18.md

Results of GPT-4 on HumanEval/18

Date: 18 July, 2023
Model: gpt-4
Temperature: 0.2
Passing runs: 10
Total runs: 10
Success rate: 100%

Prompt

Tokens: 114

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: 2023-07-18 18:50:51
Duration: 15.0s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 2

Start time: 2023-07-18 18:50:40
Duration: 10.97s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 3

Start time: 2023-07-18 18:50:27
Duration: 13.09s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 4

Start time: 2023-07-18 18:50:17
Duration: 10.18s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 5

Start time: 2023-07-18 18:50:03
Duration: 13.75s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 6

Start time: 2023-07-18 18:49:48
Duration: 14.45s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 7

Start time: 2023-07-18 18:49:34
Duration: 14.47s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 8

Start time: 2023-07-18 18:49:22
Duration: 11.52s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 9

Start time: 2023-07-18 18:49:11
Duration: 10.93s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count

✅ Run 10

Start time: 2023-07-18 18:49:00
Duration: 11.08s
Completion tokens: 142
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 = start = 0
    while start < len(string):
        pos = string.find(substring, start)
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count