Related
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);
I have one array which contains values array1 = [a,b,c,d,e,f,g] and another array contains index of array1 and the value to be replaced array2 = [[2,u],[3,x],[6,z]].
Now I want to replace value of array1 with respect to array2 please not array2 consist of [array1_position, value_to_be_replaced]
Now my new arra1 should look like this array1 = [a,b,u,x,e,f,z].
I can do this with for loop but its again time consuming. what trick can i use to replace the value quickly. I am just learning about arrays and have a little knowledge on it.
I don't think there is any shorthand for such a verbose question.
Something on top of my mind is
array2.forEach((el) => {
array1[el[0]] = el[1];
})
you could do this
array2.forEach(arr => {
const [index, replaced] = arr;
array1[index] = replaced;
});
You can take fromEntries of the other array then just map it:
var array1 = ['a','b','c','d','e','f','g'];
var array2 = [[2,'u'],[3,'x'],[6,'z']];
var d=Object.fromEntries(array2);
var result = array1.map((val, i)=> d[i] || val);
console.log(result);
With an ordered array2 by indices, you could take a closure over an index of this array and map the values from eithe the first array or the replacements array.
var array1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g'],
array2 = [[2, 'u'], [3, 'x'], [6, 'z']],
result = array1.map(
(j => (v, i) => i === array2[j][0] ? array2[j++][1] : v)
(0)
);
console.log(result);
I have two arrays, id like to return the items from array1 if they appear as much as the length of array2
I understand how to do this in python, but I cant figure out how to do it in javascript
arr1 = ['a', 'b', 'c', 'c']
arr2 = ['one', 'two']
arr3 = []
for i in arr1:
if arr1.count(i) == len(arr2):
arr3.append(i)
desired result would be ['c']
Can someone please help me write this in javascript?
You can use reduce and filter
Here idea is :-
First use elements of arr1 as key on obj, if key is already there increment it's value, else set it to zero,
Now take the key's of obj and filter if it's value is equal to length of arr2
let arr1 = ['a', 'b', 'c', 'c']
let arr2 = ['one', 'two']
let obj = arr1.reduce((op,inp)=>{
let key = inp.toLowerCase()
op[key] = op[key] || 0
op[key]++
return op
},{})
let final = Object.keys(obj).filter(key=>{
return obj[key] === arr2.length
})
console.log(final)
Note :- Here i ignored case, if you want both case to be different than you can remove this line
let key = inp.toLowerCase()
Nested loops could do that. One for the for, and one for the count:
var arr1 = ['a', 'b', 'c', 'c'];
var arr2 = ['one', 'two'];
var arr3 = [];
var unique = new Set(arr1);
var len2 = arr2.length;
for(var i of unique){
var count = 0;
for(var j of arr1)
if(j === i)
count++;
if(count == len2)
arr3.push(i);
}
console.log(arr3);
Side remark: I think that Python code would result in ['c','c'], though I have not actually tried.
First use reduce() to create object of whose keys will elements of arr1 and value will be their count. Then filter() its keys according to given condition.
let arr1 = ['a', 'b', 'c', 'c']
let arr2 = ['one', 'two']
const obj = arr1.reduce((ac,a) => (ac[a] = ac[a] + 1 || 1, ac),{});
let res = Object.keys(obj).filter(k => obj[k] === arr2.length);
console.log(res)
i would like to split an array into multiple chunks but applying a function to it to decide how to create the chunks.
For example, if i have an array of letters, numbers or letters and numbers, apply a function to the array to split it into array of arrays of the previous categories.
let arr = ['a', 'b', '1', '2', 'a1', 'a2', 'c', '3', 'a3']
myChunkFunction(arr, myCustomSplitFunction)
// result
[['a','b','c'], ['1','2','3'], ['a1', 'a2','a3']]
Lodash has a chunk function but it splits into n chunks, also array has a slice function but you need to specify the begin and the end so how could i split with a custom function.
Try doing this
let arr = ['a', 'b', '1', '2', 'a1', 'a2', 'c', '3', 'a3']
const splitFn = (str) => Number.isInteger(+str) ? 0 : str.length == 1 ? 1 : 2
const myChunkFunction = (arr, fn) => arr.reduce((r,c) => {
let t = fn(c)
r[t] = [...r[t], c]
return r
}, [[],[],[]])
console.log(myChunkFunction(arr, splitFn))
Hint
The key to the answer is to, somehow, reorganize the source array such that all the elements with the same key will be in the same place.
The easiest way I can think to solve it is by using hash-map. Each element in the hash-map will be a different array containing all the elements with the same key.
Try it for your self before you keep reading and see the full solution.
The implementation
As you can see, I solved it as functional as possible. To avoid mutations, I used reduce to iterate over the source array and put each element in the hashmap (by generating a key from the element).
I recreate the final hash-map over and over using shallow copy. Finally, I convert the hash-map to an array of array (because that was your demand) using Object.values
const splitArrayByKey = extractKey => array => {
const arraysByKey_obj = array.reduce((hashMapOfArrays,element)=> {
const key = extractKey(element);
// if we already added an element with the same key,
// then we add the current element to there.
// else, we create a new key and put the current element there.
if(hashMapOfArrays.hasOwnProperty(key))
return {
...hashMapOfArrays,
[key]: [...hashMapOfArrays[key],element]
};
return {
...hashMapOfArrays,
[key]: [element]
};
},{});
// transform the arraysByKey_obj to an array of arrays:
return Object.values(arraysByKey_obj);
};
// example 1:
const result1 = splitArrayByKey(element=>element)([1,2,3,1,2,3]);
console.log(result1);
console.log('------------------');
// example 2:
const result2 = splitArrayByKey(element=>element.id)([{id:1,x:1},{id:{},x:2},{id:"id",x:3},{id:1,x:4}]);
console.log(result2);
Here is a way to do this via ES6:
let arr = ['a', 'b', '1', '2', 'a1', 'a2', 'c', '3', 'a3']
const splitFn = (str) => Number.isInteger(+str) ? 0 : str.length == 1 ? 1 : 2
const myChunkFunction = (arr, fn) => arr.reduce((r,c) => {
let t = fn(c)
r[t] = [...r[t], c]
return r
}, [[],[],[]])
console.log(myChunkFunction(arr, splitFn))
The typeFn plays the role of filtering the elements to number, string with 1 length and other. That output is used by the myChunkFunction to place the element in the right array.
You could do something like this with less control and in one line with reduce and ES6 array spread:
let arr = ['a', 'b', '1', '2', 'a1', 'a2', 'c', '3', 'a3']
const result = arr.reduce((r,c) =>
(Number.isInteger(+c) ? r[0] = [...r[0], c] :
c.length == 1 ? r[1] = [...r[1], c] : r[2] = [...r[2], c], r), [[],[],[]])
console.log(result)
You start with [[],[],[]] and fill each of the sub arrays based on number, length of the string == 1, other lenghts.
You could wrap that in a function.
const arr = ['a', 'b', '1', '2', 'a1', 'a2', 'c', '3', 'a3'];
const getClassification = function(x){
const hasNumber = x.split('').some(x => parseFloat(x));
const hasChar = x.split('').some(x => !parseFloat(x));
if(!parseFloat(x) && (!hasNumber && hasChar)) return 0;
else if(parseFloat(x)) return 1;
else return 2;
}
const myChunkFunction = function(arr, classifier){
let jaggedArray = [[], [], []];
arr.forEach(x => {
jaggedArray[classifier(x)].push(x);
})
return jaggedArray;
}
console.log(myChunkFunction(arr, getClassification));
I think this satisfies.
Is it possible to sort and rearrange an array that looks like this:
itemsArray = [
['Anne', 'a'],
['Bob', 'b'],
['Henry', 'b'],
['Andrew', 'd'],
['Jason', 'c'],
['Thomas', 'b']
]
to match the arrangement of this array:
sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ]
Unfortunately, I don’t have any IDs to keep track on. I would need to priority the items-array to match the sortingArr as close as possible.
Update:
Here is the output I’m looking for:
itemsArray = [
['Bob', 'b'],
['Jason', 'c'],
['Henry', 'b'],
['Thomas', 'b']
['Anne', 'a'],
['Andrew', 'd'],
]
Any idea how this can be done?
One-Line answer.
itemsArray.sort(function(a, b){
return sortingArr.indexOf(a) - sortingArr.indexOf(b);
});
Or even shorter:
itemsArray.sort((a, b) => sortingArr.indexOf(a) - sortingArr.indexOf(b));
Something like:
items = [
['Anne', 'a'],
['Bob', 'b'],
['Henry', 'b'],
['Andrew', 'd'],
['Jason', 'c'],
['Thomas', 'b']
]
sorting = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
result = []
sorting.forEach(function(key) {
var found = false;
items = items.filter(function(item) {
if(!found && item[1] == key) {
result.push(item);
found = true;
return false;
} else
return true;
})
})
result.forEach(function(item) {
document.writeln(item[0]) /// Bob Jason Henry Thomas Andrew
})
Here's a shorter code, but it destroys the sorting array:
result = items.map(function(item) {
var n = sorting.indexOf(item[1]);
sorting[n] = '';
return [n, item]
}).sort().map(function(j) { return j[1] })
If you use the native array sort function, you can pass in a custom comparator to be used when sorting the array. The comparator should return a negative number if the first value is less than the second, zero if they're equal, and a positive number if the first value is greater.
So if I understand the example you're giving correctly, you could do something like:
function sortFunc(a, b) {
var sortingArr = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
return sortingArr.indexOf(a[1]) - sortingArr.indexOf(b[1]);
}
itemsArray.sort(sortFunc);
Case 1: Original Question (No Libraries)
Plenty of other answers that work. :)
Case 2: Original Question (Lodash.js or Underscore.js)
var groups = _.groupBy(itemArray, 1);
var result = _.map(sortArray, function (i) { return groups[i].shift(); });
Case 3: Sort Array1 as if it were Array2
I'm guessing that most people came here looking for an equivalent to PHP's array_multisort (I did) so I thought I'd post that answer as well. There are a couple options:
1. There's an existing JS implementation of array_multisort(). Thanks to #Adnan for pointing it out in the comments. It is pretty large, though.
2. Write your own. (JSFiddle demo)
function refSort (targetData, refData) {
// Create an array of indices [0, 1, 2, ...N].
var indices = Object.keys(refData);
// Sort array of indices according to the reference data.
indices.sort(function(indexA, indexB) {
if (refData[indexA] < refData[indexB]) {
return -1;
} else if (refData[indexA] > refData[indexB]) {
return 1;
}
return 0;
});
// Map array of indices to corresponding values of the target array.
return indices.map(function(index) {
return targetData[index];
});
}
3. Lodash.js or Underscore.js (both popular, smaller libraries that focus on performance) offer helper functions that allow you to do this:
var result = _.chain(sortArray)
.pairs()
.sortBy(1)
.map(function (i) { return itemArray[i[0]]; })
.value();
...Which will (1) group the sortArray into [index, value] pairs, (2) sort them by the value (you can also provide a callback here), (3) replace each of the pairs with the item from the itemArray at the index the pair originated from.
this is probably too late but, you could also use some modified version of the code below in ES6 style. This code is for arrays like:
var arrayToBeSorted = [1,2,3,4,5];
var arrayWithReferenceOrder = [3,5,8,9];
The actual operation :
arrayToBeSorted = arrayWithReferenceOrder.filter(v => arrayToBeSorted.includes(v));
The actual operation in ES5 :
arrayToBeSorted = arrayWithReferenceOrder.filter(function(v) {
return arrayToBeSorted.includes(v);
});
Should result in arrayToBeSorted = [3,5]
Does not destroy the reference array.
function sortFunc(a, b) {
var sortingArr = ["A", "B", "C"];
return sortingArr.indexOf(a.type) - sortingArr.indexOf(b.type);
}
const itemsArray = [
{
type: "A",
},
{
type: "C",
},
{
type: "B",
},
];
console.log(itemsArray);
itemsArray.sort(sortFunc);
console.log(itemsArray);
Why not something like
//array1: array of elements to be sorted
//array2: array with the indexes
array1 = array2.map((object, i) => array1[object]);
The map function may not be available on all versions of Javascript
ES6
const arrayMap = itemsArray.reduce(
(accumulator, currentValue) => ({
...accumulator,
[currentValue[1]]: currentValue,
}),
{}
);
const result = sortingArr.map(key => arrayMap[key]);
More examples with different input arrays
I would use an intermediary object (itemsMap), thus avoiding quadratic complexity:
function createItemsMap(itemsArray) { // {"a": ["Anne"], "b": ["Bob", "Henry"], …}
var itemsMap = {};
for (var i = 0, item; (item = itemsArray[i]); ++i) {
(itemsMap[item[1]] || (itemsMap[item[1]] = [])).push(item[0]);
}
return itemsMap;
}
function sortByKeys(itemsArray, sortingArr) {
var itemsMap = createItemsMap(itemsArray), result = [];
for (var i = 0; i < sortingArr.length; ++i) {
var key = sortingArr[i];
result.push([itemsMap[key].shift(), key]);
}
return result;
}
See http://jsfiddle.net/eUskE/
var sortedArray = [];
for(var i=0; i < sortingArr.length; i++) {
var found = false;
for(var j=0; j < itemsArray.length && !found; j++) {
if(itemsArray[j][1] == sortingArr[i]) {
sortedArray.push(itemsArray[j]);
itemsArray.splice(j,1);
found = true;
}
}
}
http://jsfiddle.net/s7b2P/
Resulting order: Bob,Jason,Henry,Thomas,Anne,Andrew
In case you get here needing to do this with an array of objects, here is an adaptation of #Durgpal Singh's awesome answer:
const itemsArray = [
{ name: 'Anne', id: 'a' },
{ name: 'Bob', id: 'b' },
{ name: 'Henry', id: 'b' },
{ name: 'Andrew', id: 'd' },
{ name: 'Jason', id: 'c' },
{ name: 'Thomas', id: 'b' }
]
const sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ]
Object.keys(itemsArray).sort((a, b) => {
return sortingArr.indexOf(itemsArray[a].id) - sortingArr.indexOf(itemsArray[b].id);
})
let a = ['A', 'B', 'C' ]
let b = [3, 2, 1]
let c = [1.0, 5.0, 2.0]
// these array can be sorted by sorting order of b
const zip = rows => rows[0].map((_, c) => rows.map(row => row[c]))
const sortBy = (a, b, c) => {
const zippedArray = zip([a, b, c])
const sortedZipped = zippedArray.sort((x, y) => x[1] - y[1])
return zip(sortedZipped)
}
sortBy(a, b, c)
For getting a new ordered array, you could take a Map and collect all items with the wanted key in an array and map the wanted ordered keys by taking sifted element of the wanted group.
var itemsArray = [['Anne', 'a'], ['Bob', 'b'], ['Henry', 'b'], ['Andrew', 'd'], ['Jason', 'c'], ['Thomas', 'b']],
sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ],
map = itemsArray.reduce((m, a) => m.set(a[1], (m.get(a[1]) || []).concat([a])), new Map),
result = sortingArr.map(k => (map.get(k) || []).shift());
console.log(result);
I hope that I am helping someone, but if you are trying to sort an array of objects by another array on the first array's key, for example, you want to sort this array of objects:
const foo = [
{name: 'currency-question', key: 'value'},
{name: 'phone-question', key: 'value'},
{name: 'date-question', key: 'value'},
{name: 'text-question', key: 'value'}
];
by this array:
const bar = ['text-question', 'phone-question', 'currency-question', 'date-question'];
you can do so by:
foo.sort((a, b) => bar.indexOf(a.name) - bar.indexOf(b.name));
This is what I was looking for and I did for sorting an Array of Arrays based on another Array:
It's On^3 and might not be the best practice(ES6)
function sortArray(arr, arr1){
return arr.map(item => {
let a = [];
for(let i=0; i< arr1.length; i++){
for (const el of item) {
if(el == arr1[i]){
a.push(el);
}
}
}
return a;
});
}
const arr1 = ['fname', 'city', 'name'];
const arr = [['fname', 'city', 'name'],
['fname', 'city', 'name', 'name', 'city','fname']];
console.log(sortArray(arr,arr1));
It might help someone
I had to do this for a JSON payload I receive from an API, but it wasn't in the order I wanted it.
Array to be the reference array, the one you want the second array sorted by:
var columns = [
{last_name: "last_name"},
{first_name: "first_name"},
{book_description: "book_description"},
{book_id: "book_id"},
{book_number: "book_number"},
{due_date: "due_date"},
{loaned_out: "loaned_out"}
];
I did these as objects because these will have other properties eventually.
Created array:
var referenceArray= [];
for (var key in columns) {
for (var j in columns[key]){
referenceArray.push(j);
}
}
Used this with result set from database. I don't know how efficient it is but with the few number of columns I used, it worked fine.
result.forEach((element, index, array) => {
var tr = document.createElement('tr');
for (var i = 0; i < referenceArray.length - 1; i++) {
var td = document.createElement('td');
td.innerHTML = element[referenceArray[i]];
tr.appendChild(td);
}
tableBody.appendChild(tr);
});
let sortedOrder = [ 'b', 'c', 'b', 'b' ]
let itemsArray = [
['Anne', 'a'],
['Bob', 'b'],
['Henry', 'b'],
['Andrew', 'd'],
['Jason', 'c'],
['Thomas', 'b']
]
a.itemsArray(function (a, b) {
let A = a[1]
let B = b[1]
if(A != undefined)
A = A.toLowerCase()
if(B != undefined)
B = B.toLowerCase()
let indA = sortedOrder.indexOf(A)
let indB = sortedOrder.indexOf(B)
if (indA == -1 )
indA = sortedOrder.length-1
if( indB == -1)
indB = sortedOrder.length-1
if (indA < indB ) {
return -1;
} else if (indA > indB) {
return 1;
}
return 0;
})
This solution will append the objects at the end if the sorting key is not present in reference array
const result = sortingArr.map((i) => {
const pos = itemsArray.findIndex(j => j[1] === i);
const item = itemsArray[pos];
itemsArray.splice(pos, 1);
return item;
});
this should works:
var i,search, itemsArraySorted = [];
while(sortingArr.length) {
search = sortingArr.shift();
for(i = 0; i<itemsArray.length; i++) {
if(itemsArray[i][1] == search) {
itemsArraySorted.push(itemsArray[i]);
break;
}
}
}
itemsArray = itemsArraySorted;
You could try this method.
const sortListByRanking = (rankingList, listToSort) => {
let result = []
for (let id of rankingList) {
for (let item of listToSort) {
if (item && item[1] === id) {
result.push(item)
}
}
}
return result
}
with numerical sortingArr:
itemsArray.sort(function(a, b){
return sortingArr[itemsArray.indexOf(a)] - sortingArr[itemsArray.indexOf(b)];
});
This seems to work for me:
var outputArray=['10','6','8','10','4','6','2','10','4','0','2','10','0'];
var template=['0','2','4','6','8','10'];
var temp=[];
for(i=0;i<template.length;i++) {
for(x=0;x<outputArray.length;x++){
if(template[i] == outputArray[x]) temp.push(outputArray[x])
};
}
outputArray = temp;
alert(outputArray)
Use the $.inArray() method from jQuery. You then could do something like this
var sortingArr = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
var newSortedArray = new Array();
for(var i=sortingArr.length; i--;) {
var foundIn = $.inArray(sortingArr[i], itemsArray);
newSortedArray.push(itemsArray[foundIn]);
}
Use intersection of two arrays.
Ex:
var sortArray = ['a', 'b', 'c', 'd', 'e'];
var arrayToBeSort = ['z', 's', 'b', 'e', 'a'];
_.intersection(sortArray, arrayToBeSort)
=> ['a', 'b', 'e']
if 'z and 's' are out of range of first array, append it at the end of result
this.arrToBeSorted = this.arrToBeSorted.sort(function(a, b){
return uppthis.sorrtingByArray.findIndex(x => x.Id == a.ByPramaeterSorted) - uppthis.sorrtingByArray.findIndex(x => x.Id == b.ByPramaeterSorted);
});
You can do something like this:
function getSorted(itemsArray , sortingArr ) {
var result = [];
for(var i=0; i<arr.length; i++) {
result[i] = arr[sortArr[i]];
}
return result;
}
You can test it out here.
Note: this assumes the arrays you pass in are equivalent in size, you'd need to add some additional checks if this may not be the case.
refer link
refer