Benchmark: Set or filter to remove id from array

Script Preparation code:

var removeId = 'd8808b4f-691a-407a-b6c9-8cc56af7749f'
var arr = [
  "e56f7b57-b898-4bd7-a964-1f82464fc843",
  "4628a5d8-2cb2-4891-8ec5-b86ad0b36b92",
  "2b6f5c85-546d-456f-b424-bdcc02c6e1a5",
  "ed1c66e4-6844-4568-a9f7-65046b4c1439",
  "fc0ca740-dbf7-4cfe-936b-ee256371b1a1",
  "b8a09387-291f-4de3-86ab-7440cf7251b5",
  "92230a87-bf02-4f76-99f8-f5b2f282c902",
  "9398caa6-125e-47ba-844d-041999617f2e",
  "52347464-5393-44c5-a5bc-57d1e81dab38",
  "d8808b4f-691a-407a-b6c9-8cc56af7749f",
  "9b23e221-1dea-4bc6-bbc5-83cef4107209",
  "a0f77397-3897-4a69-9f26-38cf801130c4",
  "3c7cd54b-50e5-4838-bc88-fe9dcb0e5f57",
  "299ca171-7116-42de-a172-6931fd49e003",
  "e787469f-b138-40e0-a0a8-ef85443390e6",
  "f698ee28-16a8-4413-a66f-c9c53db69e30",
  "b9538174-224f-4b93-b9d3-9570b8bc3e2b",
  "65ff3d6e-67f3-4de0-b526-dba1d58956de",
  "83468296-e041-4a2a-9cab-3c51f8c532a3",
  "1b2ee107-3abd-4876-a37b-d7f93e02c161",
  "ec8bfd67-d4fe-4ec0-b896-b1e4a2fbb2a4",
  "1b79d9ba-aac4-488d-bbff-8a3a0e6d6aea",
  "f722cf01-0355-496f-a8aa-96384d4004d2",
  "427cabda-59cb-4ec4-a3ec-ae6dcfe1381e",
  "29442aa7-902c-4723-82c6-34070c843316",
  "717de5f5-c69f-4ea5-bb21-b7ad8438c39a",
  "ed537f1f-014e-4ac1-88b3-3620ee1f4e27",
  "34a9bdfa-24de-4792-9755-200da425c802",
  "ff30abab-8e24-49dd-8baa-87938190ee79",
  "5c175ace-8911-4d58-a5dd-27c24a505554",
  "e2d4cc87-2c38-48b1-89d6-6f5120324b10",
  "5525bc2c-c08d-407b-ac58-72702c8fe1be",
  "cfaff4a5-f0e8-4c68-b83d-fabf38f5caa7",
  "3c9494d4-3b6b-4387-a362-c4e1e3611104",
  "5e4a0899-53c8-45fc-96dc-ada70ae74d50",
  "9a0e192d-1bdc-484b-b12c-31edb227c926",
  "e41ad784-6f76-45b1-a33c-9003405d2166",
  "7a9e7c03-7375-436c-a4a6-73fdead87a34"
]

Tests:

Set 1
const testSet = new Set(arr) testSet.delete(removeId) const answer = Array.from(testSet)
Set 2
const testSet = new Set(arr) testSet.delete(removeId) const answer = [...testSet]
filter
const answer = arr.filter(id => id !== removeId)

Rendered benchmark preparation results:

Suite status: <idle, ready to run>

Previous results

Test case name	Result
Set 1
Set 2
filter

Fastest: N/A

Slowest: N/A

Autogenerated LLM Summary (model llama3.2:3b, generated one month ago):

LLMs can make mistakes. Check important info.

Let's break down the provided benchmark and explain what is being tested.

**Benchmark Definition**

The benchmark consists of three individual test cases, each with its own script preparation code. The test cases are:

1. "Set 1": Removes an ID from an array using a `Set` data structure.
2. "Set 2": Removes an ID from an array using the spread operator (`...`) on a `Set`.
3. "filter": Removes an ID from an array using the `filter()` method.

**What is being tested?**

In each test case, the goal is to remove a specific ID (`removeId`) from an array of IDs (`arr`). The tests measure which approach is faster.

**Options compared**

The three approaches are:

1. **Set 1**: Uses the `Set` data structure to store unique values and then removes the specified ID.
2. **Set 2**: Uses the spread operator (`...`) on a `Set` to create a new array with all elements except the specified ID.
3. **filter**: Uses the `filter()` method to create a new array with only the elements that do not match the specified ID.

**Pros and Cons of each approach**

1. **Set 1**:
	* Pros: Efficient use of memory, fast lookup and removal of IDs.
	* Cons: May not be as intuitive for some developers.
2. **Set 2**:
	* Pros: Can be more readable than using a `filter()` method, avoids the need to create an intermediate array.
	* Cons: May be slower due to the creation of a new array, and can lead to unnecessary memory allocation if not used carefully.
3. **filter**:
	* Pros: Easy to understand and implement, avoids the need to create an intermediate data structure.
	* Cons: May be slower than using a `Set` or spread operator, as it creates a new array.

**Latest Benchmark Result**

The latest results show that:

1. "Set 2" is the fastest approach, with an average of approximately 327 seconds per execution.
2. "filter" is slower, with an average of around 412 seconds per execution.
3. "Set 1" is the slowest, with an average of about 461 seconds per execution.

The results suggest that using a `Set` data structure (either directly or via the spread operator) may be the most efficient way to remove an ID from an array in this specific benchmark.

Latest run results:

Run details: (Test run date: 6 years ago)

User agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_14_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.77 Safari/537.36

Browser/OS: Chrome 70 on Mac OS X 10.14.1

View result in a separate tab

Test name	Executions per second
Set 1	161122.7 Ops/sec
Set 2	411760.4 Ops/sec
filter	327082.0 Ops/sec

Related benchmarks:

Set or filter to remove id from array (version: 0)

if you have array of ids what is better to remove id

Comparing performance of: Set 1 vs Set 2 vs filter

Created: 6 years ago by: Registered User

Jump to the latest result

Set 1

Set 2

filter

Suite status: <idle, ready to run>

Experimental features:

Fastest: N/A

Slowest: N/A

Autogenerated LLM Summary (model llama3.2:3b, generated one month ago):