Let's say that I have this array.
const arr = [['grass', 'water'], ['fire', 'ground'], ['fairy', 'mage'], ['fighter', 'fire']];
So what I want is to concatenate arr[0] with arr[2], arr[1] with arr[3].
This has to be done trough some array method or loop, since I don't know how much of elements I will have.
Result should look like this
const arr = [['grass', 'water', 'fairy', 'mage'], ['fire', 'ground', 'fighter', 'fire']];
If you mod the current index with half the length of the array, then you would get corresponding index in the resultant array.
Index in original array
0 1 2 3 4 5
Index in resultant array
0 1 2 0 1 2 // After taking modulus with 3 i.e. half of the array length
Solution
I've used the Nullish coalescing operator to assign the res array with empty array if the desired position is empty.
Then push all the elements into this position and finally return the resultant.
NOTE: This solution only works for arrays of even length.
const arr = [
["grass", "water"],
["fire", "ground"],
["fairy", "mage"],
["fighter", "fire"],
["water", "ice"],
["ground", "rock"],
];
const result = arr.reduce(
(r, el, i) => (
(r[i % (arr.length / 2)] ??= []), r[i % (arr.length / 2)].push(...el), r
),
[]
);
console.log(result);
??= at times but this seems like a good usage. The solution is good - I like it. But a bit non-idiomatic. Makes it harder for some to figure it out. I'd write out the full body for clarity like this, as it's harder to read it as an expression. You can then just add the single expression body separately in the answer just to show it off. Not sure how to keep the logic but make it easier to comprehend. Maybe a conditional operator but I really dislike those. So, good solution but a bit niche.You can do something like this:
const arr = [['grass', 'water'], ['fire', 'ground'], ['fairy', 'mage'], ['fighter', 'fire']];
const newArr = [];
for (let i = 0; i < arr.length/2; i++){
newArr.push(arr[i].concat(arr[i + 2]));
}
console.log(newArr);Here are few options that work for even numbered arrays.
Iterate from the start to half the array.
Join the item at the current index with the one that is equal distance from the middle of the array:
[0, 1, 2, 3]
^ ^
[0, 1, 2, 3]
^ ^
Add each of these to a new array. The original and all its members are untouched.
function* slice(start, end, arr) {
if (start < 0)
start = arr.length + start;
if (end < 0)
end = arr.length + start;
for (let i = start; i < end; i++) {
yield arr[i];
}
}
const arr = [['grass', 'water'], ['fire', 'ground'], ['fairy', 'mage'], ['fighter', 'fire']];
const middle = arr.length / 2;
const result = Array.from(
slice(0, middle, arr),
(first, i) => first.concat(arr[middle + i])
);
console.log(result);const arr = [['grass', 'water'], ['fire', 'ground'], ['fairy', 'mage'], ['fighter', 'fire']];
const middle = arr.length / 2;
const result = arr
.slice(0, middle)
.map((first, i) => first.concat(arr[middle + i]));
console.log(result);To avoid intermediate arrays from .slice() you can use a generator function and pass it through Array.from() supplying a mapping function to generate an array:
const arr = [['grass', 'water'], ['fire', 'ground'], ['fairy', 'mage'], ['fighter', 'fire']];
const middle = arr.length / 2;
const result = arr
.slice(0, middle)
.map((first, i) => first.concat(arr[middle + i]));
console.log(result);Start before the middle of the array and go backwards.
Remove the last item in the array. Add its members to the item at the current index.
[0, 1, 2, 3]
^ ^
remove last: 3
combine with: 1
[0, 13, 2]
^ ^
remove last: 2
combine with: 0
[02, 13]
This is all done in-place. Both arr and its members are modified.
const arr = [['grass', 'water'], ['fire', 'ground'], ['fairy', 'mage'], ['fighter', 'fire']];
const middle = arr.length / 2;
for (let i = middle - 1; i >= 0; i--) {
const first = arr[i];
const second = arr.pop();
first.push(...second);
}
console.log(arr);