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splice twice VS shift + pop 5k
(version: 0)
5k list
Comparing performance of:
splice twice vs shift + pop
Created:
6 years ago
by:
Guest
Jump to the latest result
Script Preparation code:
var list = []; for (var i = 0; i < 500; i++) { list.push(i); }
Tests:
splice twice
list = list.splice(1, 1); list = list.splice(list.length - 1, 1)
shift + pop
list.shift(); list.pop();
Rendered benchmark preparation results:
Suite status:
<idle, ready to run>
Run tests (2)
Previous results
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Test case name
Result
splice twice
shift + pop
Fastest:
N/A
Slowest:
N/A
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Autogenerated LLM Summary
(model
llama3.2:3b
, generated one year ago):
Let's dive into the world of JavaScript microbenchmarks and explore what's being tested in this benchmark. **Benchmark Definition** The benchmark definition json represents a single test case. It has several important fields: * **Name**: A unique name for the benchmark, which can be used to identify it later. * **Description**: A brief description of the benchmark, providing context about what is being measured. * **Script Preparation Code**: A JavaScript code snippet that initializes a list and populates it with numbers from 0 to 499. This is done before each test case execution. * **Html Preparation Code**: An empty string, indicating that no additional HTML code needs to be generated for this benchmark. **Individual Test Cases** The benchmark definition also includes an array of individual test cases: 1. **splice twice**: Measures the performance of removing two elements from a list using `list.splice(1, 1)` and then removing another element at the end of the list (`list.splice(list.length - 1, 1)`). 2. **shift + pop**: Compares the performance of removing an element from both ends of a list simultaneously using `list.shift()` followed by `list.pop()`. **Approaches Compared** The benchmark is testing two different approaches for removing elements from a list: * **splice twice**: Uses `list.splice(1, 1)` to remove one element and then `list.splice(list.length - 1, 1)` to remove another element. This approach can lead to some overhead due to the repeated calls to `splice`. * **shift + pop**: Removes an element from both ends of the list simultaneously using `list.shift()` followed by `list.pop()`. This approach is generally more efficient because it only requires a single call to `length` and two calls to `pop`. **Pros and Cons** Here's a brief overview of each approach: * **splice twice**: Pros: Simple and easy to understand. Cons: More overhead due to repeated calls to `splice`. * **shift + pop**: Pros: More efficient, as it only requires a single call to `length` and two calls to `pop`. Cons: May be less intuitive for developers who are not familiar with this approach. **Library Usage** In this benchmark, no JavaScript libraries are explicitly mentioned. However, the test cases assume that the `list` variable is already initialized and populated with numbers from 0 to 499. **Special JS Features or Syntax** The benchmark does not use any special JavaScript features or syntax beyond what's typically used in everyday coding. It simply focuses on measuring the performance of list manipulation operations. **Alternative Approaches** Other alternatives for removing elements from a list include: * **forEach() + unshift()**: Removes an element and then adds it to the beginning of the list using `list.forEach((element) => { ... })` followed by `list.unshift(element)`. * **map() + push()**: Removes an element and then adds it to the end of the list using `list.map((_, index) => { if (index === 0) return undefined; return element; })` followed by `list.push(element)`. Keep in mind that these alternative approaches might have different performance characteristics compared to `splice twice` or `shift + pop`.
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