Asked  7 Months ago    Answers:  5   Viewed   61 times

Assuming I have the following:

var array = 
    [
        {"name":"Joe", "age":17}, 
        {"name":"Bob", "age":17}, 
        {"name":"Carl", "age": 35}
    ]

What is the best way to be able to get an array of all of the distinct ages such that I get an result array of:

[17, 35]

Is there some way I could alternatively structure the data or better method such that I would not have to iterate through each array checking the value of "age" and check against another array for its existence, and add it if not?

If there was some way I could just pull out the distinct ages without iterating...

Current inefficent way I would like to improve... If it means that instead of "array" being an array of objects, but a "map" of objects with some unique key (i.e. "1,2,3") that would be okay too. Im just looking for the most performance efficient way.

The following is how I currently do it, but for me, iteration appears to just be crummy for efficiency even though it does work...

var distinct = []
for (var i = 0; i < array.length; i++)
   if (array[i].age not in distinct)
      distinct.push(array[i].age)

 Answers

12

If this were PHP I'd build an array with the keys and take array_keys at the end, but JS has no such luxury. Instead, try this:

var flags = [], output = [], l = array.length, i;
for( i=0; i<l; i++) {
    if( flags[array[i].age]) continue;
    flags[array[i].age] = true;
    output.push(array[i].age);
}
Tuesday, June 1, 2021
 
eek
answered 7 Months ago
eek
10

If you have an array such as

var people = [
  { "name": "bob", "dinner": "pizza" },
  { "name": "john", "dinner": "sushi" },
  { "name": "larry", "dinner": "hummus" }
];

You can use the filter method of an Array object:

people.filter(function (person) { return person.dinner == "sushi" });
  // => [{ "name": "john", "dinner": "sushi" }]

In newer JavaScript implementations you can use a function expression:

people.filter(p => p.dinner == "sushi")
  // => [{ "name": "john", "dinner": "sushi" }]

You can search for people who have "dinner": "sushi" using a map

people.map(function (person) {
  if (person.dinner == "sushi") {
    return person
  } else {
    return null
  }
}); // => [null, { "name": "john", "dinner": "sushi" }, null]

or a reduce

people.reduce(function (sushiPeople, person) {
  if (person.dinner == "sushi") {
    return sushiPeople.concat(person);
  } else {
    return sushiPeople
  }
}, []); // => [{ "name": "john", "dinner": "sushi" }]

I'm sure you are able to generalize this to arbitrary keys and values!

Friday, June 11, 2021
 
o_flyer
answered 6 Months ago
79

Using the object literal is exactly what I would do. A lot of people miss this technique a lot of the time, opting instead for typical array walks as the original code that you showed. The only optimization would be to avoid the arr.length lookup each time. Other than that, O(n) is about as good as you get for uniqueness and is much better than the original O(n^2) example.

function unique(arr) {
    var hash = {}, result = [];
    for ( var i = 0, l = arr.length; i < l; ++i ) {
        if ( !hash.hasOwnProperty(arr[i]) ) { //it works with objects! in FF, at least
            hash[ arr[i] ] = true;
            result.push(arr[i]);
        }
    }
    return result;
}

// * Edited to use hasOwnProperty per comments

Time complexities to summarize

  f()    | unsorted | sorted | objects | scalar | library
____________________________________________________________
unique   |   O(n)   |  O(n)  |   no    |  yes   |    n/a
original |  O(n^2)  | O(n^2) |   yes   |  yes   |    n/a
uniq     |  O(n^2)  |  O(n)  |   yes   |  yes   | Prototype
_.uniq   |  O(n^2)  |  O(n)  |   yes   |  yes   | Underscore

As with most algorithms, there are trade offs. If you are only sorting scalar values, you're modifications to the original algorithm give the most optimal solution. However, if you need to sort non-scalar values, then using or mimicking the uniq method of either of the libraries discussed would be your best choice.

Wednesday, July 28, 2021
 
Raef
answered 5 Months ago
62

Just for the sake of simplicity I will assume your elements have a class.

Example on jsFiddle

So, I would grab all elements:

var elements = document.getElementsByClassName("square");

Then I would create an array with the IDs

var ids = [];

for (var i = 0; i < elements.length; ++i)
{
    ids.push(elements[i].getAttribute("id"));
}

And then generate random numbers on the length of the array

var random = roundingFunction(Math.random() * ids.length);

Then retrieve the element at the index and remove from the array

var elementID = ids.splice(random, 1);

And repeat.

Saturday, August 7, 2021
 
Pupil
answered 4 Months ago
28

Try converting the inner arrays to a string, then filter the dupes and parse the string again.

let x = [[1, 2], [3, 4], [1, 2]];

var unique = x.map(ar=>JSON.stringify(ar))
  .filter((itm, idx, arr) => arr.indexOf(itm) === idx)
  .map(str=>JSON.parse(str));

console.log(unique);
Tuesday, August 17, 2021
 
Nate
answered 4 Months ago
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