How to concat arrays alternatively every 2 items? JS - javascript

I have:
array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9];
array2 = ['a', 'b', 'c'];
I got the concat of arrays working, but only one at a time:
var alternatingArrayResult = [array1, array2].reduce(function (r, a) {
return a.forEach(function (a, i) {
return (r[i] = r[i] || []).push(a);
}), r;
}, []).reduce(function (a, b) {
return a.concat(b);
});
// => alternatingArrayResult = [1, 'a', 2, 'b', 3, 'c', 4, 5, 6, 7, 8, 9];
I want to add two items from array 1, then one from array 2 - and so on.
Desired output:
result = [1, 2, 'a', 3, 4, 'b', 5, 6, 'c', 7, 8, 9];
var array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9];
var array2 = ['a', 'b', 'c'];
var alternatingResultArray = [array1, array2].reduce(function (r, a) {
return a.forEach(function (a, i) {
return (r[i] = r[i] || []).push(a);
}), r;
}, []).reduce(function (a, b) {
return a.concat(b);
});
console.log(alternatingResultArray);

You could get a lenght and iterate the parts for pushing.
const
array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9],
array2 = ['a', 'b', 'c'],
result = [];
for (i = 0, l = Math.max(Math.ceil(array1.length / 2), array2.length); i < l; i++) {
result.push(
...array1.slice(i * 2, (i + 1) * 2),
...array2.slice(i, i + 1),
);
}
console.log(...result);
A more abstact version with size of the wanted subarrays.
const
array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9],
array2 = ['a', 'b', 'c'],
data = [array1, array2],
sizes = [2, 1]
result = [];
for (i = 0, l = Math.max(...data.map(({ length }, i) => Math.ceil(length / sizes[i]))); i < l; i++) {
data.forEach((a, j) => result.push(...a.slice(i * sizes[j], (i + 1) * sizes[j])));
}
console.log(...result);

Use Array.map with a thisArg for the desired result (see MDN).
thisArg (Optional) Value to use as this when executing callback.
let array1 = [1, 2, 3, 4, 5, 6, 7, 8, 9];
let array2 = ['a', 'b', 'c'];
let arrayConcatted = array1
.flatMap( function(v, i) {
// ^ not arrow to keep thisArg in scope
const arr2Value = i && i % 2 === 0 && this.length && this.shift();
return arr2Value ? [arr2Value, v] : v;
}, array2.slice())
// ^ 'this', a copy of array2
console.log(JSON.stringify(arrayConcatted));
// IE
var arrConcatIE = [];
var arr2Clone = array2.slice();
for (var i=0; i < array1.length; i +=1) {
const arr2Value = i && i % 2 === 0 && arr2Clone.length && arr2Clone.shift();
if (arr2Value) {
arrConcatIE.push(arr2Value);
}
arrConcatIE.push(array1[i]);
}
console.log(JSON.stringify(arrConcatIE));

Related

Find array element that is true for ALL elements in the second array

I'm trying to find the smallest common multiple from one array by comparing it to the values in the other by using % num === 0. The answer should be 6, but because each number in the first array is true at least once, they all get returned. How do I find the value that is only true and never false?
let arr = [1, 2, 3]
function test(arr) {
let x = [1, 2, 3, 4, 5, 6, 7, 8]
for (let n of arr) {
return x.filter(k => k % n === 0)
}
}
console.log(test(arr))
You need loop over the x array and return the first element that gets divided by every value in arr.
let arr = [1, 2, 3];
function test(arr) {
let x = [1, 2, 3, 4, 5, 6, 7, 8];
for (let n of x) {
if (arr.every((a) => n % a === 0)) {
return n;
}
}
}
console.log(test(arr));
You can also simply the solution using Array.prototype.find.
const
arr = [1, 2, 3],
test = (arr) =>
[1, 2, 3, 4, 5, 6, 7, 8].find((n) => arr.every((a) => n % a === 0));
console.log(test(arr));
Note: If x is not sorted, then you will have to sort it first.
const arr = [1, 2, 3],
test = (arr) =>
[8, 7, 6, 5, 4, 3, 2, 1]
.sort((a, b) => a - b)
.find((n) => arr.every((a) => n % a === 0));
console.log(test(arr));
Update based on OP's comment
You can use Array.prototype.filter and Array.prototype.some.
const arr = [1, 2, 3],
test = (arr) =>
[1, 2, 3, 4, 5, 6, 7, 8].filter((n) => arr.some((a) => n / a === 2));
console.log(test(arr));
If x is sorted can use find and every
let arr = [1, 2, 3]
let x = [1, 2, 3, 4, 5, 6, 7, 8, 12]
let res = x.find(n => arr.every(a => n%a === 0))
console.log(res)
if unsorted x
let arr = [1, 2, 3]
let x = [1, 12, 6, 4, 2, 7, 8]
let res = [...x].sort((a,b)=> a-b).find(n => arr.every(a => n%a === 0))
console.log(res)
Filter and intersect
let arr = [1, 2, 3]
let x = [1, 2, 3, 4, 5, 6, 7, 8]
function test(arr) {
let common = []
arr.forEach(n => common.push(x.filter(k => k % n === 0)))
return common.reduce((acc, cur) => acc.filter(e => cur.includes(e)));
}
console.log(test(arr))
Late to the party answer
You could solved this with one line of code using Array.reduce. Here we initialize it with an empty array and then use a ternary operator to append values that pass the remainder test. Note that we don't need to check for zero (c % n === 0) because we treat the result as a boolean.
x.reduce((p,c) => c % n ? p : p.concat([c]), [])
// TEST
const x = [1, 2, 3, 4, 5, 6, 7, 8];
[1, 2, 3].forEach(n => {
console.log(
"n =" + n,
x.reduce((p,c) => c % n ? p : p.concat([c]), [])
);
});

How to merge two arrays in JavaScript loop

Is there a better way to do this? Faster or more readable? Please share your approach.
const a = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
const b = [ 'a', 'b', 'c', 'd', 'e', 'f' ]
let i = 0
let j = 1
while (true) {
const item = b[i]
if (!item) break
a.splice(j, 0, item)
j += 2
i++
}
// expected output [0, 'a', 1, 'b', 2, 'c', 3, 'd', 4, 'e', 5, 'f', 6, 7, 8, 9]
You could iterate the array by the minimum of both array lengths and take the rest by slicing the arrays from the minimum length.
function merge(a, b) {
const
result = [];
l = Math.min(a.length, b.length);
for (let i = 0; i < l; i++) result.push(a[i], b[i]);
result.push(...a.slice(l), ...b.slice(l));
return result;
}
console.log(...merge([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], ['a', 'b', 'c', 'd', 'e', 'f']));
You can use either recursion:
const Nil =
Symbol();
const intercalate =
([x=Nil, ...xn], [y=Nil, ...yn], ret=[]) =>
x === Nil && y === Nil
? ret
: intercalate(xn, yn, ret.concat( x === Nil ? [] : x
, y === Nil ? [] : y
));
Or Array#flatMap:
const intercalate =
(xn, yn) =>
xn.flatMap((x, i) =>
i >= yn.length
? [x]
: [x, yn[i]]);
My approach:
function mergeBetween(a, b) {
let i = 0
let j = 1
while (true) {
const item = b[i]
if (!item) break
a.splice(j, 0, item)
j += 2
i++
}
return a
}
const a = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
const b = [ 'a', 'b', 'c', 'd', 'e', 'f' ]
mergeBetween(a, b) //[0, 'a', 1, 'b', 2, 'c', 3, 'd', 4, 'e', 5, 'f', 6, 7, 8, 9]
This would be my approach, if speed is your game, then you should stick to for loops... But I would suggest avoiding premature optimization in general... Not sure if that is what you meant by "faster" either...
const a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
const b = ["a", "b", "c", "d", "e", "f"];
// check which is longer
const longer = a.length > b.length ? a : b;
const shorter = a.length < b.length ? a : b;
// declare a new variable to hold the combined array
let newArray = [];
for (let i = 0; i < longer.length; i++)
newArray =
i < shorter.length
? [...newArray, longer[i], shorter[i]]
: [...newArray, longer[i]];
console.log(newArray)

Remove a specific amount of elements in the array

i want to specify how many 1 elements I want to remove within that array:
const array = [1, 1, 2, 3, 1, 5];
I tried like that:
const array = [1, 1, 2, 3, 1, 5];
const i = array.indexOf(1);
if (i > - 1) {
array.splice(i, 1);
}
but this remove just first element 1 in array
Try this, Array.prototype.filter will create a new array and will not mutate or change the array on which it is called.
const arr = [1, 1, 2, 3, 1, 5];
const newArr = arr.filter((item) => item !== 1);
console.log(newArr);
const removeArrayItem = (array, item) => array.filter((i) => i !== item);
console.log(removeArrayItem([1, 2, 2, 2, 2, 3, 4, 5], 2));
console.log(removeArrayItem([1, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8, 8, 8], 8));
// or using reduce
const removeDuplicateItem = (arr, itemToRemove, count) =>
arr.reduce(
(accumulator, currentItem) => {
if (
accumulator.count &&
accumulator.count > 0 &&
currentItem === itemToRemove
) {
accumulator.count -= 1;
return accumulator;
}
accumulator.result.push(currentItem);
return accumulator;
},
{
count,
result: [],
},
).result;
// This will remove 5 number twos
console.log(removeDuplicateItem([1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8, 8, 8], 2, 5));
You need to iterate the count and check if an item is available for removing.
function remove(array, value, count) {
while (count--) {
const i = array.indexOf(value);
if (i === -1) break;
array.splice(i, 1);
}
}
const array = [1, 1, 2, 3, 1, 5];
remove(array, 1, 2);
console.log(array);
If you want to remove only the amount of 1s you specify,
You can use a for loop with 2 conditions,
one for array size and the second for the number of times 1 was found and removed.
const array = [1, 1, 2, 3, 1, 5];
const counter = 2;
for(let i = 0 , n = 0; i < array.length && n < counter; i++) {
if(array[i] === 1) {
array.splice(i, 1);
i -= 1;
n++;
}
}
console.log(array);
You can use a while loop to remove N number of elements with this value:
const removeNInstancesOfX = (array, n, x) => {
let i = null, count = 0, arr = [...array];
while ((i = arr.indexOf(1)) !== -1 && count++ < n)
arr.splice(i, 1);
return arr;
}
let array = [1, 1, 2, 3, 1, 5];
array = removeNInstancesOfX(array, 2, 1);
console.log(...array);

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 can I check it an array contains all elements from another array, including count, in JavaScript?

I have two arrays:
arr1 = [1, 2, 3, 1, 2, 3, 4]
arr2 = [1, 3, 1, 1]
arr3 = [1, 1, 2, 2, 3]
Using arr1 as the baseline, arr2 does not match because it contains three 1's whereas arr1 only has two. arr3, however should return true because it has elements from arr1.
I tried
if(_.difference(arr2, arr1).length === 0)
But this does not take into account the number of occurrences
You could count all value from the first array and iterate the second with every and return early if a value is not given or zero, otherwise decrement the count and go on.
function check(a, b) {
var count = a.reduce((o, v) => (o[v] = (o[v] || 0) + 1, o), {});
return b.every(v => count[v] && count[v]--);
}
var arr1 = [1, 2, 3, 1, 2, 3, 4],
arr2 = [1, 3, 1, 1],
arr3 = [1, 1, 2, 2, 3];
console.log(check(arr1, arr2)); // false
console.log(check(arr1, arr3)); // true
You can try to loop through second array and compare t against main array if value found make main array cell to false and set flag as true
arr1 = [1, 2, 3, 1, 2, 3, 4]
arr2 = [1, 3, 1, 1]
arr3 = [1, 1, 2, 2, 3]
function checkArray(compareMe, toThis){
var flag = false;
for(var i = 0; i < toThis.length; i++){
flag = false;
for(var j = 0; j < compareMe.length; j++){
if(compareMe[j] == toThis[i]){
compareMe[j] = false;
flag = true;
break;
}
}
}
return flag;
}
console.log(checkArray(arr1, arr2));
console.log(checkArray(arr1, arr3));
Try this solution:
const arr1 = [1, 2, 3, 1, 2, 3, 4],
arr2 = [1, 3, 1, 1],
arr3 = [1, 1, 2, 2, 3],
arr4 = [1, 2, 3, 1, 2, 3, 4, 1];
function containsFn(src, trg) {
const srcCp = src.slice();
for(let i = 0; i < trg.length; i++) {
const index = srcCp.indexOf(trg[i]);
if(index > - 1) {
srcCp.splice(index, 1);
} else {
return false;
}
}
return true;
}
console.log(containsFn(arr1, arr2));
console.log(containsFn(arr1, arr3));
console.log(containsFn(arr1, arr4));
Looks like I'm already late to the party but this would be a recursive solution:
arr1 = [1, 2, 3, 1, 2, 3, 4]
arr2 = [1, 3, 1, 1]
arr3 = [1, 1, 2, 2, 3]
function findAllIn(find, search) {
if (find.length == 0) {
return true;
}
i = search.indexOf(find[0]);
if (i == -1) {
return false;
}
return findAllIn(find.slice(1,find.length), search.slice(0,i).concat(search.slice(i+1, search.length)));
}
console.log(findAllIn(arr2, arr1)); // false
console.log(findAllIn(arr3, arr1)); // true
This should do the trick
Not efficient but I think it is easy to understand
const count = (x, xs) =>
xs.reduce((y, xi) => y + (xi === x ? 1 : 0), 0)
const isInclusive = (xs, ys) =>
xs.every((xi) => count(xi, xs) >= count(xi, ys))
const arr1 = [1, 2, 3, 1, 2, 3, 4]
const arr2 = [1, 3, 1, 1]
const arr3 = [1, 1, 2, 2, 3]
console.log(isInclusive(arr1, arr2))
console.log(isInclusive(arr1, arr3))
Based on this answer: https://stackoverflow.com/a/4026828/3838031
You can do this:
Array.prototype.diff = function(a) {
return this.filter(function(i) {return a.indexOf(i) < 0;});
};
arr1 = [1, 2, 3, 1, 2, 3, 4]
arr2 = [1, 3, 1, 1]
arr3 = [1, 1, 2, 2, 3]
console.log((arr1.diff(arr2).length === 0) ? "True" : "False");
console.log((arr1.diff(arr3).length === 0) ? "True" : "False");
console.log((arr2.diff(arr1).length === 0) ? "True" : "False");
console.log((arr2.diff(arr3).length === 0) ? "True" : "False");
console.log((arr3.diff(arr1).length === 0) ? "True" : "False");
console.log((arr3.diff(arr2).length === 0) ? "True" : "False");

Categories