We Keep Coding

Shuffle nested arrays in Javascript

I'm trying to sort multiple arrays within an array (which also has to be shuffled). A simplified example is:
let toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
...
];
const shuffled = shuffle(toShuffle);
// outout would look something like:
// [
// [8, 6, 5, 7, 9],
// [4, 3, 1, 5, 2],
// [19, 26, 10, 67],
// ...
// ]
This needs to be flexible, so any number of arrays with any amount of values should be valid.
Here is what I've tried:
function shuffle(a) {
for (let e in a) {
if (Array.isArray(a[e])) {
a[e] = shuffle(a[e]);
} else {
a.splice(e, 1);
a.splice(Math.floor(Math.random() * a.length), 0, a[e]);
}
}
return a;
}
console.log("Shuffled: " + shuffle([
[1, 2, 3, 4, 5],
[5, 4, 3, 2, 1]
]))
But it's not working as intended. Is their an easier way to do this? Or is my code correct and just buggy.

You can use Array.from() to create a new shallow-copied array and then to shuffle Array.prototype.sort() combined with Math.random()
Code:
const toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
]
const shuffle = a => Array.from(a).sort(() => .5 - Math.random())
const result = toShuffle.map(shuffle)
console.log('Shuffled:', JSON.stringify(result))
console.log('To shuffle:', JSON.stringify(toShuffle))

You almost got it. The problem is that you are removing one item from an array, instead of capturing the removed item and them placing in a random position:
let toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
];
function shuffle(a) {
a = [...a]; //clone array
for (let e in a) {
if (Array.isArray(a[e])) {
a[e] = shuffle(a[e]);
} else {
a.splice(~~(Math.random() * a.length), 0, a.splice(e, 1)[0]);
}
}
return a;
}
console.log(JSON.stringify(shuffle(toShuffle)))
console.log(JSON.stringify(toShuffle))
[EDIT]
The original code did not shuffle the parent array, if you need shuffle everything recursively, you can use this:
let toShuffle = [
[1, 2, 3, 4, 5],
[9, 8, 7, 6, 5],
[10, 67, 19 ,27]
];
function shuffle(a) {
a = a.map(i => Array.isArray(i) ? shuffle(i) : i); //clone array
a.sort(i => ~~(Math.random() * 2) - 1); //shuffle
return a;
}
console.log("shuffled", JSON.stringify(shuffle(toShuffle)))
console.log("original", JSON.stringify(toShuffle))

Trying to find a method to refactor my function to make it more efficient

I have this function as seen below that groups an array of numbers based on a length parameter. The length represents the max length of each sub-array. What I am trying to figure out, is a method I can take to shift x => x % 2 out of the declaration of the result variables and into the function. The only thing I could think of would be a callback, but I am not sure how I could do this. Any help with this is appreciated and if you notice any other redundant code please let me know.
function myFunc(arr, length, fnc) {
groups = [];
result = [];
for (let val of arr) {
x = fnc(val);
if (!groups[x]) {
groups[x] = [];
}
if (!groups[x].length) {
result .push(groups[x]);
}
groups[x].push(val);
if (groups[x].length === length) {
groups[x] = [];
}
}
return result ;
}
//examples
const result1 = myFunc([1, 2, 3, 4], 2, x => x % 2)
console.log(result1) //[[1, 3], [2, 4]]
const result2 = myFunc([1, 2, 3, 4, 5, 6, 7], 4, x => x % 2)
console.log(result2) //[[1, 3, 5, 7], [2, 4, 6]]
const result3 = myFunc([1, 2, 3, 4, 5], 1, x => x % 2)
console.log(result3) //[[1], [2], [3], [4], [5]]
const result4 = myFunc([1, 2, 3, 4, 5, 6], 4, x => x % 2)
console.log(result4) //[[1, 3, 5], [2, 4, 6]]
What Id like to achieve is only needing to call the array and the size of the sub-arrays that Id like to create. Just a brief understanding of what is happening, the arrays are based off seeing whether they can make a full array to the size of "length," and any overflow is pushed to another sub-array. An example of this looks like this:
console.log(myfunc([1,2,3,4,5,6,7,8,9,10],3))
This would return [[1,3,5][2,4,6][7,9][8,10]]
So if someone could assist me in removing the fnc parameter from the console.log statements and placing it into the function, it would be very helpful

The issue you're having is that you know this parameter will always be the same, so there's
no need to write it every time; it can just be baked into the function.
At its most simplest, you can literally just take fnc out of the parameter list and declare it manually as a variable, always setting it to the same value:
function myFunc(arr, length) {
let fnc = x => x % 2;
groups = [];
result = [];
for (let val of arr) {
x = fnc(val);
if (!groups[x]) {
groups[x] = [];
}
if (!groups[x].length) {
result .push(groups[x]);
}
groups[x].push(val);
if (groups[x].length === length) {
groups[x] = [];
}
}
return result ;
}
//examples
const result1 = myFunc([1, 2, 3, 4], 2)
console.log(result1) //[[1, 3], [2, 4]]
const result2 = myFunc([1, 2, 3, 4, 5, 6, 7], 4)
console.log(result2) //[[1, 3, 5, 7], [2, 4, 6]]
const result3 = myFunc([1, 2, 3, 4, 5], 1)
console.log(result3) //[[1], [2], [3], [4], [5]]
const result4 = myFunc([1, 2, 3, 4, 5, 6], 4)
console.log(result4) //[[1, 3, 5], [2, 4, 6]]
At this point there really isn't almost any point to fnc being a function though. It's the same operation every time. Just take that operation and paste it in place of the function call. Also, you should always use let or const to declare things, don't just assign them with no declaration keyword - that makes them globals and they might interact with your global namespace in an unexpected way. Finally, I urge you to rename your function to something that describes what it does. This helps everyone reading your code understand it more intuitively, including your future self.
This is what that all would look like:
function splitArray(arr, length) {
let groups = [];
let result = [];
for (let val of arr) {
let x = val % 2; //this does the same thing
if (!groups[x]) {
groups[x] = [];
}
if (!groups[x].length) {
result.push(groups[x]);
}
groups[x].push(val);
if (groups[x].length === length) {
groups[x] = [];
}
}
return result;
}
//examples
const result1 = splitArray([1, 2, 3, 4], 2)
console.log(result1) //[[1, 3], [2, 4]]
const result2 = splitArray([1, 2, 3, 4, 5, 6, 7], 4)
console.log(result2) //[[1, 3, 5, 7], [2, 4, 6]]
const result3 = splitArray([1, 2, 3, 4, 5], 1)
console.log(result3) //[[1], [2], [3], [4], [5]]
const result4 = splitArray([1, 2, 3, 4, 5, 6], 4)
console.log(result4) //[[1, 3, 5], [2, 4, 6]]

Get item what's not on the 2nd, 3rd arrays - JS

I am trying to create a function that will get the items that cannot be seen on the 2nd or 3rd and upcoming arrays passed within the function.
Right now my function gets only the similar items. How can I make it get the difference (w/c are the items that doesn't exist to the 2nd and 3rd and proceeding arrays.
const callM = function(arrays) {
arrays = Array.prototype.slice.call(arguments);
let result = [];
for(let i = 1; i < arrays.length; i++){
for(let x = 0; x < arrays[i].length; x++){
if(arrays[0].includes(arrays[i][x])){
result.push(arrays[i][x]);
}
}
}
return result;
};
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10])); // -> must be [1, 3, 4]
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // -> must be [3,4]
The logic right now is a bit off as it gets the opposite. How do i fix this?
Also is there a way to do this using Higher Order functions such as reduce or filter?
Thanks!

I'd think about this differently. As the difference between two sets: array 0 and array 1...n
To get array 0, just shift it off the top
const arr0 = arrays.shift()
Ref: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/shift
This removes the first array from arrays
Next we combine the remaining arrays
const arrN = arrays.reduce(function(prev, curr) {
return prev.concat(curr)
})
Ref: http://www.jstips.co/en/javascript/flattening-multidimensional-arrays-in-javascript/
Unneeded, handled by includes as mentioned by #Phil
Next filter duplicates from arrN by comparing with itself
const unique = arrN.filter(function(elem, index, self) {
return index == self.indexOf(elem);
})
Ref: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
Next filter with includes to find the difference (or union)
const diff = arr0.filter(function(item) {
return !arrN.includes(item))
}
Full snippet:
function callM(arrays) {
const arr0 = arrays.shift()
const arrN = arrays.reduce(function(prev, curr) {
return prev.concat(curr)
})
//const unique = arrN.filter(function(elem, index, self) {
// return index == self.indexOf(elem)
//})
return arr0.filter(function(item) {
return !arrN.includes(item)
})
}
console.log(callM([[1, 2, 3, 4, 5], [5, 2, 10]]))
console.log(callM([[1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8]]))
of course ES6 would be easier. ;)
const callM = (first, ...rest) => {
const arrays = [].concat(...rest)
return first.filter(item => !arrays.includes(item))
}
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10]))
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8]))

A short solution for small and medium sized arrays:
// Return elements in array but not in filters:
function difference(array, ...filters) {
return array.filter(el => !filters.some(filter => filter.includes(el)));
}
// Example:
console.log(difference([1, 2, 3, 4, 5], [5, 2, 10])); // [1, 3, 4]
console.log(difference([1, 2, 3, 4, 5], [5, 1, 10], [7, 2, 8])); // [3, 4]
For large inputs, consider creating a Set from all filters and filtering in linear time using set.has(el).
In order to fix your implementation, you could label the outer for-loop and continue from there whenever a filter contains one of the array elements. Only when all filters pass without match, you push the array element into the result:
// Return elements in array but not in filters:
function difference(array, ...filters) {
const result = [];
loop: for (const el of array) {
for (const filter of filters) {
if (filter.includes(el)) continue loop;
}
result.push(el);
}
return result;
}
// Example:
console.log(difference([1, 2, 3, 4, 5], [5, 2, 10])); // [1, 3, 4]
console.log(difference([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // [3,4]

If you're willing to use Underscore, you can do this in one line of code:
console.log(_.difference([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8]))
https://jsfiddle.net/o1zuaa6m/

You can use array#reduce to create object lookup of all the other array excluding the first array. Then use array#filter to get the values which are not present in the object lookup
var callM = (first, ...rest) => {
var combined = rest
.reduce((res,arr) => res.concat(arr))
.reduce((o, v) => {
o[v] = true;
return o;
},{});
return first
.filter(v => !combined[v]);
}
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10])); // -> must be [1, 3, 4]
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // -> must be [3,4]

The "proper" way to exclude values is usually to use a lookup hash set with the values to exclude:
const callM = (a, ...b) => (b = new Set(b.concat.apply(...b)), a.filter(v => !b.has(v)))
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10])); // [1, 3, 4]
console.log(callM([1, 2, 3, 4, 5], [5, 2, 10], [7, 1, 8])); // [3, 4]

Array map on arrays not updating array

I have an array of arrays
[
[1,3,5,7,8,8],
[1,3,5,7,8,8],
[1,3,5,7,8,8],
[1,3,5,7,8,8],
[1,3,5,7,8,8]
]
I am trying to insert a value between each item. So I have this:
let reelList = [
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8]
]
reelList.map(reel => {
// Adds the separator (works)
let v = separate(reel, '-')
console.log(v)
return v
})
function separate(arr, value) {
return arr.reduce((result, element, index, array) => {
result.push(element)
index < array.length - 1 && result.push(value)
return result
}, []);
}
// Logs the new list to the console (doesn't work)
console.log(reelList)
When I log the values after I run the separate function they are separated, however, when I display reelList they are not separated. Why is that?

The map() function returns a new array. It doesn't modify the existing one.
You would need to set the results of reelList.map() to something else.
let reelList = [
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8]
]
const finalResult = reelList.map(reel => {
// Adds the separator (works)
let v = separate(reel, '-')
console.log(v)
return v
})
function separate(arr, value) {
return arr.reduce((result, element, index, array) => {
result.push(element)
index < array.length - 1 && result.push(value)
return result
}, []);
}
// Logs the new list to the console (doesn't work)
console.log(finalResult);
If you want to edit the array in place, instead of using map(), use a forEach() with a callback that has a second and third parameter, which are index and array. Then you can update the array with the new values as you go.
let reelList = [
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8]
]
reelList.forEach((reel, index, arr) => {
// Adds the separator (works)
let v = separate(reel, '-')
arr[index] = v;
console.log(v)
})
function separate(arr, value) {
return arr.reduce((result, element, index, array) => {
result.push(element)
index < array.length - 1 && result.push(value)
return result
}, []);
}
// Logs the new list to the console (doesn't work)
console.log(reelList);

You could map a new array with a calculated lenght and take the result of the calculated index ot the dash.
var array = [[1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8]],
result = array.map(a => Array.from({ length: a.length * 2 - 1 }, (_, i) => a[i / 2] || '-'));
console.log(result);

JS how to reform this irregular object

I have this object which its keys are guaranteed sorted and will be used for the operation. And each of its value is a 2d array.
var obj = {
"0": [
[0, 1], [0, 3], [0, 4]
],
"1": [
[1, 2], [1, 3]
],
"2": [
[2, 3], [2, 5]
],
"3": [
[3, 4], [3, 6]
],
"5": [
[5, 6]
],
"6": [
[6, 5]
]
}
I am trying to concatenate them and for each of its last value of the array is the next index of the object. So, my expected result is an array like this,
The pattern is, I have to find a way from 0 which is the first index of obj, to the last index which is 6 by using the values in each of it and linking its last array value to the next object. If that makes sense.
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 6]
[0, 1, 2, 5, 6]
[0, 1, 3, 4, 5, 6]
[0, 1, 3, 4]
[0, 1, 3, 6]
[0, 3, 4, 5, 6]
[0, 3, 6]
[0, 4]
This is my code so far, as I don't know how to proceed further..
var result = [];
for (var key in obj) {
var myarr = obj[key];
for (var i = 0; i < myarr.length; i++) {
result.push(myarr[i])
}
}
Any idea or feedback is welcome.
Edit
One of the expected result was [0, 1, 2, 3, 4, 5, 6], here's the step by step explanation.
The obj key starts from 0 and ends in 6, I have to form a way from 0 to 6 with the arrays in its value.
Starts from obj[0], the first array returns [0, 1], save this to res. (res is now [0, 1])
The last value of array in res is 1, now find the next value in obj[1]
obj[1] has two arrays, and ends with 2 or 3.. So it's possible to append with both of them, so it can be [0, 1, 2] or [0, 1, 3]. In this case, get the first one which is [1, 2] and append the last value to res. (res is now [0, 1, 2]).
The last value of array in res is now 2, now find the next value in obj[2].
obj[2] has two arrays, and ends with 3, or 5.. It's possible to append with both of them, so it can be [0, 1, 2, 3] or [0, 1, 2, 5]. In this case, get the first one which is [2, 3] and append the last value to res. (res is now [0, 1, 2, 3])
The last value of array in res is now 3, now find the next value in obj[3].
Repeat step 4 or 6. (res is now [0, 1, 2, 3, 4]).
The last value of array in res is now 4, now find the next value in obj[4].
Repeat step 4 or 6. (res is now [0, 1, 2, 3, 4, 5]).
The last value of array in res is now 5, now find the next value in obj[5].
Now value 6 is found which should be the end of iteration if you look at the step 4. Repeat step 4 or 6. (res is now [0, 1, 2, 3, 4, 5, 6]).
Repeat from step 1, and form another way to do it, with no duplicates of [0, 1, 2, 3, 4, 5 ,6].

This is a proposal, with a single extra output, mentioned below.
[
[0, 1, 2, 3, 4, 5, 6],
[0, 1, 2, 3, 6],
[0, 1, 2, 5, 6],
[0, 1, 3, 4, 5, 6], /* extended from below */
[0, 1, 3, 4], /* original result */
[0, 1, 3, 6],
[0, 3, 4, 5, 6], /* extended from below */
[0, 3, 4], /* extra line, line should not be in result */
[0, 3, 6], /* but follows the same building rule than above */
[0, 4]
]
Basically this solution is building a tree with the given information about linked nodes.
If some nodes are not contiguous, a backtracking is made for the missing links, with the above function for nodes, checkNodes or with iterPath, to walk the actual collected nodes for missing items.
function getParts(value, path, nodes) {
function checkNodes(a) {
if (a[1] === value + 1) {
getParts(a[1], path.concat(a[1]), nodes);
return true;
}
}
function iterPath(k) {
return (object[k] || []).some(function (a) {
return path[path.length - 1] + 1 === a[1] || iterPath(a[1]);
});
}
value = value || 0;
path = path || [value];
nodes = nodes || [];
if (object[value]) {
object[value].forEach(function (a, i, aa) {
if (a[1] === lastKey) {
parts.push(path.concat(a[1]));
return;
}
getParts(a[1], path.concat(a[1]), nodes.concat(aa.slice(i + 1)));
});
return;
}
if (nodes.some(checkNodes)) {
return;
}
path.slice(1).some(iterPath) && getParts(path[path.length - 1] + 1, path.concat(path[path.length - 1] + 1), nodes);
parts.push(path);
}
var object = {
0: [[0, 1], [0, 3], [0, 4]],
1: [[1, 2], [1, 3]],
2: [[2, 3], [2, 5]],
3: [[3, 4], [3, 6]],
5: [[5, 6]],
6: [[6, 5]]
},
lastKey = 6,
parts = [];
getParts();
parts.forEach(function (a) { console.log(JSON.stringify(a)); });
.as-console-wrapper { max-height: 100% !important; top: 0; }

Well, I was sitting on this for some time now, and sharing across my take on the problem:
The input object can be considered as an adjacency list of a tree:
var obj={0:[[0,1],[0,3],[0,4]],1:[[1,2],[1,3]],2:[[2,3],[2,5]],3:[[3,4],[3,6]],5:[[5,6]],6:[[6,5]]};
and the following as the result required, which is in fact, as I see it, the list of all root-to-leaf paths of the tree:
[0,1,2,3,4]
[0,1,2,3,6]
[0,1,2,5,6]
[0,1,3,4]
[0,1,3,6]
[0,3,4]
[0,3,6]
[0,4]
a little different than the result set mentioned in the question which is the below:
[0,1,2,3,4,5,6]
[0,1,2,3,6]
[0,1,2,5,6]
[0,1,3,4,5,6]
[0,1,3,4]
[0,1,3,6]
[0,3,4,5,6]
[0,3,6]
[0,4]
The difference between the results is only the question whether 4 and 6 are leaf nodes
Solution:
So I assume that for our Tree here:
0 is the root node
4 and 6 are the leaf nodes
See code below - I created a tree first, and from that listed out all the root to leaf paths:
// removed "6:[[6,5]]" as 6 is a 'leaf' of the tree
var obj={0:[[0,1],[0,3],[0,4]],1:[[1,2],[1,3]],2:[[2,3],[2,5]],3:[[3,4],[3,6]],5:[[5,6]]};
var availableNodes = Object.keys(obj);
var tree = availableNodes.reduce(function(hash) {
return function(prev, curr) {
hash[curr] = hash[curr] || {};
hash[curr].children = hash[curr].children || [];
obj[curr].forEach(function(element) {
hash[element[1]] = hash[element[1]] || {};
hash[element[1]].children = hash[element[1]].children || [];
hash[curr].rootPath = hash[curr].rootPath || [];
hash[curr].children.push({value: element[1],children: hash[element[1]].children});
});
curr && prev.push({value: curr,children: hash[curr].children});
return prev;
};
}(Object.create(null)), []);
//console.log(JSON.stringify(tree));
var result = [];
function rootToLeafPaths(node, path) {
path.push(+node.value);
if (node.children.length === 0) {
result.push(Array.from(path));
path.pop();
} else {
node.children.forEach(function(element) {
rootToLeafPaths(element, path);
});
path.pop();
}
}
rootToLeafPaths(tree[0], []);
console.log(JSON.stringify(result));
.as-console-wrapper{top:0;max-height:100%!important;}