111.md

Results of GPT-4 on HumanEval/111

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

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 18:11:45
Duration: 20.58s
Completion tokens: 289
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('') == {}

    """
    # Remove spaces from the string
    test = test.replace(" ", "")
    
    # Create a dictionary to store the count of each letter
    count_dict = {}
    for letter in test:
        if letter in count_dict:
            count_dict[letter] += 1
        else:
            count_dict[letter] = 1
    
    # Find the maximum count
    max_count = max(count_dict.values()) if count_dict else 0
    
    # Create a new dictionary to store the letters with the maximum count
    max_count_dict = {k: v for k, v in count_dict.items() if v == max_count}
    
    return max_count_dict

✅ Run 2

Start time: 2023-07-18 18:11:13
Duration: 31.57s
Completion tokens: 373
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('') == {}

    """
    # Split the string into a list of letters
    letters = test.split()

    # Create a dictionary to store the count of each letter
    letter_counts = {}

    # Iterate over the list of letters
    for letter in letters:
        # If the letter is already in the dictionary, increment its count
        if letter in letter_counts:
            letter_counts[letter] += 1
        # If the letter is not in the dictionary, add it with a count of 1
        else:
            letter_counts[letter] = 1

    # Find the maximum count
    max_count = max(letter_counts.values(), default=0)

    # Create a new dictionary to store the letters with the maximum count
    max_count_letters = {}

    # Iterate over the letter counts
    for letter, count in letter_counts.items():
        # If the count is equal to the maximum count, add the letter to the new dictionary
        if count == max_count:
            max_count_letters[letter] = count

    # Return the new dictionary
    return max_count_letters

✅ Run 3

Start time: 2023-07-18 18:10:43
Duration: 29.99s
Completion tokens: 333
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('') == {}

    """
    # Split the string into a list of letters
    letters = test.split()
    
    # Create a dictionary to store the count of each letter
    letter_counts = {}
    
    # Iterate over the list of letters
    for letter in letters:
        # If the letter is already in the dictionary, increment its count
        if letter in letter_counts:
            letter_counts[letter] += 1
        # If the letter is not in the dictionary, add it with a count of 1
        else:
            letter_counts[letter] = 1
    
    # Find the maximum count
    max_count = max(letter_counts.values(), default=0)
    
    # Create a new dictionary to store the letters with the maximum count
    max_count_letters = {letter: count for letter, count in letter_counts.items() if count == max_count}
    
    return max_count_letters

✅ Run 4

Start time: 2023-07-18 18:10:18
Duration: 24.46s
Completion tokens: 333
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('') == {}

    """
    # Split the string into a list of letters
    letters = test.split()
    
    # Create a dictionary to store the count of each letter
    letter_count = {}
    
    # Iterate over the list of letters
    for letter in letters:
        # If the letter is already in the dictionary, increment its count
        if letter in letter_count:
            letter_count[letter] += 1
        # If the letter is not in the dictionary, add it with a count of 1
        else:
            letter_count[letter] = 1
    
    # Find the maximum count
    max_count = max(letter_count.values(), default=0)
    
    # Create a new dictionary to store the letters with the maximum count
    max_count_letters = {letter: count for letter, count in letter_count.items() if count == max_count}
    
    return max_count_letters

✅ Run 5

Start time: 2023-07-18 18:09:53
Duration: 25.19s
Completion tokens: 341
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('') == {}

    """
    # Split the string into a list of letters
    letters = test.split()
    # Create a dictionary to store the count of each letter
    letter_count = {}
    # Iterate over the list of letters
    for letter in letters:
        # If the letter is already in the dictionary, increment its count
        if letter in letter_count:
            letter_count[letter] += 1
        # If the letter is not in the dictionary, add it with a count of 1
        else:
            letter_count[letter] = 1
    # Find the maximum count
    max_count = max(letter_count.values(), default=0)
    # Create a new dictionary to store the letters with the maximum count
    max_count_letters = {letter: count for letter, count in letter_count.items() if count == max_count}
    # Return the dictionary of letters with the maximum count
    return max_count_letters

✅ Run 6

Start time: 2023-07-18 18:09:23
Duration: 30.26s
Completion tokens: 332
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('') == {}

    """
    # Create a dictionary to store the frequency of each letter
    freq_dict = {}
    # Split the string into a list of letters
    letters = test.split()
    # Iterate over the list of letters
    for letter in letters:
        # If the letter is already in the dictionary, increment its count
        if letter in freq_dict:
            freq_dict[letter] += 1
        # If the letter is not in the dictionary, add it with a count of 1
        else:
            freq_dict[letter] = 1
    # Find the maximum frequency
    max_freq = max(freq_dict.values()) if freq_dict else 0
    # Create a new dictionary to store the letters with the maximum frequency
    max_freq_dict = {k: v for k, v in freq_dict.items() if v == max_freq}
    return max_freq_dict

✅ Run 7

Start time: 2023-07-18 18:08:55
Duration: 27.73s
Completion tokens: 333
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('') == {}

    """
    # Create a dictionary to store the count of each letter
    count_dict = {}
    # Split the string into a list of letters
    letters = test.split()
    # Iterate over each letter in the list
    for letter in letters:
        # If the letter is already in the dictionary, increment its count
        if letter in count_dict:
            count_dict[letter] += 1
        # If the letter is not in the dictionary, add it with a count of 1
        else:
            count_dict[letter] = 1
    # Find the maximum count
    max_count = max(count_dict.values()) if count_dict else 0
    # Create a new dictionary to store the letters with the maximum count
    max_count_dict = {k: v for k, v in count_dict.items() if v == max_count}
    return max_count_dict

✅ Run 8

Start time: 2023-07-18 18:08:33
Duration: 22.15s
Completion tokens: 330
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('') == {}

    """
    # Split the string into a list of letters
    letters = test.split()
    # Create a dictionary to store the count of each letter
    count_dict = {}
    # Iterate over the list of letters
    for letter in letters:
        # If the letter is already in the dictionary, increment its count
        if letter in count_dict:
            count_dict[letter] += 1
        # If the letter is not in the dictionary, add it with a count of 1
        else:
            count_dict[letter] = 1
    # Find the maximum count
    max_count = max(count_dict.values()) if count_dict else 0
    # Create a new dictionary to store the letters with the maximum count
    max_dict = {k: v for k, v in count_dict.items() if v == max_count}
    return max_dict

✅ Run 9

Start time: 2023-07-18 18:08:04
Duration: 28.92s
Completion tokens: 285
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('') == {}

    """
    # split the string into a list of letters
    letters = test.split()
    # create a dictionary to store the count of each letter
    letter_count = {}
    for letter in letters:
        if letter in letter_count:
            letter_count[letter] += 1
        else:
            letter_count[letter] = 1
    # find the maximum count
    max_count = max(letter_count.values(), default=0)
    # create a new dictionary to store the letters with the maximum count
    max_count_letters = {letter: count for letter, count in letter_count.items() if count == max_count}
    return max_count_letters

✅ Run 10

Start time: 2023-07-18 18:07:45
Duration: 18.75s
Completion tokens: 229
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 {}
    else:
        test = test.split()
        dic = {}
        for i in test:
            if i in dic:
                dic[i] += 1
            else:
                dic[i] = 1
        max_val = max(dic.values())
        return {k: v for k, v in dic.items() if v == max_val}