arr.reverse()]
}

document.getElementsByTagName("pre")[0]
.textContent = JSON.stringify(largestOfFour(input), null, 2);

<pre></pre>

Most efficient method for organizing a nested array (comparing output versus sorting and using math.max)

Answer №1

Answer №2

Answer №3

Answer №4

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