matrix to array with every other row reversed - javascript

I have to convert matrix to array. For example
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
]
should convert to [1, 2, 3, 6, 5, 4, 7, 8, 9 ].
I write a code
function sortMatrix (matrix) {
let newArr = [];
for(let i = 0; i < matrix.length; i++)
{
newArr = newArr.concat(matrix[i]);
}
return newArr
}
but output is [1, 2, 3, 4, 5, 6, 7, 8, 9 ]. How can I do so that on the second line, it counts from the end


To fix your code - reverse sub-arrays at odd indexes.
Note: since Array.reverse() mutates the original array, we need to shallow clone it using array spread.
function sortMatrix(matrix) {
let newArr = [];
for (let i = 0; i < matrix.length; i++) {
const arr = matrix[i];
newArr = newArr.concat(i % 2 ? [...arr].reverse() : arr);
}
return newArr;
}
const matrix = [
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
];
const result = sortMatrix(matrix);
console.log(result);
Another option is to use Array.flatMap(), and reverse sub-arrays at odd indexes.
function sortMatrix(matrix) {
return matrix.flatMap((arr, i) => i % 2 ? [...arr].reverse() : arr);
}
const matrix = [
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
];
const result = sortMatrix(matrix);
console.log(result);


You can use reduce to create it.
const arr = [
[1,2,3],
[4,5,6],
[7,8,9]
]
const flatArray = (arr) => arr.reverse().reduce((acc, curr)=> {
acc.push(...curr.reverse());
return acc;
},[])
console.log(flatArray(arr))


You can directly use the flat function
array = [
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
]
console.log(array.flat());

Related

How to convert the below array [duplicate]

I have an array like so: [1, 2, 3, 4, 5, 6, 7, 9, 10]. I need to chunk it into different size chunks, yet with a simple pattern of: 4, 3, 3, 3, 4, 3, 3, 3 like so:
[
[ // four
1,
2,
3,
4
],
[ // three (1/3)
5,
6,
7
],
[ // three (2/3)
8,
9,
10
],
[ // three (3/3)
11,
12,
13
],
[ // four
14,
15,
16,
17
],
[ // three (1/3)
18,
19,
20
], // and so on..
]
I have tried with this code I have customized:
const arr; // my array of values
const chuncked = arr.reduce((acc, product, i) => {
if (i % 3) {
return acc;
} else if (!didFourWayReduce) {
didFourWayReduce = true;
fourWayReduces++;
if ((fourWayReduces - 1) % 2) { // only make every second a "4 row"
return [...acc, arr.slice(i, i + 3)];
} else {
return [...acc, arr.slice(i, i + 4)];
}
} else {
didFourWayReduce = false;
return [...acc, arr.slice(i, i + 3)];
}
}, []);
And it works, almost, expect that the first chunk of threes (1/3) have the last element of the chunk with 4. So 1 key is repeated every first chunk of three. Like so:
[
[
1,
2,
3,
4
],
[
4, // this one is repeated, and it shouldn't be
5,
6
]
]

You could take two indices, one for the data array and one for sizes. Then slice the array with a given length and push the chunk to the chunks array.
Proceed until end of data.
var data = Array.from({ length: 26 }, (_, i) => i + 1),
sizes = [4, 3, 3, 3],
i = 0,
j = 0,
chunks = [];
while (i < data.length) chunks.push(data.slice(i, i += sizes[j++ % sizes.length]));
console.log(chunks);
.as-console-wrapper { max-height: 100% !important; top: 0; }

const arr = Array.from({ length: 100 }, (_, i) => i);
const copy = [...arr];
const sizes = [4, 3, 3, 3];
const result = [];
let i = 0;
while (i <= arr.length && copy.length) {
result.push(copy.splice(0, sizes[i % sizes.length]));
i++;
}
console.log(result);

A recursive approach is fairly elegant:
const chunks = (xs, [s, ...ss]) =>
xs.length ? [xs .slice (0, s), ... chunks (xs .slice (s), [...ss, s])] : []
const data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
const sizes = [4, 3, 3, 3]
console .log (chunks (data, sizes))
.as-console-wrapper { max-height: 100% !important; top: 0; }
By replacing [s, ...ss] with [...ss, s], we pass a cycled version of the sizes array, so that for instance, [4, 3, 3, 3] becomes [3, 3, 3, 4]. This makes it easy to parse step-by-step.

Mod operator to check if it should be 4 or 3. Use two arrays just to make it easier (can be done with one)
const groupIt = arr => arr.reduce(({
group,
out
}, v, i) => {
var max = out.length % 4 === 0 ? 4 : 3
group.push(v)
if (group.length === max || i === arr.length - 1) {
out.push(group)
group = []
}
return {
group,
out
}
}, {
group: [],
out: []
}).out
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x,i) => i + 1)
console.log(groupIt(test))
with just one:
const groupIt = arr => arr.reduce((out, v, i) => {
var max = (out.length - 1) % 4 === 0 ? 4 : 3
out[out.length - 1].push(v)
if (out[out.length - 1].length === max) {
out.push([])
}
return out
}, [[]])
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x, i) => i + 1)
console.log(groupIt(test))

This answer is similar to that of Nina Scholz, but uses a for loop, which I personally find more clear.
const arr = Array.from({length: 100}, (_, i) => i + 1);
const sizes = [4, 3, 3, 3];
const result = [];
for (let i = 0, j = 0; i < arr.length; i += sizes[j], j = (j + 1) % sizes.length) {
result.push(arr.slice(i, i + sizes[j]));
}
console.log(result);

Split a two-dimensional array of specified length

Split the original array into a two-dimensional array of the specified length
list => source array
columns => columns number
targetList => two-dimensional array
const list = [1,2,3,4,5,6,7,8,9,10]
const columns = 4;
const targetList = [ [1,2,3], [4,5,6], [7,8,9], [10] ];
const columns = 5;
const targetList = [ [1,2], [3,4], [5,6], [7,8], [9,10] ];
const columns = 6;
const targetList = [ [1,2], [3,4], [5,6], [7,8], [9], [10] ];
const list = [1,2,3,4,5,6]
const columns = 4;
const targetList = [ [1,2], [3,4], [5], [6] ];
const list = [1,2,3,4]
const columns = 5;
const targetList = [ [1], [2], [3], [4] ];

You can use Array.prototype.reduce and transform the given list to the desired grid.
Push the a new row to the resultant grid if any of the following conditions meet:
If there are no rows currently.
If the last row is filled i.e. no more columns can be added to the last row, then add a new row with the current item.
If items remaining to be pushed becomes equal to rows remaining to be created.
const transform = (list, rows) =>
list.reduce((t, l, i) => {
if (
!t.length ||
t.at(-1).length >= Math.ceil(list.length / rows) ||
list.length - i === rows - t.length
) {
t.push([]);
}
t.at(-1).push(l);
return t;
}, []);
console.log(transform([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 4));
console.log(transform([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5));
console.log(transform([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 6));
console.log(transform([1, 2, 3, 4, 5, 6], 4));
console.log(transform([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 4));
console.log(transform([1, 2, 3, 4], 5));
Other relevant documentations:
Array.prototype.at
Array.prototype.push

Split array into different size chunks (4, 3, 3, 3, 4, 3, 3, 3, etc)

I have an array like so: [1, 2, 3, 4, 5, 6, 7, 9, 10]. I need to chunk it into different size chunks, yet with a simple pattern of: 4, 3, 3, 3, 4, 3, 3, 3 like so:
[
[ // four
1,
2,
3,
4
],
[ // three (1/3)
5,
6,
7
],
[ // three (2/3)
8,
9,
10
],
[ // three (3/3)
11,
12,
13
],
[ // four
14,
15,
16,
17
],
[ // three (1/3)
18,
19,
20
], // and so on..
]
I have tried with this code I have customized:
const arr; // my array of values
const chuncked = arr.reduce((acc, product, i) => {
if (i % 3) {
return acc;
} else if (!didFourWayReduce) {
didFourWayReduce = true;
fourWayReduces++;
if ((fourWayReduces - 1) % 2) { // only make every second a "4 row"
return [...acc, arr.slice(i, i + 3)];
} else {
return [...acc, arr.slice(i, i + 4)];
}
} else {
didFourWayReduce = false;
return [...acc, arr.slice(i, i + 3)];
}
}, []);
And it works, almost, expect that the first chunk of threes (1/3) have the last element of the chunk with 4. So 1 key is repeated every first chunk of three. Like so:
[
[
1,
2,
3,
4
],
[
4, // this one is repeated, and it shouldn't be
5,
6
]
]

You could take two indices, one for the data array and one for sizes. Then slice the array with a given length and push the chunk to the chunks array.
Proceed until end of data.
var data = Array.from({ length: 26 }, (_, i) => i + 1),
sizes = [4, 3, 3, 3],
i = 0,
j = 0,
chunks = [];
while (i < data.length) chunks.push(data.slice(i, i += sizes[j++ % sizes.length]));
console.log(chunks);
.as-console-wrapper { max-height: 100% !important; top: 0; }

const arr = Array.from({ length: 100 }, (_, i) => i);
const copy = [...arr];
const sizes = [4, 3, 3, 3];
const result = [];
let i = 0;
while (i <= arr.length && copy.length) {
result.push(copy.splice(0, sizes[i % sizes.length]));
i++;
}
console.log(result);

A recursive approach is fairly elegant:
const chunks = (xs, [s, ...ss]) =>
xs.length ? [xs .slice (0, s), ... chunks (xs .slice (s), [...ss, s])] : []
const data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
const sizes = [4, 3, 3, 3]
console .log (chunks (data, sizes))
.as-console-wrapper { max-height: 100% !important; top: 0; }
By replacing [s, ...ss] with [...ss, s], we pass a cycled version of the sizes array, so that for instance, [4, 3, 3, 3] becomes [3, 3, 3, 4]. This makes it easy to parse step-by-step.

Mod operator to check if it should be 4 or 3. Use two arrays just to make it easier (can be done with one)
const groupIt = arr => arr.reduce(({
group,
out
}, v, i) => {
var max = out.length % 4 === 0 ? 4 : 3
group.push(v)
if (group.length === max || i === arr.length - 1) {
out.push(group)
group = []
}
return {
group,
out
}
}, {
group: [],
out: []
}).out
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x,i) => i + 1)
console.log(groupIt(test))
with just one:
const groupIt = arr => arr.reduce((out, v, i) => {
var max = (out.length - 1) % 4 === 0 ? 4 : 3
out[out.length - 1].push(v)
if (out[out.length - 1].length === max) {
out.push([])
}
return out
}, [[]])
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x, i) => i + 1)
console.log(groupIt(test))

This answer is similar to that of Nina Scholz, but uses a for loop, which I personally find more clear.
const arr = Array.from({length: 100}, (_, i) => i + 1);
const sizes = [4, 3, 3, 3];
const result = [];
for (let i = 0, j = 0; i < arr.length; i += sizes[j], j = (j + 1) % sizes.length) {
result.push(arr.slice(i, i + sizes[j]));
}
console.log(result);

How to arrange items in an array in all possible ways?

Items in an array can be arranged in ascending order using sort() method in JavaScript but how to arrange them in all possible ways and show them in our web page.

You describe permutations, one way to implement it:
function permutations(arr, r=[]) {
if (arr.length === 0) {
console.log(r)
} else {
const first = arr[0]
for (let i = 0; i <= r.length; i++) {
permutations(arr.slice(1), r.slice(0, i).concat([first]).concat(r.slice(i)))
}
}
}
permutations([1, 2, 3])
OUTPUT
[ 3, 2, 1 ]
[ 2, 3, 1 ]
[ 2, 1, 3 ]
[ 3, 1, 2 ]
[ 1, 3, 2 ]
[ 1, 2, 3 ]

javascript: Trying to flatten array only one level

I am trying to write a function to flatten an array. I have part of the function working and I need help in the other half.
flatten: function(anyArray, singleLevel) {
if (singleLevel == true) {
flatArray = Array.prototype.concat.apply([], anyArray);
return flatArray;
}
flatArray = Array.prototype.concat.apply([], anyArray);
if (flatArray.length != anyArray.length) {
flatArray = someObject.array.flatten(flatArray);
}
return flatArray;
}
if I type
.flatten([[[1],[1,2,3,[4,5],4],[2,3]]], true);
I want it to flatten only one level:
[[1],[1,2,3,[4,5],4],[2,3]]

Modern JavaScript allows us to handle this very easily using a variety of techniques
Using Array.prototype.flat -
const arr =
[ [ 1 ], [ 2, 3, [ 4, 5, [ 6 ] ] ], [ 7, [ 8, 9 ] ] ]
const flatArr =
arr.flat(1) // 1 is default depth
console.log(JSON.stringify(arr))
console.log(JSON.stringify(flatArr))
// [[1],[2,3,[4,5,[6]]],[7,[8,9]]]
// [1,2,3,[4,5,[6]],7,[8,9]]
Using Array.prototype.flatMap -
const arr =
[ [ 1 ], [ 2, 3, [ 4, 5, [ 6 ] ] ], [ 7, [ 8, 9 ] ] ]
const flatArr =
arr.flatMap(x => x)
console.log(JSON.stringify(arr))
console.log(JSON.stringify(flatArr))
// [[1],[2,3,[4,5,[6]]],[7,[8,9]]]
// [1,2,3,[4,5,[6]],7,[8,9]]
Using a spread argument to Array.prototype.concat
const arr =
[ [ 1 ], [ 2, 3, [ 4, 5, [ 6 ] ] ], [ 7, [ 8, 9 ] ] ]
const flatArr =
[].concat(...arr)
console.log(JSON.stringify(arr))
console.log(JSON.stringify(flatArr))
// [[1],[2,3,[4,5,[6]]],[7,[8,9]]]
// [1,2,3,[4,5,[6]],7,[8,9]]
Older version of JavaScript (ECMAScript 5 and below) can use techniques like Function.prototype.apply -
var arr =
[ [ 1 ], [ 2, 3, [ 4, 5, [ 6 ] ] ], [ 7, [ 8, 9 ] ] ]
var flatArr =
Array.prototype.concat.apply([], arr)
console.log(JSON.stringify(arr))
console.log(JSON.stringify(flatArr))
// [[1],[2,3,[4,5,[6]]],[7,[8,9]]]
// [1,2,3,[4,5,[6]],7,[8,9]]
Using Array.prototype.reduce -
var arr =
[ [ 1 ], [ 2, 3, [ 4, 5, [ 6 ] ] ], [ 7, [ 8, 9 ] ] ]
var flatArr =
arr.reduce((r, a) => r.concat(a), [])
console.log(JSON.stringify(arr))
console.log(JSON.stringify(flatArr))
// [[1],[2,3,[4,5,[6]]],[7,[8,9]]]
// [1,2,3,[4,5,[6]],7,[8,9]]
Using a primitive for loop -
var arr =
[ [ 1 ], [ 2, 3, [ 4, 5, [ 6 ] ] ], [ 7, [ 8, 9 ] ] ]
var flatArr =
[]
for (var i = 0; i < arr.length; i = i + 1)
flatArr = flatArr.concat(arr[i])
console.log(JSON.stringify(arr))
console.log(JSON.stringify(flatArr))
// [[1],[2,3,[4,5,[6]]],[7,[8,9]]]
// [1,2,3,[4,5,[6]],7,[8,9]]

The concat array method expects one or more arrays as arguments, whose elements will be appended:
[1].concat([2, 3], [4]) // [1, 2, 3, 4]
So if you are using apply, that will flatten another level:
[].concat.apply([1], [[2], [3]]) // === [1].concat([2], [3])
So you can either use push instead of concat, or call (or just direct invocation) instead of apply to get only a single flattening level.

if you use ES6/ES2015 you can use spread operator. Something like this
console.log(...[[[1],[1,2,3,[4,5],4],[2,3]]])

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