Traversing through rows of a two-dimensional array that stores arrays of varying lengths

Question

Traversing through rows of a two-dimensional array that stores arrays of varying lengths

I have developed a function that extracts all elements from a 2D array by rows and converts them into a 1D array. The input array can have varying numbers of columns and rows.

For example:

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

This will return:

[1, 5, 9, 2, 6, 10, 3, 7, 11, 4, 8, 12]

The function I have created is as follows:

convertList = (list) => {
    let result = [];
    let listTotalEntries = R.sum(R.map(R.length)(list));
    let mod = R.modulo(R.__, list.length);

        let counterRow = -1;

        for (let i = 0; i < listTotalEntries; i++) {
            if (mod(i) === 0) {
                counterRow++;
            }
            if (list[mod(i)][counterRow]) {
                result.push(list[mod(i)][counterRow]);
                console.log(list[mod(i)][counterRow]);
            }
        }
        console.log(result);
        return result;
};

Inquiry: Currently, this function only works with square matrices. How can I modify it to handle arrays with varying lengths?

For instance:

let arr = [
    [1, 2],
    [],
    [9, 10, 11, 12]
];

Desired output:

[1, 9, 2, 10, 11, 12]

Thank you for your assistance!

Muff

javascript arrays multidimensional-array ramda.js

Answer 1

Answer №1

There was a ramda.js tag included in this content. Utilizing Ramda makes it simple, as there are two functions that come in handy:

const convertList = compose(flatten, transpose);
convertList(arr); //=> [1, 9, 2, 10, 11, 12]

transpose reverses a matrix over its main diagonal, essentially changing rows to columns and vice versa. The flatten function converts a list of lists into a single list. By using compose in this manner, it effectively creates the equivalent of

list => flatten(transpose(list))

.

To witness this process firsthand, visit the Ramda REPL.

Answer 2

There was a ramda.js tag included in this content. Utilizing Ramda makes it simple, as there are two functions that come in handy:

const convertList = compose(flatten, transpose);
convertList(arr); //=> [1, 9, 2, 10, 11, 12]

transpose reverses a matrix over its main diagonal, essentially changing rows to columns and vice versa. The flatten function converts a list of lists into a single list. By using compose in this manner, it effectively creates the equivalent of

list => flatten(transpose(list))

.

To witness this process firsthand, visit the Ramda REPL.

Answer 3

Answer №2

My recommendation is to tackle the arrays one step at a time.

var arr1 = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]],
    arr2 = [[1, 2], [], [9, 10, 11, 12]];

function single(array) {
    var result = [],
        maximum = Math.max.apply(null, array.map(function (a) { return a.length; })),
        index = 0, j,
        length = array.length;

    while (index < maximum) {
        for (j = 0; j < length ; j++) {
            index in array[j] && result.push(array[j][index]);
        }
        index++;
    }
    return result;
}

document.write('<pre>' + JSON.stringify(single(arr1), 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(single(arr2), 0, 4) + '</pre>');

Answer 4

My recommendation is to tackle the arrays one step at a time.

var arr1 = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]],
    arr2 = [[1, 2], [], [9, 10, 11, 12]];

function single(array) {
    var result = [],
        maximum = Math.max.apply(null, array.map(function (a) { return a.length; })),
        index = 0, j,
        length = array.length;

    while (index < maximum) {
        for (j = 0; j < length ; j++) {
            index in array[j] && result.push(array[j][index]);
        }
        index++;
    }
    return result;
}

document.write('<pre>' + JSON.stringify(single(arr1), 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(single(arr2), 0, 4) + '</pre>');

Answer 5

Answer №3

Have you experimented with this straightforward approach?

const singleDimensionArr = arr.reduce((prev, current) => prev.concat(current));

For instance:

[
    [1, 2],
    [],
    [9, 10, 11, 12]
].reduce((prev, current) => prev.concat(current));

will produce [1, 2, 9, 10, 11, 12]

Note:

After receiving feedback from the original poster below, it seems that concatenation needs to occur column-wise.

const max = Math.max(...arr.map(a => a.length));
const finalArr = [];
for (let i = 0; i < max; i++) {
   for (let j = 0; j < arr.length; j++) {
       arr[j][i] ? finalArr.push(arr[j][i]) : "";
   }
}
console.log(finalArr);

Answer 6

Have you experimented with this straightforward approach?

const singleDimensionArr = arr.reduce((prev, current) => prev.concat(current));

For instance:

[
    [1, 2],
    [],
    [9, 10, 11, 12]
].reduce((prev, current) => prev.concat(current));

will produce [1, 2, 9, 10, 11, 12]

Note:

After receiving feedback from the original poster below, it seems that concatenation needs to occur column-wise.

const max = Math.max(...arr.map(a => a.length));
const finalArr = [];
for (let i = 0; i < max; i++) {
   for (let j = 0; j < arr.length; j++) {
       arr[j][i] ? finalArr.push(arr[j][i]) : "";
   }
}
console.log(finalArr);

Answer 7

Answer №4

This illustration demonstrates the creation of a large sparse array by placing each element in its corresponding position in a square-like arrangement. It then eliminates any null values that may have been generated where no input item exists.

let arr = [
    [1, 2],
    [],
    [9, 10, 11, 12]
];

var output = arr.reduce(function(outputArray, innerArray, index, mainArray) {
  for (var subIndex=0; subIndex<innerArray.length; subIndex++){
    outputArray[mainArray.length * subIndex + index] = innerArray[subIndex];
  }
  return outputArray;
},[]).filter(function(element) {
  return element !== null;
});

alert(JSON.stringify(output));

Answer 8

This illustration demonstrates the creation of a large sparse array by placing each element in its corresponding position in a square-like arrangement. It then eliminates any null values that may have been generated where no input item exists.

let arr = [
    [1, 2],
    [],
    [9, 10, 11, 12]
];

var output = arr.reduce(function(outputArray, innerArray, index, mainArray) {
  for (var subIndex=0; subIndex<innerArray.length; subIndex++){
    outputArray[mainArray.length * subIndex + index] = innerArray[subIndex];
  }
  return outputArray;
},[]).filter(function(element) {
  return element !== null;
});

alert(JSON.stringify(output));

Traversing through rows of a two-dimensional array that stores arrays of varying lengths

Answer №1

Answer №2

Answer №3

Answer №4

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