I'm trying to write a function, that accepts an array of arrays as argument and filtering this array under certain condition. I will explain this condition later.
For example I have such array:
const arr = [
[1, 1, 20],
[2, 1, 15],
[3, 1.5, 15],
[4, 1, 15],
[5, 1, 20],
[6, 1.5, 15],
[7, 1, 25],
[8, 1, 15],
[9, 0, 15],
[10, 0, 15],
[11, 0, 15],
]
and the condition. I want to filter it by the value of the third column, but only if it's at least three in a sequence. The function should accept two arguments (or more if needed): value and array
So for example
const filterByValue = (array, val) => { //definition
return filtredArr
}
const newArr = filterByValue(array,15)
and now newArr should equal:
[
[2, 1, 15],
[3, 1.5, 15],
[4, 1, 15],
[8, 1, 15],
[9, 0, 15],
[10, 0, 15],
[11, 0, 15],
]
For now I only made:
const filterByValue = (arr, value) => {
const newArr = arr.filter((elem, index) => elem[2] === value)
return newArr
}
but this function returns
[
[2, 1, 15],
[3, 1.5, 15],
[4, 1, 15],
[6, 1.5, 15],
[8, 1, 15],
[9, 0, 15],
[10, 0, 15],
[11, 0, 15],
]
there shouldn't be [6,1.5,15]. I have no idea how it could be done to return only fragments of an array that contain at least three internal arrays in a row. Maybe any of you have any idea?
edit more explanation
This sequence means to me that I want only output which internal arrays that contain a set value in the third value (value argument in a function), but that they also have internal arrays that follow each other (at least three). Let's assume that the function accepts arr (first array in my post) and value = 15. The first internal array contains 20 as the third value, so it falls off. The second is okey, the third also and the fourth also - and should return in this situation 2,3,4 internal board (because they follow three in succession). Then the fifth value is 20, the sixth value is 15, but the next (seventh) value 25 again, so the function should not take into account the sixth table (because it is not followed by at least two tables that have the value 15). In the table 8,9,10,11 we have the value 15, and these are four tables (i.e. at least three). Therefore, the function should return an array containing the following tables 2,3,4 and 8,9,10,11. The sixth array contains the value 15, but not followed by two more that would have this value as well. The expected output is the second board in my post.
const filterByValue = (arr,value) => {
let newArray = [];
let cache = [];
arr.forEach(elem => {
if (elem[2] === value) {
cache.push(elem);
} else {
if (cache.length >= 3) {
newArray = newArray.concat(cache);
}
cache = [];
}
});
if (cache.length >= 3) {
newArray = newArray.concat(cache);
}
return newArray;
}
You could take a closure over an index and check if the values are in a sequence of three.
const
array = [[1, 1, 20], [2, 1, 15], [3, 1.5, 15], [4, 1, 15], [5, 1, 20], [6, 1.5, 15], [7, 1, 25], [8, 1, 15], [9, 0, 15], [10, 0, 15], [11, 0, 15]],
value = 15,
result = array.filter(
(index => ({ 2: v }, i, a) => {
if (v !== value) return false;
if (index === i || a[i + 1][2] === value && a[i + 2][2] === value) {
index = i + 1;
return true;
}
})
(-1)
);
console.log(result);
Try with filter and conditions based on whether match the value or not.
const arr = [
[1, 1, 20],
[2, 1, 15],
[3, 1.5, 15],
[4, 1, 15],
[5, 1, 20],
[6, 1.5, 15],
[7, 1, 25],
[8, 1, 15],
[9, 0, 15],
[10, 0, 15],
[11, 0, 15],
];
const filterByValue = (array, val) =>
array.filter((ar, i, items) => {
if (ar[2] !== val) {
return true;
} else {
return items[i - 1][2] === val || items[i + 1][2] === val;
}
});
console.log(filterByValue(arr, 15));
Without using the cache array.
Check if element contains number 15 if yes increment the counter if that counter reaches at least 3 it means you can store those 3 element into a new list. Use the the difference between the current index and the counter to get from the first sequence element to the last one. Repeat the process until end of the array is reached.
const arr = [
[1, 1, 20],
[2, 1, 15],
[3, 1.5, 15],
[4, 1, 15],
[5, 1, 20],
[6, 1.5, 15],
[7, 1, 25],
[8, 1, 15],
[9, 0, 15],
[10, 0, 15],
[11, 0, 15]
];
let cnt = 0;
let list = [];
for (let i = 0; i < arr.length; i++) {
// check if elements meet criteria
while (i < arr.length && arr[i][arr[i].length - 1] === 15) {
++i;
++cnt;
}
// acumulate into your array if at least 3 elm
if (cnt >= 3) {
for (let j = i - cnt; j < i; j++) {
list.push(arr[j]);
}
}
cnt = 0;
}
console.log(list)
Related
I need to find Max values for any number of arrays within the function. For example, user can add 1 array or 10 arrays. Output should show max values for each of the arrays.
Currently, my code works only if 1 array is provided.
Please let me know what is wrong.
function getMaxs(args){
array = args;
var max = array[0];
for(i = 0; i < array.length; i++){
if(max < array[i]){
max = array[i];
}
}
console.log(max);
return max;
}
getMaxs([5, 6, 7], [18, 19, 20], [5, 7, 3, 10]); // output is 7
getMaxs([5, 6, 7]); // output is 7
getMaxs([18, 19, 20], [5, 10, 74, 394]); // output is 20
If you need a max for every provided array:
const getMaxs = (...args) => args.map(arr => Math.max(...arr));
console.log( getMaxs([5, 6, 7], [18, 19, 20], [5, 7, 3, 10]) ); // [7, 20, 10]
console.log( getMaxs([5, 6, 7]) ); // [7]
console.log( getMaxs([18, 19, 20], [5, 10, 74, 394]) ); // [20, 394]
If you don't care which sub-array the result comes from, you can use Array.prototype.flat:
function getMaxs(array) {
const flatArray = array.flat();
return Math.max(...flatArray);
}
const arr = [[5, 6, 7], [18, 19, 20], [5, 7, 3, 10]];
const max = getMaxs(arr);
console.log(max);
I believe you're looking for rest parameters.
Something along these lines:
function getMaxs(...arrays) {
return Math.max(...arrays.flat());
}
// usage
getMaxs([5, 6, 7], [18, 19, 20], [5, 7, 3, 10]); // output is 20
getMaxs([5, 6, 7]); // output is 7
getMaxs([18, 19, 20], [5, 10, 74, 394]); // output is 394
Related: The arguments object in case you can't use rest parameters. In that case, you probably don't support flat either, which in that case I'll recommend you to look for a polyfill somewhere.
You can use the .flat() function, so the arrays would merge and become one, like this:
function getMaxs(args){
array = args.flat();
//it will now see it as one big array
....
}
But, the .flat() function may not work on older browsers, so you may also do this:
array = args.reduce((acc, val) => acc.concat(val), [])
The effect will be the same.
Iraklis solution is fine, but it suppose you give an array of arrays. However, if you want to pass arrays as separated arguments, you could use the Spread syntax and Rest parameters:
function getMaxs(...arrays) {
const max = Math.max(...arrays.flat())
console.log(max) // DEBUG
return max
}
getMaxs([5, 6, 7], [18, 19, 20], [5, 7, 3, 10])
getMaxs([5, 6, 7])
getMaxs([18, 19, 20], [5, 10, 74, 394])
Of course, you should be careful about browser compatibility:
function getMaxs(...args) {
array = args.flat();
var max = array[0];
for (i = 0; i < array.length; i++) {
if (max < array[i]) {
max = array[i];
}
}
// console.log(max);
return max;
}
console.log(getMaxs([5, 6, 7], [18, 19, 20], [5, 7, 3, 10]));
console.log(getMaxs([5, 6, 7]));
console.log(getMaxs([18, 19, 20], [5, 10, 74, 394]));
the above one will be the solution for your question. You can check the below screenshots for your reference.
For this input i have added screen shots for your reference how it will process the code
getMaxs([5, 6, 7], [18, 19, 20], [5, 7, 3, 10])
now it will take the whole
[![enter image description here][1]][1]
//I am trying to learn/understand to create a one-dimensional array which contains, in exact order, the array indices used to access a given number within a multi dimensional array
var multiDimensionalArray = [1, 2, 3, [4, 5, 6, [7, 8, 9, [10, 11, 12, [13, 14, [15]]]]]];
// to access 15...
fifteen = multiDimensionalArray[3][3][3][3][2][0];
// Define the variable 'indexArray' here:
indexArray = fifteen.join();
//join method does not work for me. I have tried concat, slice, indexOf methods. Can't seem to be able to find the solution for this part. HELP!
// This will log your variables to the console
console.log(fifteen); //15
console.log(indexArray); //[3, 3, 3, 3, 2, 0]
You can use a recursive function :
var multiDimensionalArray = [1, 2, 3, [4, 5, 6, [7, 8, 9, [10, 11, 12, [13, 14, [15]]]]]];
const getIndex = (arr, num) => {
const path = [];
for (let i = 0; i < arr.length; i++) {
if (Array.isArray(arr[i])) {
path.push(i);
path.push(getIndex(arr[i], num));
break;
}
if (arr[i] === num) {
path.push(i);
break;
}
}
return path.flat();
};
const indexArray = getIndex(multiDimensionalArray, 15);
console.log(indexArray);
Another approach with a recursive function:
var multiDimensionalArray = [1, 2, 3, [4, 5, 6, [7, 8, 9, [10, 11, 12, [13, 14, [15]]]]]];
const find = (node, needle) => {
const index = node.indexOf(needle);
const nextIndex = node.findIndex(Array.isArray);
return [].concat(...
(index !== -1)
? [index]
: [nextIndex, find(node[nextIndex], needle)]
);
};
const result = find(multiDimensionalArray, 15);
console.log(result);
You could take a recursive approach and return either an empty array fro a not found value or the indices.
function getIndices(array, value) {
var subIndices = [],
index = array.findIndex(v => v === value || Array.isArray(v) && (subIndices = getIndices(v, value)).length);
return index === -1
? []
: [index, ...subIndices];
}
var array = [1, 2, 3, [4, 5, 6, [7, 8, 9, [10, 11, 12, [13, 14, [15]]]]]]
console.log(getIndices(array, 15));
console.log(getIndices(array, 42));
Problem description:
The idea is to insert into existing intervals new interval which doesn't merge with existing intervals but fills the missing gaps between intervals. (This is not the interval merging problem)
For example, inserting interval [0, 7] to intervals [[0, 1], [3, 5]] would result new intervals with gaps filled [[0, 1], [1, 3], [3, 5], [5, 7]].
Interval range is already sorted smallest to larger [[0, 1], [3, 5]].
My current solution is a bit "broken", I ended up using too many if checks to cover some special cases which makes everything more complex then needed. I am looking for better ways to simplify the condition part. In the bottom of the code there are test cases included, also cases where my solution fails.
The test cases where my algorithm is failing and producing wrong results:
assert.deepEqual( // Broken
insertIntervalSec([[1, 5], [7, 10]], [4, 12]),
[[1, 5], [5, 7], [7, 10], [10, 12]],
);
assert.deepEqual(insertIntervalSec([[1, 1]], [1, 3]), [[1, 3]]); // Broken
function isOverLapping(a, b) {
return Math.max(a[0], b[0]) <= Math.min(a[1], b[1]);
}
function insertIntervalSec(arr, interval) {
const result = [];
let i = 0;
const contains = (a, b) => {
return a[0] >= b[0] && a[1] <= b[1]
};
if (arr.length <= 0) {
result.push(interval);
return result;
}
if (arr.length === 1 && contains(interval, arr[0])) {
result.push(interval);
return result;
}
// Start point
if (interval[1] >= arr[0][0] && isOverLapping(interval, arr[0])) {
result.push([interval[0], arr[0][0]]);
} else if (interval[1] <= arr[0][0]) {
result.push([interval[0], Math.min(interval[1], arr[0][0])]);
}
while (i < arr.length) {
const current = arr[i];
result.push(arr[i]);
if (!contains(interval, arr[i]) && isOverLapping(arr[i], interval)) {
const next = arr[i + 1];
// Special handling for the last item
if (next !== undefined) {
if (interval[1] > current[1]) {
result.push([current[1], next[0]]);
}
} else {
if (interval[0] <= current[0] && interval[1] <= current[1]) {
// TODO: No action
} else if (interval[0] >= current[0] || interval[1] >= current[0]) {
result.push([current[1], interval[1]]);
}
}
}
i++;
}
// End point
const len = arr.length;
const last = arr[len - 1];
if (last[1] <= interval[0] && !isOverLapping(last, interval)) {
result.push(interval);
}
return result;
}
assert.deepEqual(
insertIntervalSec([[1, 5], [10, 15], [20, 25]], [12, 27]),
[[1, 5], [10, 15], [15, 20], [20, 25], [25, 27]]
);
assert.deepEqual(
insertIntervalSec([[1, 5], [10, 15], [20, 25]], [-3, 0]),
[[-3, 0], [1, 5], [10, 15], [20, 25]]
);
assert.deepEqual(
insertIntervalSec([[1, 5], [10, 15], [20, 25]], [-3, 3]),
[[-3, 1], [1, 5], [10, 15], [20, 25]]
);
assert.deepEqual(
insertIntervalSec([[0, 5], [10, 15], [20, 25]], [15, 15]),
[[0, 5], [10, 15], [20, 25]]
);
assert.deepEqual(
insertIntervalSec([[0, 5], [10, 15], [20, 25]], [20, 21]),
[[0, 5], [10, 15], [20, 25]]
);
assert.deepEqual(
insertIntervalSec([[0, 5], [10, 15], [20, 25]], [26, 27]),
[[0, 5], [10, 15], [20, 25], [26, 27]]
);
assert.deepEqual(
insertIntervalSec([[0, 5], [10, 15], [20, 25]], [25, 27]),
[[0, 5], [10, 15], [20, 25], [25, 27]]
);
assert.deepEqual(insertIntervalSec([], [25, 27]), [[25, 27]]);
assert.deepEqual(insertIntervalSec([[1, 1]], [1, 1]), [[1, 1]]);
assert.deepEqual( // Broken
insertIntervalSec([[1, 5], [7, 10]], [4, 12]),
[[1, 5], [5, 7], [7, 10], [10, 12]],
);
assert.deepEqual(insertIntervalSec([[1, 1]], [1, 3]), [[1, 3]]); // Broken
assert.deepEqual(
insertIntervalSec2([[5, 5]], [6, 6]),
[[5, 5], [6, 6]]
);
assert.deepEqual(
insertIntervalSec2([[1, 3]], [6, 6]),
[[1, 3], [6, 6]]
);
With the exception of the last test case (see comment on question), this passes all the tests. The basic idea is you just keep track of start variable that indicated where how much of the inserted range you have used. This allows you to narrow it down to three cases:
the inserted interval fits entirely before the the current item
the current item in the iteration fits completely before the inserted interval
the item in the iteration overlaps.
After iterating the items, you can check if the inserted range has anything left to insert:
function insertIntervalSec(arr, insert) {
let start = insert[0]
let res = []
for (i = 0; i < arr.length; i++) {
let a = arr[i]
// smaller item in range
if (a[0] <= start) {
res.push(a)
start = Math.max(a[1], start)
continue
}
// moved past inserted interval add rest of arr
if (start >= insert[1]) {
res.push(...arr.splice(i))
break
}
// fill in spaces
let end = Math.min(insert[1], a[0])
res.push([start, end], a)
start = a[1]
}
// clean up left over range
if (start < insert[1]) res.push([start, insert[1]])
return res
}
console.log(insertIntervalSec([ [1, 5],[10, 15],[20, 25]], [-2, 27]))
I have multiple arrays in a main/parent array like this:
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
here are the array's for simpler reading:
[1, 17]
[1, 17]
[1, 17]
[2, 12]
[5, 9]
[2, 12]
[6, 2]
[2, 12]
[2, 12]
I want to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable. So in this example, var repeatedArrays would be [1, 17] and [2, 12].
So this should be the final result:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]
I found something similar here but it uses underscore.js and lodash.
How could I it with javascript or even jquery (if need be)?
Try this
array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))
console.log(JSON.stringify(r));
Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.
Explanation
The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.
UPDATE
Even shorter version based on #ken comment can be written (it should always work with arrays of numbers). The original longer version of #ken code is in snippet and shows how #ken uses in clever way second argument of .filter to avoid usage global variable r.
array.filter(a=>!(2-(this[a]=++this[a]|0)))
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
var r= array.filter(a=>!(2-(this[a]=++this[a]|0)), {})
console.log(JSON.stringify(r));
You could take a Map with stringified arrays and count, then filter by count and restore the arrays.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = Array
.from(array.reduce(
(map, array) =>
(json => map.set(json, (map.get(json) || 0) + 1))
(JSON.stringify(array)),
new Map
))
.filter(([, count]) => count > 2)
.map(([json]) => JSON.parse(json));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Filter with a map at wanted count.
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(map => a =>
(json =>
(count => map.set(json, count) && !(2 - count))
(1 + map.get(json) || 1)
)
(JSON.stringify(a))
)
(new Map)
);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Unique!
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
result = array.filter(
(s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
(new Set)
);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
You can use Object.reduce, Object.entries for this like below
var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let res = Object.entries(
array.reduce((o, d) => {
let key = d.join('-')
o[key] = (o[key] || 0) + 1
return o
}, {}))
.flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : [])
console.log(res)
OR may be just with Array.filters
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
let temp = {}
let res = array.filter(d => {
let key = d.join('-')
temp[key] = (temp[key] || 0) + 1
return temp[key] == 3
})
console.log(res)
For a different take, you can first sort your list, then loop through once and pull out the elements that meet your requirement. This will probably be faster than stringifying keys from the array even with the sort:
var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res = []
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)
You could also do this with a single Array.reduce where you would only push to a result property if the length is equal to 3:
var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];
console.log(array.reduce((r,c) => {
let key = c.join('-')
r[key] = (r[key] || 0) + 1
r[key] == 3 ? r.result.push(c) : 0 // if we have a hit push to result
return r
}, { result: []}).result) // print the result property
ES6:
const repeatMap = {}
array.forEach(arr => {
const key = JSON.stringify(arr)
if (repeatMap[key]) {
repeatMap[key]++
} else {
repeatMap[key] = 1
}
})
const repeatedArrays = Object.keys(repeatMap)
.filter(key => repeatMap[key] >= 3)
.map(key => JSON.parse(key))
I have nested array data and I would like to extract all nested arrays to be siblings of their parent. I am pretty close, but I am getting an extra empty array in the results and I cannot figure out where it is coming from or how to get rid of it.
Note: I would really like to understand why this is happening and how to get rid of it in my function, and not just a .filter(arr => arr.length) on my results list.
This is my attempt so far:
var arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
];
// Desired Output
// [
// [1, 2, 5],
// [3, 4],
// [6],
// [7, 8, 9],
// [10, 11],
// [12, 13],
// [14, 15],
// [16, 17],
// [4, 4]
// [1]
// [1, 1]
// ]
function extractArrays (arr) {
return arr.reduce((res, curr) => {
if (Array.isArray(curr)) {
res = res.concat(extractArrays(curr));
}
else {
res[0].push(curr);
}
return res;
}, [[]]);
}
console.log(extractArrays(arrs));
// Results:
// [
// [], <-- Where is this coming from?
// [ 1, 2, 5 ],
// [ 3, 4 ],
// [ 6 ],
// [ 7, 8, 9 ],
// [ 10, 11 ],
// [ 12, 13 ],
// [], <-- Also here
// [ 14, 15 ],
// [ 16, 17 ],
// [ 4, 4 ],
// [ 1 ],
// [ 1, 1 ]
// ]
.as-console-wrapper {
max-height: 100% !important;
}
Element like [[14, 15], [16, 17]] will introduce a [] after recursion. This should be handled by checking length.
var arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
];
function extractArrays (arr, acc=[]) {
if (arr.length == 0 ) return acc;
let pure = arr.filter(elm => !Array.isArray(elm));
if (pure.length > 0) {
acc.push(pure);
}
acc.concat(arr.filter(elm => Array.isArray(elm)).map(elm => extractArrays(elm, acc)));
return acc;
}
console.log(extractArrays(arrs));
You can try the following code
var arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[
[14, 15],
[16, 17]
], // <-- added additional test case
[
[1], 4, [1, 1], 4
]
];
function extractArrays(arr) {
return arr.reduce((res, curr, i) => {
if (Array.isArray(curr)) {
res = res.concat(extractArrays(curr));
} else {
let index = 0;
for (let j = 0; j <= i; j++) {
if (!Array.isArray(arr[j])) {
res[index] ? res[index].push(curr) : res.push([curr]);
break;
} else {
index++;
}
}
}
return res;
}, []); // <-- no initial empty array inside here
}
console.log(extractArrays(arrs));
I just wanted to share my approach to this problem, I enjoyed trying to solve it, in my case I also passed an array to the extractArrays method, in order to make easier to capture and filter every array inside the arrs param.
let result = [];
extractArrays(arrs, result);
console.log(result);
function extractArrays(arr, result) {
let newResult = arr.reduce((acc, curr) => {
if (Array.isArray(curr)) {
extractArrays(curr, result);
} else {
acc.push(curr);
}
return acc;
}, []);
newResult.length && result.push(newResult);
}
You can check it when you return from function. stackblitz
function extractArray(arr) {
const res = arr.reduce((res, curr) => {
if(!Array.isArray(curr)){
return [[...res[0], curr], ...res.slice(1)]
}
return [...res, ...extractArray(curr)]
}, [[]]);
return res[0].length ? res : res.slice(1);
}
EDIT: More performant function (check stackblitz link)
function extractFaster(arr) {
let res = [0];
function recExtract(arr) {
let hasNonArrayElm = false;
let index = res.length -1;
arr.forEach(curr => {
if (!Array.isArray(curr)) {
hasNonArrayElm ? res[index].push(curr) : res.splice(index, 0, [curr]);
hasNonArrayElm = true;
return;
}
recExtract(curr);
});
}
recExtract(arr);
res.splice(-1, 1)
return res;
}
EDIT: The answer below the line is a great way to flatten arrays, but I suggested it because I misunderstood this question. I will leave it in case it benefits someone to know, but in order to keep an accurate record, I'll also update my answer to address the problem posed in the question.
The accepted answer seems sufficient enough, but I'll try my hand at it. I would use Array.reduce to cover everything with one swoop, and inside use Array.filter to separate the normal items from the array items, then use the spread ... operator on the nested arrays so everything gets pushed to the same level, after recursively calling the same extract function on all nested arrays. Honestly, the explanation might be harder to understand than the code, have a look:
const data = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
]
const extractChildArrays = arrs => arrs.reduce((acc, cur) => {
const nestedArrs = cur.filter(a => Array.isArray(a))
const normalItems = cur.filter(a => !Array.isArray(a))
acc.push(normalItems, ...extractChildArrays(nestedArrs))
return acc
}, [])
console.log(extractChildArrays(data))
UPDATE: Array.flat() is now accepted as part of the spec and it's supported by all modern browsers except Edge.
In ES6 there is actually an experimental array method called flat(). As of the writing of this answer, it's only compatible with Chrome, but a polyfill might be worth looking into, because it's SO EASY!
The first parameter of flat() is depth, so with the help of another answer, you can easily figure that out dynamically.
const data = arrs = [
[1, 2, [3, 4], 5],
[6, [7, 8, 9, [10, 11]]],
[12, 13],
[[14, 15], [16, 17]],
[[1], 4, [1, 1], 4]
]
const flattenArray = arr => {
const getDepth = a => Array.isArray(a) ?
1 + Math.max(...a.map(getDepth)) : 0
return arr.flat(getDepth(arr))
}
console.log(flattenArray(data))