Question: I have this array that have a nested array that I want to loop to find a specific value.
arr = [['123456','234567'], ['345678']];
specificValue = '123456';
The output that I want is to find out if there's a value same with specificvalue and return this value?
I tried
arr.filter(id => id === specificValue);
Thanks for the help
Let's keep going with your attempt.
The problem is your array is nested so array.filter will not work
Use array.flat with array.filter:
let arr = [['123456','234567'], ['345678']];
let specificValue = '123456';
console.log(arr.flat().filter(i=>i==specificValue))
Try this code
const arr = [['123456','234567'], ['345678']];
const result = arr.flat(Infinity).find(val => val === "123456");
console.log(result);
You can learn more about array.flat() method here
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/flat
Since flat() takes a depth as an argument, to be able to search in an indefinite number of nested arrays you could try good old recursion:
const findRecursive = (array, target) => {
for (let i = 0; i < array.length; i++) {
if (array[i] === target) {
return array[i];
} else if (Array.isArray(array[i])) {
return findRecursive(array[i], target);
}
}
}
console.log(findRecursive(arr, specificValue));
EDIT: Abdelrhman Mohamed has corrected me, you can specify an indefinite amount of nested arrays using array.flat(Infinity)
I'm building my own map method to be as close as the native map method.
Since the native map pushes(i think) the changed values into a new array, it still keeps the empty slots. I wasn't able to find a solution to push an empty slot into an array, like this example below.
[1, 2, 3].push(some code) // [1, 2, 3, empty]
I tried pushing an array with one empty item prefixed with a spread operator arr.push(...(new Array(1))) or arr.push(...[,]) but that just pushes undefined.
I solved my problem by not using push and instead assigning values to the array index that way skipped indices will be set to empty.
But I'm writing this post to see if anyone knows that if it's possible to use the push method to push an empty slot to an array.
No, it's not possible, not with the push method. empty can only exist if the array has a certain length, but a whole number property of the array does not exist at some index. This is called a sparse array, and cannot be created with push (or other array methods, if they're called on and with non-sparse arrays).
The only way to do so would be to assign to an index for which a lower index doesn't exist yet.
Look at the results for the below two snippets in your browser console, not the snippet console:
const arr = [];
arr[1] = 'a';
console.log(arr);
Or to set the .length of the array above the last index that the array has:
const arr = [];
arr.length = 1;
console.log(arr);
But the two approaches above are very weird to do and probably have no good reason to be used. Better to avoid sparse arrays entirely.
Keep in mind that an empty slot is different from undefined, which is perfectly possible to have as an array value:
const arr = [];
arr.push(undefined);
console.log(arr);
You can create an empty slot in an array by incrementing the array length:
var a = []
a.push(1)
a.length++
a.push(3)
console.log(a)
console.log(1 in a) // anything at index 1?
Alternatively, you can push something and then delete it:
var a = []
a.push(1)
a.push(2)
a.push(3)
delete a[1]
console.log(a)
console.log(1 in a) // anything at index 1?
There is no need to actually push to a new array in your implementation. You can simply do new Array(this.length) where this.length is the array you are mapping through length.
For example consider this map implementation:
if (!Array.prototype.mapIt) {
Object.defineProperty(Array.prototype, "mapIt", {
value: function(fn) {
if (this === null) {
throw new TypeError('Array.prototype.mapIt called on null or undefined');
}
if (typeof fn !== 'function') {
throw new TypeError('predicate must be a function');
}
let _array = this.filter(x => x != null) // remove empty values
let result = new Array(_array.length) // the new array we will return
for (var i = 0; i < _array.length; i++) {
result[i] = fn.call(arguments[1], _array[i], i, _array) // call the predicate
}
return result;
}
});
}
let arr = [1, 2, , , 3] // the test array
let result = arr.mapIt((c, i, a) =>
console.log(`current: ${c}`, `index: ${i}`, `array: ${a}`) || c + 2)
console.log('result: ', result)
console.log('original array: ', arr)
Hope this helps you with an gives you an idea about a possible map implementation.
Title is pretty much self explanatory...
I want to be able to find duplicated values from JavaScript array.
The array keys can be duplicated so I need to validate only the array values.
Here is an example :
var arr=[
Ibanez: 'JoeSatriani',
Ibanez: 'SteveVai',
Fender: 'YngwieMalmsteen',
Fender: 'EricJohnson',
Gibson: 'EricJohnson',
Takamine: 'SteveVai'
];
In that example:
the key is the guitar brand
the value is the guitar player name.
So:
If there is duplicated keys (like: Ibanez or Fender) as on that current example that is OK :-)
But
If there is duplicated values (like: EricJohnson or SteveVai) I'm expecting to get (return) that error:
EricJohnson,SteveVai
You can't have associative arrays in Javascript. You can create an array of objects, like:
var arr=[
{Ibanez: 'JoeSatriani'},
{Ibanez: 'SteveVai'},
{Fender: 'YngwieMalmsteen'},
{Fender: 'EricJohnson'},
{Gibson: 'EricJohnson'},
{Takamine: 'SteveVai'}
];
Then you'll need a for...in loop to go over every object in the array, create a new array of values and check that for duplicates, which is also not very straightforward - basically you'll want to sort the array and make sure no value is the same as the one after it.
var arrayOfValues = [];
arr.forEach(function(obj){
for(var prop in obj)
arrayOfValues.push(obj[prop]);
});
arrayOfValues.sort(); // by default it will sort them alphabetically
arrayOfValues.forEach(function(element,index,array){
if(array[index+1] && element==array[index+1])
alert("Duplicate value found!");
});
First of all, object keys can not be repeated.
This means that:
({
"Fender": "Jim",
"Fender": "Bob"
})["Fender"]
Would simply return: "Bob".
However, I did make a code that could allow you to find duplicates in values, but as I said, the key will have to be unique:
var arr = {
Ibanez: 'EricJohnson',
Fender: 'YngwieMalmsteen',
Gibson: 'EricJohnson',
Takamine: 'SteveVai',
"Takamine2": 'SteveVai'
};
function contains(a, obj) {
for (var i = 0; i < a.length; i++) {
if (a[i] === obj) {
return true;
}
}
return false;
}
var track = [];
var exists = [];
for (var val in arr) {
if (contains(track, arr[val])) {
exists.push(arr[val]);
} else {
track.push(arr[val])
}
}
alert(exists)
You can see it working here: http://jsfiddle.net/dr09sga6/2/
As others have commented, the example array you provided isn't a valid JavaScript array. You could, however, keep a list for each guitar type:
var mapping = {
Ibanez: ['JoeSatriani','SteveVai'],
Fender: ['YngwieMalmsteen','EricJohnson']
Gibson: ['EricJohnson'],
Takamine: ['SteveVai']
];
Or a list of each guitar/musician pair:
var pairs = [
['Ibanez','JoeSatriani'],
['Ibanez','SteveVai'],
['Fender','YngwieMalmsteen'],
['Fender','EricJohnson'],
['Gibson','EricJohnson'],
['Takamine','SteveVai']
];
Your solution is going to depend on which pattern you go with. However, in the second case it can be done in one chained functional call:
pairs.map(function(e) {return e[1]}) // Discard the brand names
.sort() // Sort by artist
.reduce(function(p,c,i,a){
if (i>0 && a[i]==a[i-1] && !p.some(function(v) {return v == c;})) p.push(c);
return p;
},[]); //Return the artist names that are duplicated
http://jsfiddle.net/mkurqmqd/1/
To break that reduce call down a bit, here's the callback again:
function(p,c,i,a){
if (i>0
&& a[i]==a[i-1]
&& !p.some(function(v) {
return v == c;
}))
p.push(c);
return p;
}
reduce is going to call our callback for each element in the array, and it's going to pass the returned value for each call into the next call as the first parameter (p). It's useful for accumulating a list as you move across an array.
Because we're looking back at the previous item, we need to make sure we don't go out of bounds on item 0.
Then we're checking to see if this item matches the previous one in the (sorted) list.
Then we're checking (with Array.prototype.some()) whether the value we've found is ALREADY in our list of duplicates...to avoid having duplicate duplicates!
If all of those checks pass, we add the name to our list of duplicate values.
I have a JavaScript array like this:
var myData=['237','124','255','124','366','255'];
I need the array elements to be unique and sorted:
myData[0]='124';
myData[1]='237';
myData[2]='255';
myData[3]='366';
Even though the members of array look like integers, they're not integers, since I have already converted each to be string:
var myData[0]=num.toString();
//...and so on.
Is there any way to do all of these tasks in JavaScript?
This is actually very simple. It is much easier to find unique values, if the values are sorted first:
function sort_unique(arr) {
if (arr.length === 0) return arr;
arr = arr.sort(function (a, b) { return a*1 - b*1; });
var ret = [arr[0]];
for (var i = 1; i < arr.length; i++) { //Start loop at 1: arr[0] can never be a duplicate
if (arr[i-1] !== arr[i]) {
ret.push(arr[i]);
}
}
return ret;
}
console.log(sort_unique(['237','124','255','124','366','255']));
//["124", "237", "255", "366"]
You can now achieve the result in just one line of code.
Using new Set to reduce the array to unique set of values.
Apply the sort method after to order the string values.
var myData=['237','124','255','124','366','255']
var uniqueAndSorted = [...new Set(myData)].sort()
UPDATED for newer methods introduced in JavaScript since time of question.
This might be adequate in circumstances where you can't define the function in advance (like in a bookmarklet):
myData.sort().filter(function(el,i,a){return i===a.indexOf(el)})
Here's my (more modern) approach using Array.protoype.reduce():
[2, 1, 2, 3].reduce((a, x) => a.includes(x) ? a : [...a, x], []).sort()
// returns [1, 2, 3]
Edit: More performant version as pointed out in the comments:
arr.sort().filter((x, i, a) => !i || x != a[i-1])
function sort_unique(arr) {
return arr.sort().filter(function(el,i,a) {
return (i==a.indexOf(el));
});
}
How about:
array.sort().filter(function(elem, index, arr) {
return index == arr.length - 1 || arr[index + 1] != elem
})
This is similar to #loostro answer but instead of using indexOf which will reiterate the array for each element to verify that is the first found, it just checks that the next element is different than the current.
Try using an external library like underscore
var f = _.compose(_.uniq, function(array) {
return _.sortBy(array, _.identity);
});
var sortedUnique = f(array);
This relies on _.compose, _.uniq, _.sortBy, _.identity
See live example
What is it doing?
We want a function that takes an array and then returns a sorted array with the non-unique entries removed. This function needs to do two things, sorting and making the array unique.
This is a good job for composition, so we compose the unique & sort function together. _.uniq can just be applied on the array with one argument so it's just passed to _.compose
the _.sortBy function needs a sorting conditional functional. it expects a function that returns a value and the array will be sorted on that value. Since the value that we are ordering it by is the value in the array we can just pass the _.identity function.
We now have a composition of a function that (takes an array and returns a unique array) and a function that (takes an array and returns a sorted array, sorted by their values).
We simply apply the composition on the array and we have our uniquely sorted array.
This function doesn't fail for more than two duplicates values:
function unique(arr) {
var a = [];
var l = arr.length;
for(var i=0; i<l; i++) {
for(var j=i+1; j<l; j++) {
// If a[i] is found later in the array
if (arr[i] === arr[j])
j = ++i;
}
a.push(arr[i]);
}
return a;
};
Here is a simple one liner with O(N), no complicated loops necessary.
> Object.keys(['a', 'b', 'a'].reduce((l, r) => l[r] = l, {})).sort()
[ 'a', 'b' ]
Explanation
Original data set, assume its coming in from an external function
const data = ['a', 'b', 'a']
We want to group all the values onto an object as keys as the method of deduplication. So we use reduce with an object as the default value:
[].reduce(fn, {})
The next step is to create a reduce function which will put the values in the array onto the object. The end result is an object with a unique set of keys.
const reduced = data.reduce((l, r) => l[r] = l, {})
We set l[r] = l because in javascript the value of the assignment expression is returned when an assignment statement is used as an expression. l is the accumulator object and r is the key value. You can also use Object.assign(l, { [r]: (l[r] || 0) + 1 }) or something similar to get the count of each value if that was important to you.
Next we want to get the keys of that object
const keys = Object.keys(reduced)
Then simply use the built-in sort
console.log(keys.sort())
Which is the set of unique values of the original array, sorted
['a', 'b']
The solution in a more elegant way.
var myData=['237','124','255','124','366','255'];
console.log(Array.from(new Set(myData)).sort((a,b) => a - b));
I know the question is very old, but maybe someone will come in handy
A way to use a custom sort function
//func has to return 0 in the case in which they are equal
sort_unique = function(arr,func) {
func = func || function (a, b) {
return a*1 - b*1;
};
arr = arr.sort(func);
var ret = [arr[0]];
for (var i = 1; i < arr.length; i++) {
if (func(arr[i-1],arr[i]) != 0)
ret.push(arr[i]);
}
}
return ret;
}
Example: desc order for an array of objects
MyArray = sort_unique(MyArray , function(a,b){
return b.iterator_internal*1 - a.iterator_internal*1;
});
No redundant "return" array, no ECMA5 built-ins (I'm pretty sure!) and simple to read.
function removeDuplicates(target_array) {
target_array.sort();
var i = 0;
while(i < target_array.length) {
if(target_array[i] === target_array[i+1]) {
target_array.splice(i+1,1);
}
else {
i += 1;
}
}
return target_array;
}
I guess I'll post this answer for some variety. This technique for purging duplicates is something I picked up on for a project in Flash I'm currently working on about a month or so ago.
What you do is make an object and fill it with both a key and a value utilizing each array item. Since duplicate keys are discarded, duplicates are removed.
var nums = [1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10];
var newNums = purgeArray(nums);
function purgeArray(ar)
{
var obj = {};
var temp = [];
for(var i=0;i<ar.length;i++)
{
obj[ar[i]] = ar[i];
}
for (var item in obj)
{
temp.push(obj[item]);
}
return temp;
}
There's already 5 other answers, so I don't see a need to post a sorting function.
// Another way, that does not rearrange the original Array
// and spends a little less time handling duplicates.
function uniqueSort(arr, sortby){
var A1= arr.slice();
A1= typeof sortby== 'function'? A1.sort(sortby): A1.sort();
var last= A1.shift(), next, A2= [last];
while(A1.length){
next= A1.shift();
while(next=== last) next= A1.shift();
if(next!=undefined){
A2[A2.length]= next;
last= next;
}
}
return A2;
}
var myData= ['237','124','255','124','366','255','100','1000'];
uniqueSort(myData,function(a,b){return a-b})
// the ordinary sort() returns the same array as the number sort here,
// but some strings of digits do not sort so nicely numerical.
function sort() only is only good if your number has same digit, example:
var myData = ["3","11","1","2"]
will return;
var myData = ["1","11","2","3"]
and here improvement for function from mrmonkington
myData.sort().sort(function(a,b){return a - b;}).filter(function(el,i,a){if(i==a.indexOf(el) & el.length>0)return 1;return 0;})
the above function will also delete empty array and you can checkout the demo below
http://jsbin.com/ahojip/2/edit
O[N^2] solutions are bad, especially when the data is already sorted, there is no need to do two nested loops for removing duplicates. One loop and comparing to the previous element will work great.
A simple solution with O[] of sort() would suffice. My solution is:
function sortUnique(arr, compareFunction) {
let sorted = arr.sort(compareFunction);
let result = sorted.filter(compareFunction
? function(val, i, a) { return (i == 0 || compareFunction(a[i-1], val) != 0); }
: function(val, i, a) { return (i == 0 || a[i-1] !== val); }
);
return result;
}
BTW, can do something like this to have Array.sortUnique() method:
Array.prototype.sortUnique = function(compareFunction) {return sortUnique(this, compareFunction); }
Furthermore, sort() could be modified to remove second element if compare() function returns 0 (equal elements), though that code can become messy (need to revise loop boundaries in the flight). Besides, I stay away from making my own sort() functions in interpreted languages, since it will most certainly degrade the performance. So this addition is for the ECMA 2019+ consideration.
The fastest and simpleness way to do this task.
const N = Math.pow(8, 8)
let data = Array.from({length: N}, () => Math.floor(Math.random() * N))
let newData = {}
let len = data.length
// the magic
while (len--) {
newData[data[len]] = true
}
var array = [2,5,4,2,5,9,4,2,6,9,0,5,4,7,8];
var unique_array = [...new Set(array)]; // [ 2, 5, 4, 9, 6, 0, 7, 8 ]
var uniqueWithSorted = unique_array.sort();
console.log(uniqueWithSorted);
output = [ 0, 2, 4, 5, 6, 7, 8, 9 ]
Here, we used only Set for removing duplicity from the array and then used sort for sorting array in ascending order.
I'm afraid you can't combine these functions, ie. you gotta do something like this:-
myData.unique().sort();
Alternatively you can implement a kind of sortedset (as available in other languages) - which carries both the notion of sorting and removing duplicates, as you require.
Hope this helps.
References:-
Array.sort
Array.unique