Divide a two-dimensional array into square sections and calculate the total of the highest value in each section using Javascript

Question

Divide a two-dimensional array into square sections and calculate the total of the highest value in each section using Javascript

I am trying to divide a matrix into square regions of specified dimensions (k) starting from the upper left corner and then calculate the sum of the maximum value in each region. Here is my progress so far.

arr = [
  [ 0, 8, 1, 1, 10, 6 ],
  [ 6, 8, 7, 0, 3, 9],
  [ 0, 7, 6, 8, 6, 5],
  [ 4, 0, 2, 7, 2, 0],
  [ 4, 4, 5, 7, 5, 1]
], l = console.log, j = JSON.stringify

result = [0, 3].map(i => arr.map(a => a.slice(i, i+3)))  // mapping over indexes to split by for segmenting

l(j(result ))

I am not very experienced with javascript but I am eager to learn. Any assistance would be greatly appreciated.

javascript arrays

Answer 1

Answer №1

While this solution may seem a bit complex, it is indeed functional. It aims to extract a specific region from the given data, but will adjust accordingly if the desired region size is not available. For example, in our case, it divides the array into two regions of 3x3 and two regions of 2x3. (Please note: this method will fail if all elements in the region are less than -100. To make it work in such cases, you need to modify the value of 'runningMax' as needed)

function getMaxValue(arr){
    var runningMax = -100;
    for(var i = 0; i < arr.length; i++) {
        runningMax = Math.max.apply(null, arr[i].concat(runningMax));
    }
    return runningMax;
}

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

var result = [0, 3].map(function(i) {
    return arr.map(function(a) {
        return a.slice(i, i + 3);
    });
});

var finalResult = [0, 3].map(function(i) {
    return result.map(function(a) {
        return a.slice(i, i + 3);
    });
});

var sum = 0;
finalResult.forEach(function(e){
    e.forEach(function(q){
        sum += getMaxValue(q);
    })
});

console.log(sum);

Answer 2

While this solution may seem a bit complex, it is indeed functional. It aims to extract a specific region from the given data, but will adjust accordingly if the desired region size is not available. For example, in our case, it divides the array into two regions of 3x3 and two regions of 2x3. (Please note: this method will fail if all elements in the region are less than -100. To make it work in such cases, you need to modify the value of 'runningMax' as needed)

function getMaxValue(arr){
    var runningMax = -100;
    for(var i = 0; i < arr.length; i++) {
        runningMax = Math.max.apply(null, arr[i].concat(runningMax));
    }
    return runningMax;
}

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

var result = [0, 3].map(function(i) {
    return arr.map(function(a) {
        return a.slice(i, i + 3);
    });
});

var finalResult = [0, 3].map(function(i) {
    return result.map(function(a) {
        return a.slice(i, i + 3);
    });
});

var sum = 0;
finalResult.forEach(function(e){
    e.forEach(function(q){
        sum += getMaxValue(q);
    })
});

console.log(sum);

Answer 3

Answer №2

You can divide your matrix into regions by utilizing the reduce() and forEach() methods

const arr = [
    [0, 8, 1, 1, 10, 6],
    [6, 8, 7, 0, 3, 9],
    [0, 7, 6, 8, 6, 5],
    [4, 0, 2, 7, 2, 0],
    [4, 4, 5, 7, 5, 1]
  ], l = console.log, j = JSON.stringify

function splitMatrix(data, n) {
  let row = 0;
  return data.reduce(function(r, a, i) {
    if (i && i % n == 0) row++;
    let col = 0;
    a.forEach(function(e, j) {
      if (j && j % n == 0) col++;
      if (!r[row]) r[row] = [];
      if (!r[row][col]) r[row][col] = [];
      r[row][col].push(e)
    })
    return r;
  }, [])
}

const s = splitMatrix(arr, 3);
console.log(j(s))

To determine the maximum value of each region, you can achieve that within the forEach loop and then utilize reduce() to sum up those values.

const arr = [
    [0, 8, 1, 1, 10, 6],
    [6, 8, 7, 0, 3, 9],
    [0, 7, 6, 8, 6, 5],
    [4, 0, 2, 7, 2, 0],
    [4, 4, 5, 7, 5, 1]
  ], l = console.log, j = JSON.stringify

function splitMatrix(data, n) {
  let row = 0;
  return data.reduce(function(r, a, i) {
    if (i && i % n == 0) row++;
    let col = 0;
    a.forEach(function(e, j) {
      if (j && j % n == 0) col++;
      if (!r[row]) r[row] = [];
      if (!r[row][col]) r[row][col] = [e];
      if (r[row][col][0] < e) r[row][col][0] = e;
    })
    return r;
  }, [])
}

const max = splitMatrix(arr, 3)
const result = [].concat(...max).reduce((r, [e]) => r + e, 0);
console.log(j(max))
console.log(result)

Answer 4

You can divide your matrix into regions by utilizing the reduce() and forEach() methods

const arr = [
    [0, 8, 1, 1, 10, 6],
    [6, 8, 7, 0, 3, 9],
    [0, 7, 6, 8, 6, 5],
    [4, 0, 2, 7, 2, 0],
    [4, 4, 5, 7, 5, 1]
  ], l = console.log, j = JSON.stringify

function splitMatrix(data, n) {
  let row = 0;
  return data.reduce(function(r, a, i) {
    if (i && i % n == 0) row++;
    let col = 0;
    a.forEach(function(e, j) {
      if (j && j % n == 0) col++;
      if (!r[row]) r[row] = [];
      if (!r[row][col]) r[row][col] = [];
      r[row][col].push(e)
    })
    return r;
  }, [])
}

const s = splitMatrix(arr, 3);
console.log(j(s))

To determine the maximum value of each region, you can achieve that within the forEach loop and then utilize reduce() to sum up those values.

const arr = [
    [0, 8, 1, 1, 10, 6],
    [6, 8, 7, 0, 3, 9],
    [0, 7, 6, 8, 6, 5],
    [4, 0, 2, 7, 2, 0],
    [4, 4, 5, 7, 5, 1]
  ], l = console.log, j = JSON.stringify

function splitMatrix(data, n) {
  let row = 0;
  return data.reduce(function(r, a, i) {
    if (i && i % n == 0) row++;
    let col = 0;
    a.forEach(function(e, j) {
      if (j && j % n == 0) col++;
      if (!r[row]) r[row] = [];
      if (!r[row][col]) r[row][col] = [e];
      if (r[row][col][0] < e) r[row][col][0] = e;
    })
    return r;
  }, [])
}

const max = splitMatrix(arr, 3)
const result = [].concat(...max).reduce((r, [e]) => r + e, 0);
console.log(j(max))
console.log(result)

Answer 5

Answer №3

When considering the example given, where the array is 6x5, it becomes evident that splitting it evenly into square regions is not possible. However, this detail is somewhat tangential to the main issue at hand. The first imperative step in approaching the problem is to devise a function capable of extracting and summing up a specific section of a matrix.

Imagine having a matrix and the need for a function that can take parameters such as x, y, width, and height to determine the area requiring summation:

const m = [
 [1,2,3,4,5],
 [2,3,4,5,6],
 [3,4,5,6,7],
 [8,9,10,11,12]
];

const sumArea = (m, x, y, w, h) => {
  let sum = 0;
  
  if (y + h > m.length) {
    throw new Error('Matrix is not tall enough for the requested area');
  }
  
  if (x + w > m[0].length) {
    throw new Error('Matrix is not wide enough for the request area');
  }
  
  for (let j = y; j < y + h; j++) {
    for (let i = x; i < x + w; i++) {
      sum += m[j][i];
    }
  }
  
  return sum;
}

console.log(sumArea(m, 0, 0, 3, 3));

This function allows traversal through any specified region within a matrix and calculates the total sum within that area. By utilizing a `for` loop from the initial point (x, y) to the final point (x + width, y + height), this task is efficiently accomplished.

An added precaution includes boundary checks to ensure the area remains within the matrix dimensions; otherwise, an error is thrown. Alternatively, you could choose to treat out-of-bounds areas as zeroes and proceed without interruption by incorporating conditions within the `for` loop.

Subsequently, to calculate the squares of all values, a straightforward iteration process is employed.

const m = [
 [1,2,3,4],
 [2,3,4,5],
 [3,4,5,6],
 [8,9,10,11]
];

const sumArea = (m, x, y, w, h) => {
  let sum = 0;
  
  if (y + h > m.length) {
    throw new Error('Matrix is not tall enough for the requested area');
  }
  
  if (x + w > m[0].length) {
    throw new Error('Matrix is not wide enough for the request area');
  }
  
  for (let j = y; j < y + h; j++) {
    for (let i = x; i < x + w; i++) {
      sum += m[j][i];
    }
  }
  
  return sum;
}

const sumSquares = (m, w, h) => {
  let result = [];
  
  for (let y = 0; y < m.length / h; y++) {
    result[y] = [];
    
    for (let x = 0; x < m[0].length / w; x++) {
      result[y][x] = sumArea(m, x * w, y * h, w, h);
    }
  }
  
  return result;
}

console.log(sumSquares(m, 2, 2));

This particular approach involves iterating through each square segment within the matrix. The number of squares corresponds to the size of the matrix divided by the width of each square. For instance, in a 4x4 matrix with 2x2 squares, there would be 2 squares across and down respectively. These values are then collated into another matrix to showcase their cumulative sum.

Answer 6

When considering the example given, where the array is 6x5, it becomes evident that splitting it evenly into square regions is not possible. However, this detail is somewhat tangential to the main issue at hand. The first imperative step in approaching the problem is to devise a function capable of extracting and summing up a specific section of a matrix.

Imagine having a matrix and the need for a function that can take parameters such as x, y, width, and height to determine the area requiring summation:

const m = [
 [1,2,3,4,5],
 [2,3,4,5,6],
 [3,4,5,6,7],
 [8,9,10,11,12]
];

const sumArea = (m, x, y, w, h) => {
  let sum = 0;
  
  if (y + h > m.length) {
    throw new Error('Matrix is not tall enough for the requested area');
  }
  
  if (x + w > m[0].length) {
    throw new Error('Matrix is not wide enough for the request area');
  }
  
  for (let j = y; j < y + h; j++) {
    for (let i = x; i < x + w; i++) {
      sum += m[j][i];
    }
  }
  
  return sum;
}

console.log(sumArea(m, 0, 0, 3, 3));

This function allows traversal through any specified region within a matrix and calculates the total sum within that area. By utilizing a `for` loop from the initial point (x, y) to the final point (x + width, y + height), this task is efficiently accomplished.

An added precaution includes boundary checks to ensure the area remains within the matrix dimensions; otherwise, an error is thrown. Alternatively, you could choose to treat out-of-bounds areas as zeroes and proceed without interruption by incorporating conditions within the `for` loop.

Subsequently, to calculate the squares of all values, a straightforward iteration process is employed.

const m = [
 [1,2,3,4],
 [2,3,4,5],
 [3,4,5,6],
 [8,9,10,11]
];

const sumArea = (m, x, y, w, h) => {
  let sum = 0;
  
  if (y + h > m.length) {
    throw new Error('Matrix is not tall enough for the requested area');
  }
  
  if (x + w > m[0].length) {
    throw new Error('Matrix is not wide enough for the request area');
  }
  
  for (let j = y; j < y + h; j++) {
    for (let i = x; i < x + w; i++) {
      sum += m[j][i];
    }
  }
  
  return sum;
}

const sumSquares = (m, w, h) => {
  let result = [];
  
  for (let y = 0; y < m.length / h; y++) {
    result[y] = [];
    
    for (let x = 0; x < m[0].length / w; x++) {
      result[y][x] = sumArea(m, x * w, y * h, w, h);
    }
  }
  
  return result;
}

console.log(sumSquares(m, 2, 2));

This particular approach involves iterating through each square segment within the matrix. The number of squares corresponds to the size of the matrix divided by the width of each square. For instance, in a 4x4 matrix with 2x2 squares, there would be 2 squares across and down respectively. These values are then collated into another matrix to showcase their cumulative sum.

Answer 7

Answer №4

    
const myArray = [
  [ 10, 5, 8, 4, 7, 1 ],
  [ 3, 7, 2, 9, 6, 0],
  [ 5, 4, 7, 1, 3, 6],
  [ 8, 2, 4, 6, 9, 5],
  [ 1, 6, 3, 4, 7, 1]
];

const sizeX = 3, sizeY = 3;

// Iterating through squares
for(let startX = 0; startX < myArray.length - sizeX; startX++){
  for(let startY = 0; startY < myArray[0].length - sizeY; startY++){
     // Finding the sum of elements in each square
     let sum = 0;
     for(let x = startX; x < startX + sizeX; x++)
        for(let y = startY; y < startY + sizeY;  y++)
            sum += myArray[x][y];
      // Performing operations with the sum
      console.log(sum)
   }
}

Answer 8

    
const myArray = [
  [ 10, 5, 8, 4, 7, 1 ],
  [ 3, 7, 2, 9, 6, 0],
  [ 5, 4, 7, 1, 3, 6],
  [ 8, 2, 4, 6, 9, 5],
  [ 1, 6, 3, 4, 7, 1]
];

const sizeX = 3, sizeY = 3;

// Iterating through squares
for(let startX = 0; startX < myArray.length - sizeX; startX++){
  for(let startY = 0; startY < myArray[0].length - sizeY; startY++){
     // Finding the sum of elements in each square
     let sum = 0;
     for(let x = startX; x < startX + sizeX; x++)
        for(let y = startY; y < startY + sizeY;  y++)
            sum += myArray[x][y];
      // Performing operations with the sum
      console.log(sum)
   }
}

Answer 9

Answer №5

When working with arrays of the same length and height, it's possible to iterate through the arrays and extract the indices for grouping and counting purposes.

var array = [[0, 8, 1, 1, 10, 6], [6, 8, 7, 0, 3, 9], [0, 7, 6, 8, 6, 5], [4, 0, 2, 7, 2, 0], [4, 4, 5, 7, 5, 1], [0, 1, 0, 1, 0, 1]],
    result = array.reduce((r, a, i) => {
        a.forEach((b, j) => {
            r[Math.floor(i / 3)] = r[Math.floor(i / 3)] || [];
            r[Math.floor(i / 3)][Math.floor(j / 3)] = (r[Math.floor(i / 3)][Math.floor(j / 3)] || 0) + b;
        });
        return r;
    }, []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 10

When working with arrays of the same length and height, it's possible to iterate through the arrays and extract the indices for grouping and counting purposes.

var array = [[0, 8, 1, 1, 10, 6], [6, 8, 7, 0, 3, 9], [0, 7, 6, 8, 6, 5], [4, 0, 2, 7, 2, 0], [4, 4, 5, 7, 5, 1], [0, 1, 0, 1, 0, 1]],
    result = array.reduce((r, a, i) => {
        a.forEach((b, j) => {
            r[Math.floor(i / 3)] = r[Math.floor(i / 3)] || [];
            r[Math.floor(i / 3)][Math.floor(j / 3)] = (r[Math.floor(i / 3)][Math.floor(j / 3)] || 0) + b;
        });
        return r;
    }, []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 11

Answer №6

It seems like your goal is to analyze all the possible 3x3 sub-matrices within the provided matrix. To clarify, in your matrix example, you are interested in examining:

[ 0, 8, 1 ]    [ 8, 1, 1 ]    [ 1, 1, 10]    [ 1, 10, 6 ]
[ 6, 8, 7 ]    [ 8, 7, 0 ]    [ 7, 0, 3 ]    [ 0, 3, 9 ]
[ 0, 7, 6 ]    [ 7, 6, 8 ]    [ 6, 8, 6 ]    [ 8, 6, 5 ]

[ 6, 8, 7 ]    [ 8, 7, 0 ]    [ 7, 0, 3 ]    [ 0, 3, 9 ]
[ 0, 7, 6 ]    [ 7, 6, 8 ]    [ 6, 8, 6 ]    [ 8, 6, 5 ]
[ 4, 0, 2 ]    [ 0, 2, 7 ]    [ 2, 7, 2 ]    [ 7, 2, 0 ]

[ 0, 7, 6 ]    [ 7, 6, 8 ]    [ 6, 8, 6 ]    [ 8, 6, 5]
[ 4, 0, 2 ]    [ 0, 2, 7 ]    [ 2, 7, 2 ]    [ 7, 2, 0]
[ 4, 4, 5 ]    [ 4, 5, 7 ]    [ 5, 7, 5 ]    [ 7, 5, 1]

Your objective appears to be identifying the maximum values (12 maximums for 12 matrices):

8    8   10   10

8    8    8    9

7    8    8    8

The total sum of these maxima is 100. If this outcome aligns with your expectations, below is a functional approach to compute it:

const arr = [
  [ 0, 8, 1, 1, 10, 6 ],
  [ 6, 8, 7, 0, 3, 9],
  [ 0, 7, 6, 8, 6, 5],
  [ 4, 0, 2, 7, 2, 0],
  [ 4, 4, 5, 7, 5, 1]
];

const sum = arr.slice(2).reduce( 
    (sum, row, i) => row.slice(2).reduce( 
        (sum, a, j) => sum + Math.max(...arr[i].slice(j,j+3), 
                                      ...arr[i+1].slice(j,j+3), 
                                      ...row.slice(j,j+3)),
        sum
    ), 0
)

console.log(sum);

Answer 12

It seems like your goal is to analyze all the possible 3x3 sub-matrices within the provided matrix. To clarify, in your matrix example, you are interested in examining:

[ 0, 8, 1 ]    [ 8, 1, 1 ]    [ 1, 1, 10]    [ 1, 10, 6 ]
[ 6, 8, 7 ]    [ 8, 7, 0 ]    [ 7, 0, 3 ]    [ 0, 3, 9 ]
[ 0, 7, 6 ]    [ 7, 6, 8 ]    [ 6, 8, 6 ]    [ 8, 6, 5 ]

[ 6, 8, 7 ]    [ 8, 7, 0 ]    [ 7, 0, 3 ]    [ 0, 3, 9 ]
[ 0, 7, 6 ]    [ 7, 6, 8 ]    [ 6, 8, 6 ]    [ 8, 6, 5 ]
[ 4, 0, 2 ]    [ 0, 2, 7 ]    [ 2, 7, 2 ]    [ 7, 2, 0 ]

[ 0, 7, 6 ]    [ 7, 6, 8 ]    [ 6, 8, 6 ]    [ 8, 6, 5]
[ 4, 0, 2 ]    [ 0, 2, 7 ]    [ 2, 7, 2 ]    [ 7, 2, 0]
[ 4, 4, 5 ]    [ 4, 5, 7 ]    [ 5, 7, 5 ]    [ 7, 5, 1]

Your objective appears to be identifying the maximum values (12 maximums for 12 matrices):

8    8   10   10

8    8    8    9

7    8    8    8

The total sum of these maxima is 100. If this outcome aligns with your expectations, below is a functional approach to compute it:

const arr = [
  [ 0, 8, 1, 1, 10, 6 ],
  [ 6, 8, 7, 0, 3, 9],
  [ 0, 7, 6, 8, 6, 5],
  [ 4, 0, 2, 7, 2, 0],
  [ 4, 4, 5, 7, 5, 1]
];

const sum = arr.slice(2).reduce( 
    (sum, row, i) => row.slice(2).reduce( 
        (sum, a, j) => sum + Math.max(...arr[i].slice(j,j+3), 
                                      ...arr[i+1].slice(j,j+3), 
                                      ...row.slice(j,j+3)),
        sum
    ), 0
)

console.log(sum);

Divide a two-dimensional array into square sections and calculate the total of the highest value in each section using Javascript

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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