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filter+pop VS reverse+find VS filter+at
(version: 0)
Comparing performance of:
revere and find vs filter and pop vs filter and at
Created:
2 years ago
by:
Guest
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Tests:
revere and find
const arr = [{name: 'jean',index: 1},{name: 'mark',index: 2},{name: 'jean',index: 3}, {name: 'mark',index: 4}] for(let i = 0; i < 10000; i++){ const item = arr.reverse().find(item => item.name === 'jean') }
filter and pop
const arr = [{name: 'jean',index: 1},{name: 'mark',index: 2},{name: 'jean',index: 3}, {name: 'mark',index: 4}] for(let i = 0; i < 10000; i++){ const item = arr.filter(item => item.name === 'jean').pop() }
filter and at
const arr = [{name: 'jean',index: 1},{name: 'mark',index: 2},{name: 'jean',index: 3}, {name: 'mark',index: 4}] for(let i = 0; i < 10000; i++){ const item = arr.filter(item => item.name === 'jean').at(-1) }
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Suite status:
<idle, ready to run>
Run tests (3)
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Test case name
Result
revere and find
filter and pop
filter and at
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Slowest:
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Autogenerated LLM Summary
(model
gemma2:9b
, generated one year ago):
This benchmark compares three different methods for finding the last object with the name 'jean' in an array of objects: **1. `reverse() + find()`:** - **Description:** This method first reverses the entire array using `reverse()`, then uses `find()` to iterate through the reversed array until it finds the first object with the name 'jean'. - **Pros:** `find()` is efficient for finding a single element. Reversing the array can potentially bring the desired element to the front, making the search faster. - **Cons:** * Reversing the entire array has a time complexity of O(n), where n is the length of the array. This can be inefficient if the array is large. * It's generally not ideal to modify the original array unless that's specifically needed. **2. `filter() + pop()`:** - **Description:** This method uses `filter()` to create a new array containing only objects with the name 'jean'. Then, it uses `pop()` to retrieve and return the last element from this filtered array. - **Pros:** * `filter()` creates a new array, leaving the original array untouched. - **Cons:** * `filter()` has a time complexity of O(n). * `pop()` removes the last element from an array. If you need to find and reuse the object later, this could be problematic. **3. `filter() + at(-1)`:** - **Description:** This method uses `filter()` to create a new array containing only objects with the name 'jean'. Then, it uses `at(-1)` to retrieve the last element of the filtered array. - **Pros:** * Similar to `filter() + pop()`, this approach avoids modifying the original array. * `at(-1)` efficiently retrieves the last element without removing it. - **Cons:** * `filter()` has a time complexity of O(n). **Other Alternatives:** - **`findIndex()`:** This method directly finds the index of the first object with the name 'jean' in the array. Once you have the index, you can access the object using array indexing (e.g., `arr[index]`). - **`forEach` Loop + Early Exit:** You could iterate through the array using a `for...of` loop and check each object's name. If you find 'jean', you can immediately exit the loop and return the object. **Choosing the Best Approach:** The best approach depends on your specific needs: * **Speed:** If speed is paramount, and the array isn't very large, `reverse() + find()` might be slightly faster due to the potential for bringing the desired element to the front. * **Preserving Original Data:** If you need to keep the original array unmodified, `filter() + at(-1)` or `filter() + pop()` are better choices. Keep in mind that small differences in performance between these methods might be negligible unless you're dealing with extremely large arrays and performance is critical.
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