ForEach loop not printing the 4 element of the array - JavaScript - javascript

Here's my code -
var a = 'abcde'.split('');
var b = 'cxefgh'.split('');
//abfgh
a.forEach((e, index) => {
console.log(e, index)
if (b.indexOf(e) > -1) {
b.splice(b.indexOf(e), 1);
a.splice(index, 1);
}
})
//Here the output -
//a 0
//b 1
//c 2
//e 3
Why this loop isn't printing d. What's the silliest thing that I am missing here?

You are changing a while iterating on it, resulting in discrepancy.
You should consider working on clone of a (created with a.slice()) and then changing the a and b, if matches are found.
var a = 'abcde'.split('');
var b = 'cxefgh'.split('');
//abfgh
a.slice().forEach((e , index)=> {
console.log(e, index)
if(b.indexOf(e) > -1) {
b.splice(b.indexOf(e) , 1); // Find where it is in `b` and remove it.
a.splice(a.indexOf(e) , 1); // Find where it is in `a` and remove it.
}
})
console.log(a); // ['a', 'b', 'd']
console.log(b); // ['x', 'f', 'g', 'h']

a.splice(index , 1); is executed when e is 'c'. index is 2 at that point, so a becomes ['a', 'b', 'd', 'e'] (the splice removes the 'c'). Then the forEach moves on with index 3, which is (now) 'e'.

Related

Better way to check if an element only exists in one array

I need help with creating a function to return the elements that are only present in one of 3 arrays, for example
let arr1 = ['a', 'b', 'c', 'a', 'b']
let arr2 = ['a', 'd', 'b', 'c']
let arr3 = ['f', 'c', 'a']
In the three arrays above, 'd' and 'f' are found only in one of the arrays (arr2 and arr3), I need to return them.
['d','f']
The arrays can be of different sizes and the returned elements must not be duplicated.
I tried to find better alternatives, but I failed and just went with the brute force approach, looping through each array and checking if the element exists in the other two arrays, but obviously, it's really slow and hard to read.
function elementsInOnlyOneArr(a1, a2, a3) {
let myArr = [];
for(let el of a1){
if(a2.includes(el) == false && a3.includes(el) == false && myArr.includes(el) == false){
myArr.push(el);
}
}
for(let el of a2){
if(a1.includes(el) == false && a3.includes(el) == false && myArr.includes(el) == false){
myArr.push(el);
}
}
for(let el of a3){
if(a2.includes(el) == false && a1.includes(el) == false && myArr.includes(el) == false){
myArr.push(el);
}
}
return myArr;
}
Assuming there are less than 32 arrays, you can do this efficiently with bitmaps. Basically, build an index key -> number where the number has the Nth bit set if the key is in the Nth array. Finally return keys whose numbers only have a single bit set (=are powers of two):
function difference(...arrays) {
let items = {}
for (let [n, a] of arrays.entries())
for (let x of a) {
items[x] = (items[x] ?? 0) | (1 << n)
}
return Object.keys(items).filter(x =>
Number.isInteger(Math.log2(items[x])))
}
let arr1 = ['a', 'b', 'c', 'a', 'b', 'z', 'z', 'z']
let arr2 = ['a', 'd', 'b', 'c']
let arr3 = ['f', 'c', 'a']
console.log(difference(arr1, arr2, arr3))
(As noted in the comments x & (x-1) === 0 would be more idiomatic to check whether x is a power of two. See How does the formula x & (x - 1) works? for explanations.)
Here's a more general approach that doesn't limit the number of arrays and doesn't require keys to be strings:
function difference(...arrays) {
let items = new Map
for (let [n, a] of arrays.entries())
for (let x of a) {
if (!items.has(x))
items.set(x, new Set)
items.get(x).add(n)
}
let result = []
for (let [x, ns] of items)
if (ns.size === 1)
result.push(x)
return result
}
let arr1 = ['a', 'b', 'c', 'a', 'b', 'z', 'z', 'z']
let arr2 = ['a', 'd', 'b', 'c']
let arr3 = ['f', 'c', 'a']
console.log(difference(arr1, arr2, arr3))
EDIT: Misunderstood OP and it's not an intersect, but extracting values that are unique (e.g. NOT the intersection) between the individual arrays, for that this might work:
let arr1 = ['a', 'b', 'c', 'a', 'b'];
let arr2 = ['a', 'd', 'b', 'c'];
let arr3 = ['f', 'c', 'a'];
const thereCanOnlyBeOne = function(...arrs) {
return Array.from(
arrs.reduce((map, arr) => {
new Set(arr).forEach((v) => map.set(v, map.has(v) ? map.get(v)+1 : 1));
return map;
}, new Map())
)
.filter(([value, count]) => count === 1)
.map(([value, count]) => value);
};
console.log(thereCanOnlyBeOne(arr1, arr2, arr3));
I would think #gog's answer is way more sophisticated and probably much faster, but i have a slightly hard time wrapping my head around it (call me stupid, i take it =D, EDIT: had to do some research, read/learn something about bitsets here and here), so here's the breakdown of the slightly convoluted way of doing this with a Map and array methods:
pass all arrays to be analyzed into function, order doesn't matter
Loop (i chose reduce, but any loop structure works) trough all input arrays and their values, counting up occurrences in the Map, at the end the Map will look as follows:
0: {"a" => 4}
1: {"b" => 3}
2: {"c" => 3}
3: {"d" => 1}
4: {"f" => 1}
Once done with that, we convert the Map back into an array via Array.from() creating an array of tuples:
[
["a", 4],
["b", 3],
["c", 3],
["d", 1],
["f", 1],
]
Filter that resulting array of tuples (now in the form of [<value>, <count>] to only be left with values that exactly occurred once, leaving us with:
[
["d", 1],
["f", 1],
]
Map over the filtered array to "dumb" it down into a one-dimensional array again and return the result:
["d", "f"]
WARNING: Internally this code does a ****load of loops, so call it a brute-force loop as well, it just looks "shorter" due to "sexy" ES6 array-syntax-sugar.
A slightly modified version for completeness as the Array.filter() step can be omitted (although it seems to be faster) by iterating the counter-Map once it's finalized and simply deleting Map-entries that do not have value 1.
let arr1 = ['a', 'b', 'c', 'a', 'b'];
let arr2 = ['a', 'd', 'b', 'c'];
let arr3 = ['f', 'c', 'a'];
const thereCanOnlyBeOne = function(...arrs) {
let result;
arrs
.reduce((map, arr) => {
new Set(arr).forEach((v) => map.set(v, map.has(v) ? map.get(v)+1 : 1));
return map;
}, new Map())
// the result of .reduce will be a Map!
.forEach((value, key, map) => { value !== 1 && map.delete(key); result = map; });
return Array.from(result).map(([value, count]) => value);
};
console.log(thereCanOnlyBeOne(arr1, arr2, arr3));
UPDATE: as #Nick Parsons pointed out, the previous version of the code would not output elements that were only present in one array, but multiple times.
This will produce an incorrect output if one array contains the same value multiple times and that element isn't present in any other arrays. eg, if you remove b from arr2, then only arr1 has b in it but no others do, so it should b should be included in the final result.
This can easily be solved by turning the array that is checked into a Set() (thereby reducing the arrays values to "unique" ones).
If anyone (besides me) wonders, here's a benchmark between gog's options and mine, his bitset approach is clearly the fastest, so if you are comparing less than 32 arrays, that's the most performant solution by far: https://jsben.ch/YkKSu
and if anyone prefers an ES6-ified version of gog's bitset implementation (improved by #ralphmerridew suggestion), here you go:
let arr1 = ['a', 'b', 'c', 'a', 'b'];
let arr2 = ['a', 'd', 'b', 'c'];
let arr3 = ['f', 'c', 'a'];
function onlyone(...arrays) {
return Object.entries(
arrays.reduce((map, arr, n) => {
arr.forEach((v) => map[v] = (map[v] ?? 0) | (1 << n));
return map;
}, {})
)
.filter(([value, bitmap]) => (bitmap & (bitmap-1)) == 0)
.map(([value, bitmap]) => value);
};
console.log(onlyone(arr1, arr2, arr3));
updated the benchmark with this as well, interestingly (or unexpectedly) this "slower"-looking ES6 implementation somehow beats gog's for-loop implementation by a tad, tested in chrome and firefox multiple times, as i couldn't believe it myself, thought those syntax-sugar methods slow things down slightly compared to for loops, well...good to know =)
I also tried implementing the bitset approach with BigInt() to eliminate the issue with it only being able to deal with 32 arrays (depending on the Engine with BigInt it should be possible to deal with 1 million to 1 billion arrays), unfortunately that seems to make it the slowest of all solutions (benchmark updated):
let arr1 = ['a', 'b', 'c', 'a', 'b'];
let arr2 = ['a', 'd', 'b', 'c'];
let arr3 = ['f', 'c', 'a'];
function onlyoneBigInt(...arrays) {
return Object.entries(
arrays.reduce((map, arr, n) => {
arr.forEach((v) => map[v] = (map[v] ?? 0n) | (1n << BigInt(n)));
return map;
}, {})
)
.filter(([value, bitmap]) => (bitmap & (bitmap-1n)) == 0)
.map(([value, bitmap]) => value);
};
console.log(onlyoneBigInt(arr1, arr2, arr3));
Maybe someone sees something that can be improved to make this faster?
This is really just Set operations. The method single below finds any entry in a test array that does not appear in the other arrays in the collection. Deliberately implementing this so you can test individual arrays since it's not clear in the question if you need to return the letters, or the arrays.
let arr1 = ['a', 'b', 'c', 'a', 'b']
let arr2 = ['a', 'd', 'b', 'c']
let arr3 = ['f', 'c', 'a']
// The set of arrays
let arrays = [ arr1, arr2, arr3 ]
// Finds any entries in the test array that doesn't appear in the arrays that aren't the test arrays
let singles = (test) => {
// others is the Set of all value in the other arrays
others = arrays.reduce( ( accum, elem ) => {
if (elem != test) { elem.forEach(accum.add, accum) }
return accum
}, new Set())
// find anything in the test array that the others do not have
return [...new Set(test.filter( value => ! others.has(value) ))]
}
// collect results from testing all arrays
result = []
for(const array of arrays) { result.push(...singles(array))
}
console.log(result)
Borrowing the parameter construction from #gog's excellent answer, you could also define it so that it takes a test array and an arbitrary collection of arrays to test against:
let singles = (test, ...arrays) => {
// others is the Set of all value in the other arrays
others = arrays.reduce( ( accum, elem ) => {
if (elem != test) { elem.forEach(accum.add, accum) }
return accum
}, new Set())
// find anything in the test array that the others do not have
return [...new Set(test.filter( value => ! others.has(value) ))]
}
console.log(singles(arr2, arr1, arr2, arr3))
The advantage here is that this should work with any number of arrays, while gog's answer is probably faster for a collection of less than 32 arrays (or technically any number if you were willing to extend it using BigInt, but that may lose some of the speed)
A fairly simple approach:
const inOnlyOne = (
xss,
keys = [... new Set (xss .flat ())],
uniques = xss .map (xs => new Set (xs))
) => keys .filter (k => uniques .filter (f => f .has (k)) .length == 1)
console .log (inOnlyOne ([['a', 'b', 'c', 'a', 'b'], ['a', 'd', 'b', 'c'], ['f', 'c', 'a']]))
We find the list of unique keys by flattening our array of arrays and turning that into a Set and then back into an array, convert the arrays into Sets, then filter the keys to find only those where the number of sets including that key has exactly one entry.
There is a little inefficiency here in that we check all the Sets when seeing if a number is in there. It would be easy enough to modify it to check only until we find a second Set, but the code would be more complex. I would only bother to do so if I found that this simple version was not performant enough for my needs.
One advantage of this approach is that it works for other data types than strings and numbers:
const a = {a: 1}, b = {b: 3}, c = {c: 3}, d = {d: 4}, e = {e: 5}, f = {f: 6}
inOnlyOne ([[a, b, c, a, b], [a, d, b, c], [f, c, a]])
//=> [{d: 4}, {f: 6}]
Of course that only helps if your items are shared references. If you wanted to use value equality rather than reference equality, it would be significantly more complex.
If we wanted to pass the arrays individually, rather than wrap them in a common array, this variant should work:
const inOnlyOne = (...xss) => ((
keys = [... new Set (xss .flat ())],
uniques = xss .map (xs => new Set (xs))
) => keys .filter (k => uniques .filter (f => f .has (k)) .length == 1)
) ()
The Array.prototype.includes() method seems like the way to go here.
let arr1 = ['a', 'b', 'c', 'a', 'b']
let arr2 = ['a', 'd', 'b', 'c']
let arr3 = ['f', 'c', 'a', 'f']
var arrays = [arr1,arr2,arr3];
const items = arr1.concat(arr2, arr3);
let results = [];
items.forEach(isInOneArray);
function isInOneArray(item){
let found = 0;
for (const arr of arrays){
if (arr.includes(item)){
found ++;
}
}
if (found===1 && !results.includes(item)){
results.push(item);
}
}
console.log(results);
This is a brute force iterator much like your own, but reduces the number of re-entries by removing items from the array:
function elementsInOnlyOneArr(...arrays) {
// de-dup and sort so we process the longest array first
let sortedArrays = arrays.map(ar => [...new Set(ar)]).sort((a,b) => b.length - a.length);
for (let ai1 = 0 ; ai1 < sortedArrays.length; ai1 ++) {
for(let i = sortedArrays[ai1].length - 1; i >= 0; i --){
let exists = false;
let val = sortedArrays[ai1][i];
for(let ai2 = ai1 + 1 ; ai2 < sortedArrays.length ; ai2 ++) {
let foundIndex = sortedArrays[ai2].indexOf(val);
if (foundIndex >= 0) {
exists = true;
sortedArrays[ai2].splice(foundIndex,1);
// do not break, check for match in the other arrays
}
}
// if there was a match in any of the other arrays, remove it from the first one too!
if (exists)
sortedArrays[ai1].splice(i,1);
}
}
// concat the remaining elements, they are all unique
let output = sortedArrays[0];
for(let i = 1; i < sortedArrays.length; i ++)
output = output.concat(sortedArrays[i]);
return output;
}
let arr1 = ['a', 'b', 'c', 'a', 'b']
let arr2 = ['a', 'd', 'b', 'c']
let arr3 = ['f', 'c', 'a']
console.log(elementsInOnlyOneArr(arr1,arr2,arr3));
See this fiddle: https://jsfiddle.net/4deq7xwm/
Updated - Use splice() instead of pop()
Create a collection of pairs (x,y) where x is an element (in your case, a string) and y identifies the array it comes from. Sort this in O(log n) time by x first (where n is the total number of items over all arrays). It is easy to iterate over the result and detect the desired items.
This is easily solved with the built-in .lastIndexOf() Array method:
const arr1 = ['a', 'b', 'c', 'a', 'b'];
const arr2 = ['a', 'd', 'b', 'c'];
const arr3 = ['f', 'c', 'a'];
function getValuesInOneArray(...arrays) {
const combinedArr = arrays.flat();
const result = [];
for (const value of combinedArr) {
if (combinedArr.indexOf(value) === combinedArr.lastIndexOf(value)) {
result.push(value);
}
}
return result;
}
getValuesInOneArray(arr1, arr2, arr3); // ['d', 'f']
I generally try to avoid "ninja code" for the benefit of maintainability and readability, but I couldn't resist rewriting the above getValuesInOneArray() function as a slicker arrow function.
const getValuesInOneArray = (...arrays) =>
arrays
.flat()
.filter(
(value, index, array) => array.indexOf(value) === array.lastIndexOf(value)
);
You can read more about "ninja code" (and why you should avoid it) here, on Javacript.info, but I recommend avoiding practices like this in production codebases.
Hope this helps.
function elementsInOnlyOneArr(arr1, arr2, arr3){
let arr = arr1.concat(arr2).concat(arr3);
return removeDuplicate(arr);
}
function removeDuplicate(arr){
for(each of arr){
let count = 0;
for(ch of arr){
if(each === ch){
count++;
if(count > 1){
//removing element that exist more than one
arr = arr.filter(item => item !== each);
return removeDuplicate(arr);
}
}
}
}
return arr;
}
let arr1 = ['a', 'b', 'c', 'a', 'b'];
let arr2 = ['a', 'd', 'b', 'c'];
let arr3 = ['f', 'c', 'a'];
console.log(elementsInOnlyOneArr(arr1, arr2, arr3));
Do a diff of each of the array and concat those to get the unique values only in any one of the arrays.
const arr1 = ['a', 'b', 'c', 'a', 'b'];
const arr2 = ['a', 'd', 'b', 'c'];
const arr3 = ['f', 'c', 'a'];
function diff(a1, a2, a3) {
let u1 = a1.filter(el => { return !a2.includes(el) })
.filter(el => { return !a3.includes(el) });
let u2 = a2.filter(el => { return !a1.includes(el) })
.filter(el => { return !a3.includes(el) });
let u3 = a3.filter(el => { return !a2.includes(el) })
.filter(el => { return !a1.includes(el) });
return u1.concat(u2).concat(u3);
}
/* diff them */
const adiff = diff(arr1, arr2, arr3);
console.log(adiff);

How to solve this Javascript recursion problem finding number of occurrence of letter in a array

I know there is some problem with the return statement but I cant figure out what.
let arr = ['a', 'b', 'c', 'a']
let size = arr.length - 1
let counter = 0
function findOcc(arr, size, x) {
if (size === 0) {
return counter
} else {
if (arr[size] === x) return counter++
findOcc(arr, size - 1, 'a')
}
}
console.log(findOcc(arr, size, "a"))
A simpler recursive version uses array destructuring to test the first value against the target value, adding one if it matches, and zero if it doesn't, then recurring on the remainder of the array. We stop when the value is undefined, meaning we've run out of elements in the array.
const countOcc = (target) => ([x, ...xs]) =>
x == undefined ? 0 : (x == target ? 1 : 0) + countOcc (target) (xs)
const arr = ['a', 'b', 'c', 'a']
console .log (countOcc ('a') (arr))
console .log (countOcc ('b') (arr))
console .log (countOcc ('c') (arr))
console .log (countOcc ('d') (arr))
While we could alter this to make it tail-recursive, current JS engines still do not do tail-call optimization, so it seems a bit pointless. If you were going to do this on large arrays, you would probably need to rewrite with iteration in place of recursion.
It seems you got the logic right in your head, but didn't implement it well.
So for the base condition you say if (size===0) return counter. This means that you
just return counter but arr[0] is a valid check. So what you want really when you are checking elements is arr[size-1] in the next line i.e. else
For the else condition you meant to increment the counter which is right, but you should n't return. Instead you return the value of the next call with size-1
Also your declaration of size should be let size = arr.length
Also the last argument should be x and not a inside the recursive call.
With that the code is:
let arr = ['a', 'b', 'c', 'a']
let size = arr.length
let counter = 0
function findOcc(arr,size,x) {
if (size === 0) return counter
if(arr[size-1]===x) ++counter
return findOcc(arr,size-1,x)
}
console.log(findOcc(arr, size, "a"))
If one is okay with mutating the original array, then the function may be simplified like below:
const origArr = ['a', 'b', 'c', 'a'];
// a simpler implementation of the same recursive function
function findOcc(arr, x) {
return (
arr.length // if "arr" length is 1 or more
? arr.pop() === x // ".pop()" the last elt & compare with "x"
? 1 + findOcc(arr, x) // add 1 & recurse using the shorted/mutated "arr"
: findOcc(arr, x) // do not add 1, but recurse using the mutate "arr"
: 0 // no more elements in "arr", so return 0
)
}
// using "..." spread to avoid "arr" from being changed
console.log('number of times "a" occurs is: ', findOcc([...origArr], "a"))
console.log(
`"d" occurs in ["${origArr.join('", "')}"] ${findOcc([...origArr], "d")} times...`
)

How to work re-organize array so that its sorted in columns rather than rows [duplicate]

This question already has answers here:
Transposing a 2D-array in JavaScript
(25 answers)
Closed 3 years ago.
I have some checkboxes on a page that are organized in 3 columns. However they are alphabetically sorted and go across the rows instead.
What is currently happening:
A B C
D E F
G H I
What I want to happen:
A D G
B E H
C F I
Right now I've got this code to work it out but I feel super confused by what to do when theres a remainder in the array and I don't know how to include it into the sorting without messing it up:
organizedCheckboxesArea() {
let quotient = Math.floor(this.areas.length/3);
let remainder = this.areas.length % 3;
let organizedColumns = [];
for(let i = 0 ; i < quotient; i++) {
organizedColumns.push(this.areas[i]);
organizedColumns.push(this.areas[quotient + i]);
organizedColumns.push(this.areas[(quotient * 2) + i]);
}
return organizedColumns;
},
Its frustrating me a lot because I know the solution is simple I just can't think of it.
EDIT: this is not a 2 dimensional array. it is ['A', 'B', 'C', 'D' etc...].
Also for when there is a remainder, I want it to stack to the left two columms. ie ABCDEFGHIJ would make:
A E H
B F I
C G J
D
And ABCDEFGHIJK would make:
A E I
B F J
C G K
D H
You could switch the indices and get a new array.
var data = [['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H', 'I']],
result = data.reduce((r, a, j) => {
a.forEach((v, i) => {
r[i] = r[i] || [];
r[i][j] = v;
});
return r;
}, []);
result.forEach(a => console.log(...a));
Question answered here: Transposing a 2D-array in JavaScript
array[0].map((col, i) => array.map(row => row[i]));
map calls a provided callback function once for each element in an array, in order, and constructs a new array from the results. callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values.
callback is invoked with three arguments: the value of the element, the index of the element, and the Array object being traversed. [source]
Vanila JS: live code here
let A = [
['A', 'B', 'C'],
['D', 'E', 'F'],
['G', 'H', 'I'],
];
A[0].map((col, i) => A.map(row => row[i]));
A[0].map((_, i) => A.map(row => row[i])); // This also works
console.log(A);
using Ramda:
R.transpose(A);
using lodash:
_.zip.apply(_,A);

Changing array position in javascript

I want to move to the part I want in the array [a, b, c, d, e, f]. For example, how do you move b with index 1 to e?
The array I want is:
[a, c, d, e, b, f]
You can use .splice()
.splice first argument is the start position where you want to delete. The second argument is how many elements you want to delete. Then the rest of arguments are elements you want to append at the deleted element index.
const array = ['a', 'b', 'c', 'd', 'e', 'f'];
// Delete element and save it to a variable
const el = array.splice(1, 1);
// Add deleted element to the required position
array.splice(4, 0, el[0]);
console.log(array);
I think you can object destructure to create what you are looking for. Though, I am not sure if this is the best way you can do it.
const a = ['a', 'b', 'c', 'd', 'e', 'f']
// I would just create a new array that would work like this
const newA = [
// everything before index 1
...a.slice(0, 1),
// everything after index 1 till index 4 and then the index 1
...[...a.slice(1+1,4+1), a[1]],
// everything after index 4
...a.slice(4+1)
]
console.log(newA)
You can use temp variable for temporary store data and loop from your wanted move index to target index
var d = ['a', 'b', 'c', 'd', 'e', 'f'];
function moveTo(data, from, to) {
let _tmp = data
let _move = data[from]
for(var i = from; i <= to; i++) {
if(i === to) _tmp[i] = _move
else _tmp[i] = data[i+1]
}
data = _tmp
}
moveTo(d, 1, 4)
console.log(d)
Using Array.splice:
This gives us an array containing just the element from the index you give:
arr.splice(from, 1)
We then insert this at the chosen index (Using the spread operator to expand the array)
arr.splice(to, 0, ...arr.splice(from, 1))
let arr = ['a', 'b', 'c', 'd', 'e', 'f'];
function moveElement (targetArray, from, to) {
targetArray.splice(to, 0, ...targetArray.splice(from, 1))
}
moveElement(arr, 1, 4);
console.log(arr);

Move an array element from one array position to another

I'm having a hard time figuring out how to move an element of an array. For example, given the following:
var array = [ 'a', 'b', 'c', 'd', 'e'];
How can I write a function to move the element 'd' to the left of 'b' ?
Or 'a' to the right of 'c'?
After moving the elements, the indexes of the rest of the elements should be updated. The resulting array would be:
array = ['a', 'd', 'b', 'c', 'e']
This seems like it should be pretty simple, but I can't wrap my head around it.
If you'd like a version on npm, array-move is the closest to this answer, although it's not the same implementation. See its usage section for more details. The previous version of this answer (that modified Array.prototype.move) can be found on npm at array.prototype.move.
I had fairly good success with this function:
function array_move(arr, old_index, new_index) {
if (new_index >= arr.length) {
var k = new_index - arr.length + 1;
while (k--) {
arr.push(undefined);
}
}
arr.splice(new_index, 0, arr.splice(old_index, 1)[0]);
return arr; // for testing
};
// returns [2, 1, 3]
console.log(array_move([1, 2, 3], 0, 1));
Note that the last return is simply for testing purposes: splice performs operations on the array in-place, so a return is not necessary. By extension, this move is an in-place operation. If you want to avoid that and return a copy, use slice.
Stepping through the code:
If new_index is greater than the length of the array, we want (I presume) to pad the array properly with new undefineds. This little snippet handles this by pushing undefined on the array until we have the proper length.
Then, in arr.splice(old_index, 1)[0], we splice out the old element. splice returns the element that was spliced out, but it's in an array. In our above example, this was [1]. So we take the first index of that array to get the raw 1 there.
Then we use splice to insert this element in the new_index's place. Since we padded the array above if new_index > arr.length, it will probably appear in the right place, unless they've done something strange like pass in a negative number.
A fancier version to account for negative indices:
function array_move(arr, old_index, new_index) {
while (old_index < 0) {
old_index += arr.length;
}
while (new_index < 0) {
new_index += arr.length;
}
if (new_index >= arr.length) {
var k = new_index - arr.length + 1;
while (k--) {
arr.push(undefined);
}
}
arr.splice(new_index, 0, arr.splice(old_index, 1)[0]);
return arr; // for testing purposes
};
// returns [1, 3, 2]
console.log(array_move([1, 2, 3], -1, -2));
Which should account for things like array_move([1, 2, 3], -1, -2) properly (move the last element to the second to last place). Result for that should be [1, 3, 2].
Either way, in your original question, you would do array_move(arr, 0, 2) for a after c. For d before b, you would do array_move(arr, 3, 1).
I like this way. It's concise and it works.
function arraymove(arr, fromIndex, toIndex) {
var element = arr[fromIndex];
arr.splice(fromIndex, 1);
arr.splice(toIndex, 0, element);
}
Note: always remember to check your array bounds.
Run Snippet in jsFiddle
Here's a one liner I found on JSPerf....
Array.prototype.move = function(from, to) {
this.splice(to, 0, this.splice(from, 1)[0]);
};
which is awesome to read, but if you want performance (in small data sets) try...
Array.prototype.move2 = function(pos1, pos2) {
// local variables
var i, tmp;
// cast input parameters to integers
pos1 = parseInt(pos1, 10);
pos2 = parseInt(pos2, 10);
// if positions are different and inside array
if (pos1 !== pos2 && 0 <= pos1 && pos1 <= this.length && 0 <= pos2 && pos2 <= this.length) {
// save element from position 1
tmp = this[pos1];
// move element down and shift other elements up
if (pos1 < pos2) {
for (i = pos1; i < pos2; i++) {
this[i] = this[i + 1];
}
}
// move element up and shift other elements down
else {
for (i = pos1; i > pos2; i--) {
this[i] = this[i - 1];
}
}
// put element from position 1 to destination
this[pos2] = tmp;
}
}
I can't take any credit, it should all go to Richard Scarrott. It beats the splice based method for smaller data sets in this performance test. It is however significantly slower on larger data sets as Darwayne points out.
The splice() method adds/removes items to/from an array, and returns the removed item(s).
Note: This method changes the original array. /w3schools/
Array.prototype.move = function(from,to){
this.splice(to,0,this.splice(from,1)[0]);
return this;
};
var arr = [ 'a', 'b', 'c', 'd', 'e'];
arr.move(3,1);//["a", "d", "b", "c", "e"]
var arr = [ 'a', 'b', 'c', 'd', 'e'];
arr.move(0,2);//["b", "c", "a", "d", "e"]
as the function is chainable this works too:
alert(arr.move(0,2).join(','));
demo here
My 2c. Easy to read, it works, it's fast, it doesn't create new arrays.
function move(array, from, to) {
if( to === from ) return array;
var target = array[from];
var increment = to < from ? -1 : 1;
for(var k = from; k != to; k += increment){
array[k] = array[k + increment];
}
array[to] = target;
return array;
}
Here is my one liner ES6 solution with an optional parameter on.
if (typeof Array.prototype.move === "undefined") {
Array.prototype.move = function(from, to, on = 1) {
this.splice(to, 0, ...this.splice(from, on))
}
}
Adaptation of the first solution proposed by digiguru
The parameter on is the number of element starting from from you want to move.
Here is a chainable variation of this:
if (typeof Array.prototype.move === "undefined") {
Array.prototype.move = function(from, to, on = 1) {
return this.splice(to, 0, ...this.splice(from, on)), this
}
}
[3, 4, 5, 1, 2].move(3, 0, 2) // => [1, 2, 3, 4, 5]
If you'd like to avoid prototype pollution, here's a stand-alone function:
function move(array, from, to, on = 1) {
return array.splice(to, 0, ...array.splice(from, on)), array
}
move([3, 4, 5, 1, 2], 3, 0, 2) // => [1, 2, 3, 4, 5]
And finally, here's a pure function that doesn't mutate the original array:
function moved(array, from, to, on = 1) {
return array = array.slice(), array.splice(to, 0, ...array.splice(from, on)), array
}
This should cover basically every variation seen in every other answer.
Got this idea from #Reid of pushing something in the place of the item that is supposed to be moved to keep the array size constant. That does simplify calculations. Also, pushing an empty object has the added benefits of being able to search for it uniquely later on. This works because two objects are not equal until they are referring to the same object.
({}) == ({}); // false
So here's the function which takes in the source array, and the source, destination indexes. You could add it to the Array.prototype if needed.
function moveObjectAtIndex(array, sourceIndex, destIndex) {
var placeholder = {};
// remove the object from its initial position and
// plant the placeholder object in its place to
// keep the array length constant
var objectToMove = array.splice(sourceIndex, 1, placeholder)[0];
// place the object in the desired position
array.splice(destIndex, 0, objectToMove);
// take out the temporary object
array.splice(array.indexOf(placeholder), 1);
}
This is based on #Reid's solution. Except:
I'm not changing the Array prototype.
Moving an item out of bounds to the right does not create undefined items, it just moves the item to the right-most position.
Function:
function move(array, oldIndex, newIndex) {
if (newIndex >= array.length) {
newIndex = array.length - 1;
}
array.splice(newIndex, 0, array.splice(oldIndex, 1)[0]);
return array;
}
Unit tests:
describe('ArrayHelper', function () {
it('Move right', function () {
let array = [1, 2, 3];
arrayHelper.move(array, 0, 1);
assert.equal(array[0], 2);
assert.equal(array[1], 1);
assert.equal(array[2], 3);
})
it('Move left', function () {
let array = [1, 2, 3];
arrayHelper.move(array, 1, 0);
assert.equal(array[0], 2);
assert.equal(array[1], 1);
assert.equal(array[2], 3);
});
it('Move out of bounds to the left', function () {
let array = [1, 2, 3];
arrayHelper.move(array, 1, -2);
assert.equal(array[0], 2);
assert.equal(array[1], 1);
assert.equal(array[2], 3);
});
it('Move out of bounds to the right', function () {
let array = [1, 2, 3];
arrayHelper.move(array, 1, 4);
assert.equal(array[0], 1);
assert.equal(array[1], 3);
assert.equal(array[2], 2);
});
});
You can implement some basic calculus and create a universal function for moving array elements from one position to the other.
For JavaScript it looks like this:
function magicFunction (targetArray, indexFrom, indexTo) {
targetElement = targetArray[indexFrom];
magicIncrement = (indexTo - indexFrom) / Math.abs (indexTo - indexFrom);
for (Element = indexFrom; Element != indexTo; Element += magicIncrement){
targetArray[Element] = targetArray[Element + magicIncrement];
}
targetArray[indexTo] = targetElement;
}
Check out "moving array elements" at "Gloommatter" for detailed explanation.
https://web.archive.org/web/20121105042534/http://www.gloommatter.com:80/DDesign/programming/moving-any-array-elements-universal-function.html
I've implemented an immutable ECMAScript 6 solution based off of #Merc's answer over here:
const moveItemInArrayFromIndexToIndex = (array, fromIndex, toIndex) => {
if (fromIndex === toIndex) return array;
const newArray = [...array];
const target = newArray[fromIndex];
const inc = toIndex < fromIndex ? -1 : 1;
for (let i = fromIndex; i !== toIndex; i += inc) {
newArray[i] = newArray[i + inc];
}
newArray[toIndex] = target;
return newArray;
};
The variable names can be shortened, just used long ones so that the code can explain itself.
One approach would be to create a new array with the pieces in the order you want, using the slice method.
Example
var arr = [ 'a', 'b', 'c', 'd', 'e'];
var arr2 = arr.slice(0,1).concat( ['d'] ).concat( arr.slice(2,4) ).concat( arr.slice(4) );
arr.slice(0,1) gives you ['a']
arr.slice(2,4) gives you ['b', 'c']
arr.slice(4) gives you ['e']
Another pure JS variant using ES6 array spread operator with no mutation
const reorder = (array, sourceIndex, destinationIndex) => {
const smallerIndex = Math.min(sourceIndex, destinationIndex);
const largerIndex = Math.max(sourceIndex, destinationIndex);
return [
...array.slice(0, smallerIndex),
...(sourceIndex < destinationIndex
? array.slice(smallerIndex + 1, largerIndex + 1)
: []),
array[sourceIndex],
...(sourceIndex > destinationIndex
? array.slice(smallerIndex, largerIndex)
: []),
...array.slice(largerIndex + 1),
];
}
// returns ['a', 'c', 'd', 'e', 'b', 'f']
console.log(reorder(['a', 'b', 'c', 'd', 'e', 'f'], 1, 4))
The splice method of Array might help: https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Array/splice
Just keep in mind it might be relatively expensive since it has to actively re-index the array.
I needed an immutable move method (one that didn't change the original array), so I adapted #Reid's accepted answer to simply use Object.assign to create a copy of the array before doing the splice.
Array.prototype.immutableMove = function (old_index, new_index) {
var copy = Object.assign([], this);
if (new_index >= copy.length) {
var k = new_index - copy.length;
while ((k--) + 1) {
copy.push(undefined);
}
}
copy.splice(new_index, 0, copy.splice(old_index, 1)[0]);
return copy;
};
Here is a jsfiddle showing it in action.
Array.prototype.moveUp = function (value, by) {
var index = this.indexOf(value),
newPos = index - (by || 1);
if (index === -1)
throw new Error("Element not found in array");
if (newPos < 0)
newPos = 0;
this.splice(index, 1);
this.splice(newPos, 0, value);
};
Array.prototype.moveDown = function (value, by) {
var index = this.indexOf(value),
newPos = index + (by || 1);
if (index === -1)
throw new Error("Element not found in array");
if (newPos >= this.length)
newPos = this.length;
this.splice(index, 1);
this.splice(newPos, 0, value);
};
var arr = ['banana', 'curyWurst', 'pc', 'remembaHaruMembaru'];
alert('withiout changes= '+arr[0]+' ||| '+arr[1]+' ||| '+arr[2]+' ||| '+arr[3]);
arr.moveDown(arr[2]);
alert('third word moved down= '+arr[0] + ' ||| ' + arr[1] + ' ||| ' + arr[2] + ' ||| ' + arr[3]);
arr.moveUp(arr[2]);
alert('third word moved up= '+arr[0] + ' ||| ' + arr[1] + ' ||| ' + arr[2] + ' ||| ' + arr[3]);
http://plnkr.co/edit/JaiAaO7FQcdPGPY6G337?p=preview
Here's one way to do it in an immutable way. It handles negative numbers as well as an added bonus. This is reduces number of possible bugs at the cost of performance compared to editing the original array.
const numbers = [1, 2, 3];
const moveElement = (array, from, to) => {
const copy = [...array];
const valueToMove = copy.splice(from, 1)[0];
copy.splice(to, 0, valueToMove);
return copy;
};
console.log(moveElement(numbers, 0, 2))
// > [2, 3, 1]
console.log(moveElement(numbers, -1, -3))
// > [3, 1, 2]
One approach would be to use splice() to remove the item from the array and then by using the splice() method once again, insert the removed item into the target index.
const array = ['a', 'b', 'c', 'd', 'e']
const newArray = moveItem(array, 3, 1) // move element from index 3 to index 1
function moveItem(arr, fromIndex, toIndex){
let itemRemoved = arr.splice(fromIndex, 1) // assign the removed item as an array
arr.splice(toIndex, 0, itemRemoved[0]) // insert itemRemoved into the target index
return arr
}
console.log(newArray)
In 2022, this typescript utility will work along with a unit test.
export const arrayMove = <T>(arr: T[], fromIndex: number, toIndex: number) => {
const newArr = [...arr];
newArr.splice(toIndex, 0, newArr.splice(fromIndex, 1)[0]);
return newArr;
};
const testArray = ['1', '2', '3', '4'];
describe('arrayMove', () => {
it('should move array item to toIndex', () => {
expect(arrayMove(testArray, 2, 0)).toEqual(['3', '1', '2', '4']);
expect(arrayMove(testArray, 3, 1)).toEqual(['1', '4', '2', '3']);
expect(arrayMove(testArray, 1, 2)).toEqual(['1', '3', '2', '4']);
expect(arrayMove(testArray, 0, 2)).toEqual(['2', '3', '1', '4']);
});
});
I love immutable, functional one liners :) ...
const swapIndex = (array, from, to) => (
from < to
? [...array.slice(0, from), ...array.slice(from + 1, to + 1), array[from], ...array.slice(to + 1)]
: [...array.slice(0, to), array[from], ...array.slice(to, from), ...array.slice(from + 1)]
);
Find and move an element from "n"th position to 0th position.
Eg: Find and move 'd' to 0th position:
let arr = [ 'a', 'b', 'c', 'd', 'e'];
arr = [...arr.filter(item => item === 'd'), ...arr.filter(item => item !== 'd')];
It is stated in many places (adding custom functions into Array.prototype) playing with the Array prototype could be a bad idea, anyway I combined the best from various posts, I came with this, using modern Javascript:
Object.defineProperty(Array.prototype, 'immutableMove', {
enumerable: false,
value: function (old_index, new_index) {
var copy = Object.assign([], this)
if (new_index >= copy.length) {
var k = new_index - copy.length;
while ((k--) + 1) { copy.push(undefined); }
}
copy.splice(new_index, 0, copy.splice(old_index, 1)[0]);
return copy
}
});
//how to use it
myArray=[0, 1, 2, 3, 4];
myArray=myArray.immutableMove(2, 4);
console.log(myArray);
//result: 0, 1, 3, 4, 2
Hope can be useful to anyone
This version isn't ideal for all purposes, and not everyone likes comma expressions, but here's a one-liner that's a pure expression, creating a fresh copy:
const move = (from, to, ...a) => (a.splice(to, 0, ...a.splice(from, 1)), a)
A slightly performance-improved version returns the input array if no move is needed, it's still OK for immutable use, as the array won't change, and it's still a pure expression:
const move = (from, to, ...a) =>
from === to
? a
: (a.splice(to, 0, ...a.splice(from, 1)), a)
The invocation of either is
const shuffled = move(fromIndex, toIndex, ...list)
i.e. it relies on spreading to generate a fresh copy. Using a fixed arity 3 move would jeopardize either the single expression property, or the non-destructive nature, or the performance benefit of splice. Again, it's more of an example that meets some criteria than a suggestion for production use.
const move = (from, to, ...a) =>from === to ? a : (a.splice(to, 0, ...a.splice(from, 1)), a);
const moved = move(0, 2, ...['a', 'b', 'c']);
console.log(moved)
I thought this was a swap problem but it's not. Here's my one-liner solution:
const move = (arr, from, to) => arr.map((item, i) => i === to ? arr[from] : (i >= Math.min(from, to) && i <= Math.max(from, to) ? arr[i + Math.sign(to - from)] : item));
Here's a small test:
let test = ['a', 'b', 'c', 'd', 'e'];
console.log(move(test, 0, 2)); // [ 'b', 'c', 'a', 'd', 'e' ]
console.log(move(test, 1, 3)); // [ 'a', 'c', 'd', 'b', 'e' ]
console.log(move(test, 2, 4)); // [ 'a', 'b', 'd', 'e', 'c' ]
console.log(move(test, 2, 0)); // [ 'c', 'a', 'b', 'd', 'e' ]
console.log(move(test, 3, 1)); // [ 'a', 'd', 'b', 'c', 'e' ]
console.log(move(test, 4, 2)); // [ 'a', 'b', 'e', 'c', 'd' ]
console.log(move(test, 4, 0)); // [ 'e', 'a', 'b', 'c', 'd' ]
This is a really simple method using splice
Array.prototype.moveToStart = function(index) {
this.splice(0, 0, this.splice(index, 1)[0]);
return this;
};
I ended up combining two of these to work a little better when moving both small and large distances. I get fairly consistent results, but this could probably be tweaked a little bit by someone smarter than me to work differently for different sizes, etc.
Using some of the other methods when moving objects small distances was significantly faster (x10) than using splice. This might change depending on the array lengths though, but it is true for large arrays.
function ArrayMove(array, from, to) {
if ( Math.abs(from - to) > 60) {
array.splice(to, 0, array.splice(from, 1)[0]);
} else {
// works better when we are not moving things very far
var target = array[from];
var inc = (to - from) / Math.abs(to - from);
var current = from;
for (; current != to; current += inc) {
array[current] = array[current + inc];
}
array[to] = target;
}
}
https://web.archive.org/web/20181026015711/https://jsperf.com/arraymove-many-sizes
TypeScript Version
Copied from #Merc's answer. I like that one best because it is not creating new arrays and modifies the array in place. All I did was update to ES6 and add the types.
export function moveItemInArray<T>(workArray: T[], fromIndex: number, toIndex: number): T[] {
if (toIndex === fromIndex) {
return workArray;
}
const target = workArray[fromIndex];
const increment = toIndex < fromIndex ? -1 : 1;
for (let k = fromIndex; k !== toIndex; k += increment) {
workArray[k] = workArray[k + increment];
}
workArray[toIndex] = target;
return workArray;
}
Array.move.js
Summary
Moves elements within an array, returning an array containing the moved elements.
Syntax
array.move(index, howMany, toIndex);
Parameters
index: Index at which to move elements. If negative, index will start from the end.
howMany: Number of elements to move from index.
toIndex: Index of the array at which to place the moved elements. If negative, toIndex will start from the end.
Usage
array = ["a", "b", "c", "d", "e", "f", "g"];
array.move(3, 2, 1); // returns ["d","e"]
array; // returns ["a", "d", "e", "b", "c", "f", "g"]
Polyfill
Array.prototype.move || Object.defineProperty(Array.prototype, "move", {
value: function (index, howMany, toIndex) {
var
array = this,
index = parseInt(index) || 0,
index = index < 0 ? array.length + index : index,
toIndex = parseInt(toIndex) || 0,
toIndex = toIndex < 0 ? array.length + toIndex : toIndex,
toIndex = toIndex <= index ? toIndex : toIndex <= index + howMany ? index : toIndex - howMany,
moved;
array.splice.apply(array, [toIndex, 0].concat(moved = array.splice(index, howMany)));
return moved;
}
});
I used the nice answer of #Reid, but struggled with moving an element from the end of an array one step further - to the beginning (like in a loop).
E.g. ['a', 'b', 'c'] should become ['c', 'a', 'b'] by calling .move(2,3)
I achieved this by changing the case for new_index >= this.length.
Array.prototype.move = function (old_index, new_index) {
console.log(old_index + " " + new_index);
while (old_index < 0) {
old_index += this.length;
}
while (new_index < 0) {
new_index += this.length;
}
if (new_index >= this.length) {
new_index = new_index % this.length;
}
this.splice(new_index, 0, this.splice(old_index, 1)[0]);
return this; // for testing purposes
};
As an addition to Reid's excellent answer (and because I cannot comment);
You can use modulo to make both negative indices and too large indices "roll over":
function array_move(arr, old_index, new_index) {
new_index =((new_index % arr.length) + arr.length) % arr.length;
arr.splice(new_index, 0, arr.splice(old_index, 1)[0]);
return arr; // for testing
}
// returns [2, 1, 3]
console.log(array_move([1, 2, 3], 0, 1));

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