18.md

Results of Code LLama 34B on HumanEval/18

Date: 01 September, 2023
Model: codellama-34b-instruct
Temperature: 0.2
Passing runs: 10
Total runs: 10
Success rate: 100%

Prompt

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """

The above is an incomplete Python code fragment. Return the complete and correct code with no additional text.

Correct solution

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """times = 0

    for i in range(len(string) - len(substring) + 1):
        if string[i:i+len(substring)] == substring:
            times += 1

    return times

Unit tests

METADATA = {
    'author': 'jt',
    'dataset': 'test'
}


def check(candidate):
    assert candidate('', 'x') == 0
    assert candidate('xyxyxyx', 'x') == 4
    assert candidate('cacacacac', 'cac') == 4
    assert candidate('john doe', 'john') == 1

Runs

✅ Run 1

Start time: 2023-09-01 02:48:20
Duration: 7.35s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 2

Start time: 2023-09-01 02:51:22
Duration: 7.65s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 3

Start time: 2023-09-01 02:54:17
Duration: 7.2s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 4

Start time: 2023-09-01 02:57:14
Duration: 6.38s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 5

Start time: 2023-09-01 03:00:14
Duration: 6.38s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 6

Start time: 2023-09-01 03:03:14
Duration: 7.83s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 7

Start time: 2023-09-01 03:06:18
Duration: 7.64s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 8

Start time: 2023-09-01 03:09:21
Duration: 6.67s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 9

Start time: 2023-09-01 03:12:15
Duration: 7.5s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count

✅ Run 10

Start time: 2023-09-01 03:15:22
Duration: 6.78s
Result: Pass

def how_many_times(string: str, substring: str) -> int:
    """ Find how many times a given substring can be found in the original string. Count overlaping cases.
    >>> how_many_times('', 'a')
    0
    >>> how_many_times('aaa', 'a')
    3
    >>> how_many_times('aaaa', 'aa')
    3
    """
    count = 0
    for i in range(len(string)):
        if string[i:].startswith(substring):
            count += 1
    return count