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⚡️ Speed up function sorter by 4211531.56 in PR #1 (new-sorter) #35

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@codeflash-ai codeflash-ai bot commented Jul 13, 2024

⚡️ This pull request contains optimizations for PR #1

If you approve this dependent PR, these changes will be merged into the original PR branch new-sorter.

This PR will be automatically closed if the original PR is merged.


📄 sorter() in code_to_optimize/bubble_sort.py

📈 Performance improved by 4211531.56 (42115.32 faster)

⏱️ Runtime went down from 1070554.63 to 25.42

Explanation and details

The function you provided, sorter, is already using Python's built-in sort function which has a time complexity of O(n log n), where n is a number of elements in the array. This is the fastest achievable sorting complexity for comparison-based sorts.

However, if you want to achieve a marginal speed increase, writing this in-place might help.

Here's an alternative version using list comprehension. Although this does not improve the time complexity, it gives a Pythonic touch:

def sorter(arr):
    return sorted(arr)

Again, this command returns a new sorted list and does not modify the original list. If you want to sort the list in-place, you only have the original function:

Please note that sorting time complexity cannot be improved further than O(n log n) using comparison-based sorting algorithms. To really optimize this function, you would need a guarantee about the content of your data, for example, if your array only contained integers in a particular range, then you could use counting sort or radix sort, which can have a time complexity of O(n).

Correctness verification

The new optimized code was tested for correctness. The results are listed below.

🔘 (none found) − ⚙️ Existing Unit Tests

✅ 3 Passed − 🌀 Generated Regression Tests

(click to show generated tests)
# imports
import pytest  # used for our unit tests

# function to test

def sorter(arr):
    for i in range(len(arr)):
        for j in range(len(arr) - 1):
            if arr[j] > arr[j + 1]:
                temp = arr[j]
                arr[j] = arr[j + 1]
                arr[j + 1] = temp
    return arr

# unit tests

# Test with an empty list
def test_sorter_empty():
    assert sorter([]) == []

# Test with a single-element list
def test_sorter_single_element():
    assert sorter([42]) == [42]

# Test with a two-element list
def test_sorter_two_elements():
    assert sorter([2, 1]) == [1, 2]
    assert sorter([1, 2]) == [1, 2]

# Test with sorted lists
def test_sorter_sorted_list():
    assert sorter([1, 2, 3, 4, 5]) == [1, 2, 3, 4, 5]
    assert sorter([0, 2, 4, 6, 8, 10]) == [0, 2, 4, 6, 8, 10]

# Test with reverse-sorted lists
def test_sorter_reverse_sorted_list():
    assert sorter([5, 4, 3, 2, 1]) == [1, 2, 3, 4, 5]
    assert sorter([-1, -2, -3, -4, -5]) == [-5, -4, -3, -2, -1]

# Test with lists with duplicates
def test_sorter_with_duplicates():
    assert sorter([3, 1, 2, 1, 3]) == [1, 1, 2, 3, 3]
    assert sorter([5, 5, 5, 5]) == [5, 5, 5, 5]

# Test with lists with negative numbers
def test_sorter_with_negative_numbers():
    assert sorter([-1, -3, -2, 0, 2]) == [-3, -2, -1, 0, 2]
    assert sorter([-10, 100, -50, 0]) == [-50, -10, 0, 100]

# Test with lists with varying types of numbers
def test_sorter_with_various_number_types():
    assert sorter([1.5, 2.3, 1.1, 2.0, 1.9]) == [1.1, 1.5, 1.9, 2.0, 2.3]
    assert sorter([1, 2.0, 3, 4.5]) == [1, 2.0, 3, 4.5]

# Test with large lists
def test_sorter_large_list():
    large_list = list(range(1000, 0, -1))  # 1000 to 1 in reverse order
    assert sorter(large_list) == sorted(large_list)

# Test with lists with non-numeric values
def test_sorter_non_numeric():
    with pytest.raises(TypeError):  # We expect a TypeError because sorter is not designed for non-numeric values
        sorter(['apple', 'banana', 'cherry'])

    with pytest.raises(TypeError):  # We expect a TypeError because sorter is not designed for non-numeric values
        sorter(['a', 'aa', 'aaa'])

The function you provided, sorter, is already using Python's built-in sort function which has a time complexity of O(n log n), where n is a number of elements in the array. This is the fastest achievable sorting complexity for comparison-based sorts.

However, if you want to achieve a marginal speed increase, writing this in-place might help.

Here's an alternative version using list comprehension. Although this does not improve the time complexity, it gives a Pythonic touch:

```python
def sorter(arr):
    return sorted(arr)
```

Again, this command returns a new sorted list and does not modify the original list. If you want to sort the list in-place, you only have the original function:



Please note that sorting time complexity cannot be improved further than O(n log n) using comparison-based sorting algorithms. To really optimize this function, you would need a guarantee about the content of your data, for example, if your array only contained integers in a particular range, then you could use counting sort or radix sort, which can have a time complexity of O(n).
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