111.md

Results of GPT-3.5 on HumanEval/111

Date: 18 July, 2023
Model: gpt-3.5-turbo
Temperature: 0.2
Passing runs: 9
Total runs: 10
Success rate: 90%

Prompt

Tokens: 177

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """

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

Correct solution

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """dict1={}
    list1=test.split(" ")
    t=0

    for i in list1:
        if(list1.count(i)>t) and i!='':
            t=list1.count(i)
    if t>0:
        for i in list1:
            if(list1.count(i)==t):
                
                dict1[i]=t
    return dict1

Unit tests

def check(candidate):

    # Check some simple cases
    assert candidate('a b b a') == {'a':2,'b': 2}, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate('a b c a b') == {'a': 2, 'b': 2}, "This prints if this assert fails 2 (good for debugging!)"
    assert candidate('a b c d g') == {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'g': 1}, "This prints if this assert fails 3 (good for debugging!)"
    assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, "This prints if this assert fails 4 (good for debugging!)"
    assert candidate('b b b b a') == {'b': 4}, "This prints if this assert fails 5 (good for debugging!)"
    assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, "This prints if this assert fails 6 (good for debugging!)"
    
    
    # Check some edge cases that are easy to work out by hand.
    assert candidate('') == {}, "This prints if this assert fails 7 (also good for debugging!)"
    assert candidate('a') == {'a': 1}, "This prints if this assert fails 8 (also good for debugging!)"

Runs

✅ Run 1

Start time: 2023-07-18 17:10:55
Duration: 8.89s
Completion tokens: 255
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    count_dict = {}
    max_count = 0
    
    for letter in letters:
        if letter in count_dict:
            count_dict[letter] += 1
        else:
            count_dict[letter] = 1
        
        if count_dict[letter] > max_count:
            max_count = count_dict[letter]
    
    result_dict = {}
    for letter, count in count_dict.items():
        if count == max_count:
            result_dict[letter] = count
    
    return result_dict

✅ Run 2

Start time: 2023-07-18 17:10:46
Duration: 8.57s
Completion tokens: 244
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    counts = {}
    max_count = 0

    for letter in letters:
        if letter in counts:
            counts[letter] += 1
        else:
            counts[letter] = 1

        if counts[letter] > max_count:
            max_count = counts[letter]

    result = {}
    for letter, count in counts.items():
        if count == max_count:
            result[letter] = count

    return result

✅ Run 3

Start time: 2023-07-18 17:10:37
Duration: 8.68s
Completion tokens: 239
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    if test == '':
        return {}
    
    letters = test.split()
    counts = {}
    
    for letter in letters:
        if letter in counts:
            counts[letter] += 1
        else:
            counts[letter] = 1
    
    max_count = max(counts.values())
    result = {}
    
    for letter, count in counts.items():
        if count == max_count:
            result[letter] = count
    
    return result

❌ Run 4

Start time: 2023-07-18 17:10:29
Duration: 8.0s
Completion tokens: 223
Result: Fail

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    counts = {}
    for letter in test.split():
        if letter in counts:
            counts[letter] += 1
        else:
            counts[letter] = 1
    
    max_count = max(counts.values())
    result = {}
    for letter, count in counts.items():
        if count == max_count:
            result[letter] = count
    
    return result

✅ Run 5

Start time: 2023-07-18 17:10:20
Duration: 9.25s
Completion tokens: 247
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    count = {}
    max_count = 0
    for letter in letters:
        if letter in count:
            count[letter] += 1
        else:
            count[letter] = 1
        if count[letter] > max_count:
            max_count = count[letter]
    result = {}
    for letter, letter_count in count.items():
        if letter_count == max_count:
            result[letter] = letter_count
    return result

✅ Run 6

Start time: 2023-07-18 17:10:10
Duration: 9.94s
Completion tokens: 244
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    counts = {}
    max_count = 0
    for letter in letters:
        if letter in counts:
            counts[letter] += 1
        else:
            counts[letter] = 1
        if counts[letter] > max_count:
            max_count = counts[letter]
    result = {}
    for letter, count in counts.items():
        if count == max_count:
            result[letter] = count
    return result

✅ Run 7

Start time: 2023-07-18 17:10:01
Duration: 8.71s
Completion tokens: 245
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    counts = {}
    max_count = 0
    
    for letter in letters:
        if letter in counts:
            counts[letter] += 1
        else:
            counts[letter] = 1
        
        if counts[letter] > max_count:
            max_count = counts[letter]
    
    result = {}
    for letter, count in counts.items():
        if count == max_count:
            result[letter] = count
    
    return result

✅ Run 8

Start time: 2023-07-18 17:09:52
Duration: 9.27s
Completion tokens: 252
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    count_dict = {}
    max_count = 0
    
    for letter in letters:
        if letter in count_dict:
            count_dict[letter] += 1
        else:
            count_dict[letter] = 1
        
        if count_dict[letter] > max_count:
            max_count = count_dict[letter]
    
    result = {}
    for letter, count in count_dict.items():
        if count == max_count:
            result[letter] = count
    
    return result

✅ Run 9

Start time: 2023-07-18 17:09:44
Duration: 7.79s
Completion tokens: 245
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    counts = {}
    max_count = 0
    
    for letter in letters:
        if letter in counts:
            counts[letter] += 1
        else:
            counts[letter] = 1
        
        if counts[letter] > max_count:
            max_count = counts[letter]
    
    result = {}
    for letter, count in counts.items():
        if count == max_count:
            result[letter] = count
    
    return result

✅ Run 10

Start time: 2023-07-18 17:09:34
Duration: 9.29s
Completion tokens: 244
Result: Pass

def histogram(test):
    """Given a string representing a space separated lowercase letters, return a dictionary
    of the letter with the most repetition and containing the corresponding count.
    If several letters have the same occurrence, return all of them.
    
    Example:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'b': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """
    letters = test.split()
    counts = {}
    max_count = 0
    for letter in letters:
        if letter in counts:
            counts[letter] += 1
        else:
            counts[letter] = 1
        if counts[letter] > max_count:
            max_count = counts[letter]
    result = {}
    for letter, count in counts.items():
        if count == max_count:
            result[letter] = count
    return result