i am trying to get sequence index of each array of array like I have an element of the array and that 5 arrays have multiple elements and I want to get the first index of each array then the second and so.
below is my array data
const arrData = [[1,2,4,6],[3,8,7],[12,13],[],[9,10]];
and my expected outout [1,3,12,9,2,8,13,10,4,7,6]
give me some sort of solution to fix my issue
Following might work:
const src = [[1,2,4,6],[3,8,7],[12,13],[],[9,10]],
pickByOne = arr => {
const result = [],
{length} = arr.flat()
while(result.length < length){
for(a of arr){
a.length && result.push(a.shift())
}
}
return result
}
console.log(pickByOne(src))You can also use zip from lodash. Since the function takes a sequence of arguments and you have an array, you'll need to spread the array with ...
const _ = require("lodash");
const arrData = [[1,2,4,6],[3,8,7],[12,13],[],[9,10]];
const z = _.zip(...arrData)
This will give you the following result:
[
[ 1, 3, 12, undefined, 9 ],
[ 2, 8, 13, undefined, 10 ],
[ 4, 7, undefined, undefined, undefined ],
[ 6, undefined, undefined, undefined, undefined ]
]
You can already see it's in the order that you want. Now to clean things up. You can use flatMap in its simplest form to flatten the array.
// lodash
const fm = _.flatMap(z)
// or with Array.prototype.flat()
const fm = z.flat()
This will give you a single array.
[
1, 3, 12,
undefined, 9, 2,
8, 13, undefined,
10, 4, 7,
undefined, undefined, undefined,
6, undefined, undefined,
undefined, undefined
]
Now you just need to remove all the undefined elements by filtering (keeping) all the elements that are not undefined.
// lodash
_.filter(fm, _.negate(_.isUndefined))
and you have your final result.
[1, 3, 12, 9, 2, 8, 13, 10, 4, 7, 6]
There are many ways to solve this issue so you'll need to decide based on your particular case and balance simplicity, code legibility, efficiency, etc.
_.reject(fm, _.isUndefined) instead of _.filter(fm, _.negate(_.isUndefined))Find the longest length of the sub-array using Array.map() to get the length of all sub-arrays, and Math.max(). Use Array.from() to create an array with the length of the longest length. For each index in the created array, use Array.flatMap() to get all items from the original sub-arrays. If the item is undefined, use the Nullish coalescing operator (??) to substitute it with an empty array, which Array.flatMap() would ignore when flattening the resulting array. Flatten the output of Array.from() using Array.flat() to get a single array.
const fn = arr =>
Array.from({ length: Math.max(...arr.map(o => o.length)) }, (_, i) =>
arr.flatMap(o => o[i] ?? [])
).flat();
const arrData = [[1,2,4,6],[3,8,7],[12,13],[],[9,10]];
const result = fn(arrData);
console.log(result);Please try my below solution may be it will fix your
let data = [[], [], [1, 2, 4, 6], [8, 9, 10], [], [12], [14, 15], []], index = 0;
data = data.filter(o => o.length != 0)
const result = [];
for (i = 0; i < data.length; i++) {
if (data[i].length > index) {
index = data[i].length
}
if (data[i].length > 0) {
for (j = 0; j < index; j++) {
if (data[j][i] != undefined) {
result.push(data[j][i]);
}
}
}
}
console.log("Result", result);
[1, 2, 4, 6] to [1], and result would be [1, 9, 10].Since it's a nested array, you can loop through and check if the first element of each nested array is undefined or not. If no, you can push it into the result array. I wrote the function for your ref. Please check
const arrData = [[1, 2, 4, 6], [3, 8, 7], [12, 13], [], [9, 10]];
function sorter(arr) {
var result = [];
if (arr.length > 0) {
for (var i = 0; i < arr.length; i++) {
if (arr[i].length > 0) {
if (arr[i][0] !== undefined) {
result.push(arr[i][0]);
}
}
}
}
return result;
}
console.log(sorter(arrData));
I hope this is what you expected! Please reach me if you need clarification.
You need to keep track of the index which defines the position of the elements to get in each iteration.
Use a while loop that would terminate when there are no remaining elements in any of the arrays at a certain index.
Use .map to get a list of elements in each array at the index
Remove undefined elements in this list using .filter in case one of the arrays has no element at an index
Use .isEmpty to check if this list is empty, then, terminate
const arrData = [[1,2,4,6],[3,8,7],[12,13],[],[9,10]];
let res = [], index = 0;
while(true) {
const elementsAtIndex = _.filter(
_.map(arrData, e => e[index]),
e => e!==undefined
);
if(_.isEmpty(elementsAtIndex)) break;
res = [...res, ...elementsAtIndex];
index++;
}
console.log(...res);<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.20/lodash.min.js" integrity="sha512-90vH1Z83AJY9DmlWa8WkjkV79yfS2n2Oxhsi2dZbIv0nC4E6m5AbH8Nh156kkM7JePmqD6tcZsfad1ueoaovww==" crossorigin="anonymous"></script>
Simple one without any operator whould look like....
const ipArr = [ [1, 2, 4, 6], [8, 9, 10], [], [12], [14, 15], [] ]
let maxLength = 0
for (let i = 0; i < ipArr.length; i++) {
if (ipArr[i].length > maxLength) maxLength = ipArr[i].length
}
const opArr = []
for (let j = 0; j < maxLength; j++) {
for (let k = 0; k < ipArr.length; k++) {
if (ipArr[k][j]) opArr.push(ipArr[k][j])
}
}
console.log('opArr = ', opArr)
