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Is it possible to use array.splice(argument[i], x) to edit an array for variables that meet a condition?

Key Question
-How do you use splice(argument[i], x)? Can it be used this way or am I only allowed to use numbers? ie (1, 2), (3, 0)
-I'm a little unsure of when element[i] can be used when an array is declared. So it can be used for both for loops and while loops when setting conditions? Can it be used as an argument or parameter in functions or additional methods besides splice?
What I want to do
-Write a function called "isEven".
-Given an array of numbers, "isEven" returns a new array.
-Only even numbers are outputted from the input array.
ex.
var output = isEven([1, 4, 5, 6, 10, 13]);
console.log(output); // --> [4, 6, 10]
Approach
-declare var digits to "catch" the array input.
-declare var NewArray for return of output array,
-use if condition to go through var digits and splice the variable at any given index.
-declare NewArray to the newly spliced array
function isEven(num) {
var digits = num;
var newArray = [];
digits.forEach(function(num) {
if (num[i] % 2 > 0) {
newArray = digits.splice(num[i], 1);
}
}) return newArray;
}
var ledoit = isEven([1, 4, 6]);
console.log(ledoit);

Try this:
function isEven(myArray) {
return myArray.filter(item => {
return Number.isInteger(item / 2)
})
}
Then isEven([1, 4, 5, 6, 10, 13]) will output [4,6,10]

You want to use the % operator:
var nums = [1, 4, 5, 6, 10, 13];
function getEvens(array){
for(var i=0,n,a=[],l=array.length; i<l; i++){
n = array[i];
if(n % 2 === 0)a.push(n);
}
return a;
}
console.log(getEvens(nums));
Albeit, not backward compatible, you could also do:
var nums = [1, 4, 5, 6, 10, 13];
function getEvens(array){
return array.filter(n => (n % 2 === 0));
}
console.log(getEvens(nums));

Comparing Multiple Arrays Using Reduce

pretty new to Javascript and I've tried this question about 4 times now in a span of about a month and I am still unable to solve it.
So here is the question:
Construct a function intersection that compares input arrays and returns a new array with elements found in all of the inputs. BONUS: Use reduce!
The format is:
function intersection(arrays) {
// Your Code Goes Here
}
Test Case: Should log [15, 5]
console.log('Extensions 3 Test: ' + intersection([5, 10, 15, 20], [15, 88, 1, 5, 7]/*, [1, 10, 15, 5, 20]*/));
My current solution: Works for the case of only have two items to compare, but not for the third one, I could make it so that I would loop through and compare the obtained values with the next array but I don't think I am on the right path... Also, I am not using reduce to implement it... And I am not sure if I am supposed to be using 'arguments.' Any help is appreciated! Thank you so much.
function intersection(arrays) {
array = [];
for (var i = 0; i < arguments.length; i++)
array.push(arguments[i]);
var result = [];
for(var i = 0; i < array.length - 1; i++) {
for(var j = 0; j < array[i].length; j++) {
if (array[i+1].includes(array[i][j]))
result.push(array[i][j]);
}
}
return result;
}

Although, as several suggestions said, you could use underscore, lodash, or my personal favorite, Ramda (disclaimer: I'm one of the authors), this function should be straightforward enough that you wouldn't even consider a library for it. Here's a simple version:
const intersection = (xs, ys) => xs.filter(x => ys.indexOf(x) > -1);
intersection([5, 10, 15, 20, 3], [15, 88, 3, 1, 5, 7]); //=> [5, 15, 3]
const intersectAll = (...xss) => xss.reduce(intersection);
intersectAll([5, 10, 15, 20, 3], [15, 88, 3, 1, 5, 7], [1, 10, 15, 5, 20]); //=> [5, 15]
I would think that this is all you need, at least so long as you're worried only about reference/primitive equality and don't need to consider cases where you want to know that {x: 1} and {x: 1} are the same, even though they aren't the same reference. If you do need that, you might look to Ramda's intersection function.
Note that if includes were better supported, I would recommend this version instead, as it reads better:
const intersection = (xs, ys) => xs.filter(x => ys.includes(x));
Also, if you have no need for the binary function, you can make just a variadic version of it by combining the two above:
const intersection = (...xss) => xss.reduce((xs, ys) => xs.filter(x => ys.indexOf(x) > -1));

Maybe someone will finds it useful.
As an argument to the function you can give any number of arrays of any length and the function is compact, I think ;)
const findSimilar = (...arrays) => {
return arrays.reduce((includ, current) =>
Array.from(new Set(includ.filter((a) => current.includes(a))))
);
};
console.log(
findSimilar([5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20])
);
And how it works:
Ok, first u take rest parameters(...arrays) as parameter of function, so u have
arrays = [ [5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20] ]
then in first iteration of reduce we have
includ = [5, 10, 15, 20] and current = [15, 88, 1, 5, 7]
on this two we use filter, what give us [5,15], i use Set to make shure there is no repetition and make array back (Array.from()), which is passed to the next iteration of reduce as "includ", at the next iteration we have
incude = [5,15] and current = [1, 10, 15, 5, 20] and so on ...
We can even use it like this
let result = [
[5, 10, 15, 20],
[15, 88, 1, 5, 7],
[1, 10, 15, 5, 20]
].reduce((includ, current) =>
Array.from(new Set(includ.filter((a) => current.includes(a))))
);
console.log(result);

Although not solving your problem directly, you can do what you're trying to do using the opensource library underscore.js.
_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2]
You may be able to derive inspiration from the way that's been implemented. The above is the function call to their own _.intersection function which is also dependent on other underscore.js functions as you see below:
// Produce an array that contains every item shared between all the
// passed-in arrays.
_.intersection = function(array) {
if (array == null) return [];
var result = [];
var argsLength = arguments.length;
for (var i = 0, length = array.length; i < length; i++) {
var item = array[i];
if (_.contains(result, item)) continue;
for (var j = 1; j < argsLength; j++) {
if (!_.contains(arguments[j], item)) break;
}
if (j === argsLength) result.push(item);
}
return result;
};

Here is a solution using reduce, with the empty array passed in as intersection as the initial value.
Iterate the numbers and check if each one appears in one of the subarrays.
If it doesn't, set the Boolean isPresentInAll to false.
If it does appear in all three and it's not already present in the
intersection array, then push to the intersection array.
function intersection(arrayOfArrays) {
return arrayOfArrays.reduce(function(intersection, subArray) {
subArray.forEach(function(number) {
var isPresentInAll = true;
for (var i = 0; i < arrayOfArrays.length; i++) {
if (arrayOfArrays[i].indexOf(number) === -1) {
isPresentInAll = false;
}
}
if (isPresentInAll === true && intersection.indexOf(number) === -1) {
intersection.push(number);
}
});
return intersection;
}, []);
}

I think i got the right function for you.
(Note: results are not sorted!)
var intersection = function() {
// merge deduped arrays from arguments
var arrays = Array.prototype.reduce.call(arguments, function(carry, array) {
return [].concat(carry, array.filter(function(item, index, origin) {
return origin.indexOf(item) === index;
}));
}, []);
var results = arrays.reduce(function(carry, item, index, arr) {
if(
// just select items, which have more then 1 occurance
arr.filter(function(fItem) {
return fItem === item;
}).length > 1 &&
// ... and which are not already in results
!~carry.indexOf(item)
) {
carry = [].concat(carry,item);
}
return carry;
}, []);
return results;
};

Here's a version that uses 2 reduces.
The first iterates the arrays only once to create a hashmap object to track instance counts, the second to return values where counts match number of arguments
function intersection(){
// convert arguments to array of arrays
var arrays = [].slice.call(arguments);
// create an object that tracks counts of instances and is type specific
// so numbers and strings would not be counted as same
var counts= arrays.reduce(function(a,c){
// iterate sub array and count element instances
c.forEach(function(val){
var propName = typeof val + '|' + val;
// if array value not previously encountered add a new property
a[propName] = a[propName] || {count:0, value: val};
// increment count for that property
a[propName].count++;
});
return a;
},{});
// iterate above object to return array of values where count matches total arrays length
return Object.keys(counts).reduce(function(resArr, propName){
if(counts[propName].count === arrays.length){
resArr.push(counts[propName].value);
}
return resArr;
},[]);
}
console.log(intersection([5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]))
Could use some fine tuning to make sure there are enough arguments and that they are all arrays

Here's what I came up with using vanilla javascript and one call to reduce.
function intersection(){
var arrays = [].slice.call(arguments);
var first = arrays[0];
var rest = arrays.slice(1);
return first.reduce(function(all, item, index){
var push = rest.every(function(subArray){
return subArray.indexOf(item) > -1;
});
if(push){
all.push(item);
}
return all;
},[])
}
console.log(intersection([5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]));

function intersection(arrays) {
let common = arrays.reduce(function(accumulator, currentValue) {
return accumulator.filter(function(x){
return currentValue.indexOf(x) > -1;
})
})
return common;
}

To optimize your answer that couldn't work on more than 2 subarrays and didn't use reduce, here's the code that works for however many subarrays you pass in.
function intersection(arr1, arr2, arr3){
let ans = arr1[0]; // ans = [5,10,15,20]
for(let i = 0; i < ans.length; i++){ // i = 0...3
for(let j = 1; j < arr1.length; j++){ // j = 1...2
if(!(arr1[j].includes(ans[i]))){ // if the new subarray doesn't include an element in the ans
ans.splice(i, 1); // delete the element from ans
}
}
}
return ans;
}
const arr1 = [5, 10, 15, 20];
const arr2 = [15, 88, 1, 5, 7];
const arr3 = [1, 10, 15, 5, 20];
console.log(intersection([arr1, arr2, arr3])); // should log: [5, 15]

For-loop, map and forEach Javascript

I am trying to create a deck of 52 cards. I can create it easily with double for-loop but it has O(n2) complexity. So I was trying to play with map() and forEach() array methods but things are complex with them needing to return stuffs. Here is my code below.
(function deckCreate() {
var values = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13];
var suits = ["clubs", "diamonds", "hearts", "spades"];
var newDeck = values.map(function(xValue) {
suits.forEach(function(xSuit) {
return [xSuit,xValue];
});
});
return newDeck;
}());
It gives an array of length 13 all undefined inside. I tried swapping forEach() before map() just incase but the result was the same.
The issue I found while console.log() inside those functions was that the elements were not being mapped to each other but were printed all separately. What could be the issue be?

You're not returning anything from your map function, so the implicit return value is undefined, hence your array of 13 undefined values.
suits.forEach needs to be return suits.map. This will give you an array of 13 elements, where each element is an array of four elements, where each element of the inner array is a two element [suit, value] array. You can then reduce the top-level array into the 52 element array you're after:
var newDeck = values.map(function(xValue) {
return suits.map(function(xSuit) {
return [xSuit,xValue];
});
}).reduce(function (a, b) { return a.concat(b) });

The reason you're having trouble is that you aren't returning from your outer .map() callback. Even if you did, though, [].forEach always returns undefined, regardless of what happens inside its callback.
So, since you're using forEach to iterate the inner array, you get an array of 13 undefineds from map.
What you should be using is .map() all the way down and return every step of the way:
const first = [1, 2, 3];
const second = ['a', 'b', 'c'];
const mapped = first.map(function(digit) {
return second.map(function(letter) {
return [digit, letter];
});
});
console.log(mapped);
Seeing how you're clearly trying to learn and improve yourself, I'll leave you to adjust this example to your specific case.
P.S. If you want a flattened array, have a look at [].reduce() and [].concat().

Use a simple for loop. Use suits[Math.floor(i / 13)] to get the right suit, and use the remainder operator % to get the card number for each suit:
function deckCreate() {
var values = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13];
var suits = ["clubs", "diamonds", "hearts", "spades"];
var newDeck = [];
for (var i = 0; i < 52; i++) {
newDeck.push([suits[Math.floor(i / 13)], values[i % 13]]);
}
return newDeck;
}
var result = deckCreate();
console.log(result);

I think it's better to simplify it.
We know that there are just 4 suits, so it's enough to get list by suit name:
function createDeck() {
var values = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13];
var deck = {"clubs": values.slice(), "diamonds": values.slice(), "hearts": values.slice(), "spades": values.slice()};
return deck;
}
var deck = createDeck();
console.log('CLUBS:', deck.clubs);
console.log('DIAMONDS:', deck.diamonds);
console.log('HEARTS:', deck.hearts);
console.log('SPADES:', deck.spades);
P.S. in my case I'll create a class that makes generation, iteration and etc stuff to easily use it.
function Deck() {
var values = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13];
var suits = ['clubs', 'diamonds', 'spades', 'hearts'];
this.getSuits = function() {return suits.splice();}
this.getValues = function() {return values.splice();}
var asObject;
this.asObject = function() {
if(asObject) return asObject;
asObject = {};
suits.map(function(suit) {
asObject[suit] = values.slice();
});
return asObject;
};
var asArray;
this.asArray = function() {
if(asArray) return asArray;
asArray = [];
suits.map(function(suit) {
values.map(function(value) {
asArray.push([suit, value]);
});
});
return asArray;
}
this.iterate = function(fn) {
this.asArray().map(fn);
}
}
var deck = new Deck();
deck.iterate(function(card) {
console.log('CARD: ', card[0], card[1]);
});
console.log(deck.asObject());
console.log(deck.asArray());

Compare an unspecified number of arrays for common values in javascript

I would like to know how to compare two or more -- potentially unlimited -- arrays for common values and push these values into a new array efficiently. Below I have a function that will accept unlimited arguments, but I am uncertain if this is a good place to begin. PHP appears to have a method that can do what I want called array_intersect. Does javascript offer something similar?
Note: I have found examples of how this can be done with two or so arrays, but I have not found examples of how such approaches might be applied to an unspecified number of arrays as of yet. Therefore I do not see this as a duplicate question.
To further clarify, the arrays might be filled with anything. Letters, numbers, symbols, words, you name it, it might be there.
var sampleOne = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12];
var sampleTwo = [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18];
function FindDirectRelation() {
for(var i = 0; i < arguments.length; ++i) {
console.log(arguments[i]);
};
};
var directRelation = FindDirectRelation(sampleOne, sampleTwo);
I am still a coding novice, so please ensure that everything is explained in a way that is simple enough for me to understand.

using an existing intersect that works with 2 arrays, we can chain together a common sub-set using the built-in reduce() method on an array of arrays that need intersected:
function intersect(a, b) {
var aa = {};
a.forEach(function(v) { aa[v]=1; });
return b.filter(function(v) { return v in aa; });
}
var r1=[1,2,3],
r2=[1,3,4,5],
r3=[5,1,3];
alert([r1, r2, r3].reduce(intersect)) // shows: 1,3
if you define "intersect" as just being in more than one array (not every), then it's more complex...

Check to make sure the elements in the first array are also in the remaining arrays:
function multi_intersect(a) {
var other_arrays = Array.prototype.slice.call(arguments, 1);
return a . filter(function(elt) {
return other_arrays.every(function(an) {
return an.indexOf(elt) !== -1;
});
});
}

Try using Array.prototype.filter() , Array.prototype.indexOf()
var res = sampleOne.filter(function(val) {return sampleTwo.indexOf(val) !== -1})
var sampleOne = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12];
var sampleTwo = [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18];
var arr = ["a", "b", "c"];
var arr1 = ["c", "d", "e"];
var arr2 = [2, 7];
function samples() {
var args = Array.prototype.slice.call(arguments);
var res = [];
for (var i = 0, curr, next; i < args.length; i++) {
if (args[i + 1]) {
// set `curr` to array `i`
curr = args[i];
// set `next` to array `i + 1` if it exists
next = args[i + 1]
} else {
// if at last index, set `curr` to `args` : input arrays
// flattened to single array , with element at `i` removed
curr = [].concat.apply([], args.slice(0, args.length - 1));
console.log(curr)
// set next to current index
next = args[i];
};
next = next.filter(function(val) {
return curr.indexOf(val) !== -1
// filter duplicate entries at `res`
&& res.indexOf(val) === -1
});
res = res.concat.apply(res, next);
};
return res
}
var sample = samples(sampleOne, sampleTwo, arr, arr1, arr2);
console.log(sample); // [5, 6, 7, 8, 9, 10, 11, 12, "c", 2]

Javascript: Most Efficient Way of Summing Multiple Arrays by Key

I have a JSON object returned from a web service, which is an array of objects. I need to add the "data" arrays together to form a summed array. The JSON response looks like this:
[
{
"data":[
0,3,8,2,5
],
"someKey":"someValue"
},
{
"data":[
3,13,1,0,5
],
"someKey":"someOtherValue"
}
]
There could be N amount of objects in the array. The desired output for the above example would be:
[3, 16, 9, 2, 10]
I was intending on creating an empty array variable (var arr), then looping over the objects, and for each object, loop through the "data" key and for each key increment the corresponding key in arr by the value.
Is there a more efficient way of doing this using some sort of merge function?

How about this, I believe it should work for all cases.
var data = [{
"data": [
0, 3, 8, 2, 5
],
"someKey": "someValue"
}, {
"data": [
3, 13, 1, 0, 5
],
"someKey": "someOtherValue"
}];
var datas = data.reduce(function(a, b) {
b.data.forEach(function(x, i) {
a[i] = a[i] || 0;
a[i] += x;
});
return a;
}, []);
console.log(datas);

If every object has the same data length, you can try with:
var input; // Your input data
var output = [];
for (var i = 0; i < input[0].data.length; i++) {
output[i] = input.reduce(function(prev, item) {
return +(item.data[i]) + prev;
}, 0);
}
console.log(output);
// [3, 16, 9, 2, 10]
If every object has different data size:
var input; // Your input data
var i = 0, output = [];
while (true) {
var outOfIndex = true;
var sum = input.reduce(function(prev, item) {
if (item.data[i] !== undefined) {
outOfIndex = false;
}
return +(item.data[i]) + prev;
}, 0);
if (outOfIndex) {
break;
}
output[i++] = sum;
}
console.log(output);
// [3, 16, 9, 2, 10]

Slightly less imperative solution:
//zip takes two arrays and combines them per the fn argument
function zip(left, right, fn) {
var shorter = (right.length > left.length) ? left : right;
return shorter.map(function(value, i) {
return fn(left[i], right[i]);
});
}
//assuming arr is your array of objects. Because were using
//zip, map, and reduce, it doesn't matter if the length of the
//data array changes
var sums = arr
.map(function(obj) { return obj.data; })
.reduce(function(accum, array) {
//here we want to combine the running totals w/the current data
return zip(accum, array, function(l, r) { return l + r; });
});