| 模型 | 模型答案 | 测试结果 |
|---|---|---|
| 金声玉亮2.0 | ```python def extract_freq(tuples_list): unique_tuples = set(map(tuple, map(sorted, tuples_list))) return len(unique_tuples) ``` [DONE] |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| deepseekr1 | [BEGIN] def extract_freq(lst): normalized = {tuple(sorted(t)) for t in lst} return len(normalized) [DONE] |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| gpto1 | [BEGIN] 'def extract_freq(lst): sorted_tuples = [tuple(sorted(t)) for t in lst] unique_tuples = set(sorted_tuples) return len(unique_tuples)' [DONE] |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| GPT-4o | To write a function that extracts the frequency of unique tuples in a given list, where the order of elements in the tuples does not matter, we need to consider each tuple and its possible permutations. We can use a `set` to keep track of the unique tuples we encounter. Here is the implementation of the function: ```python def extract_freq(tuples_list): unique_tuples = set() for t in tuples_list: # Add the tuple and its reversed version to the set to handle order insensitivity unique_tuples.add(tuple(sorted(t))) return len(unique_tuples) # Test cases assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)]) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)]) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)]) == 4 ``` Explanation: 1. We create an empty set `unique_tuples` to store the unique tuples. 2. For each tuple `t` in the input list `tuples_list`, we sort the elements of the tuple to ensure order insensitivity and add the sorted tuple to the set. 3. Finally, we return the length of the set, which represents the number of unique tuples. This approach ensures that tuples like `(3, 4)` and `(4, 3)` are considered the same, and only one of them is counted as unique. |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| llama3.1-70B-Instruct | def extract_freq(tup_list): freq = {} for tup in tup_list: sorted_tup = tuple(sorted(tup)) if sorted_tup in freq: freq[sorted_tup] += 1 else: freq[sorted_tup] = 1 return len(freq) |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| llama3.1-405B-Instruct | def extract_freq(lst): freq = {} for tup in lst: sorted_tup = tuple(sorted(tup)) if sorted_tup in freq: freq[sorted_tup] += 1 else: freq[sorted_tup] = 1 return len(freq) |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| DeepSeek-Coder-V2 | ```python def extract_freq(tuples_list): unique_tuples = set(map(tuple, map(sorted, tuples_list))) return len(unique_tuples) ``` [DONE] |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| Qwen2-72B-Instruct | 'def extract_freq(test_list): res = len(set(tuple(sorted(sub)) for sub in test_list)) return res' [DONE] |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| baichuan4 | To solve the task of extracting the frequency of unique tuples in a given list, regardless of their order, we can use a combination of Python's built-in data structures and functions. Specifically, we'll convert the list of tuples into a set of frozensets to eliminate duplicates while preserving the unordered nature of tuples. Then, we'll count the occurrences of each unique frozenset in the original list. Here's the implementation: ```python def extract_freq(tuple_list): # Convert each tuple to a frozenset to disregard the order frozen_tuples = [frozenset(t) for t in tuple_list] # Find unique frozensets unique_frozensets = set(frozen_tuples) # Count occurrences of each unique frozenset freq_count = len(frozen_tuples) - len(unique_frozensets) return freq_count # Test cases assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)]) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)]) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)]) == 4 ``` The `extract_freq` function first converts each tuple in the input list to a frozenset. This step is crucial because sets and frozensets do not maintain order, which means that (3, 4) and (4, 3) will be considered the same after conversion. By converting to frozensets, we can add them to a set to find unique combinations. The frequency count is then determined by subtracting the number of unique frozensets from the total number of frozensets in the list. This solution ensures that the order within tuples does not affect the counting of unique combinations, fulfilling the requirement of the task. The provided test cases check for different scenarios, including cases with and without duplicate tuples, to confirm the correctness of the function. |
wrong_answer 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| doubao | def extract_freq(lst): return len(set(tuple(sorted(t)) for t in lst)) [DONE] |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| minimax | ```python def extract_freq(tuples_list): # Convert each tuple to a sorted tuple to ensure order doesn't matter normalized_tuples = [tuple(sorted(t)) for t in tuples_list] # Use a set to count unique tuples unique_tuples = set(normalized_tuples) return len(unique_tuples) # Test cases assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)]) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)]) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)]) == 4 ``` This function `extract_freq` normalizes each tuple by sorting its elements, which ensures that the order of elements within the tuples does not matter when counting unique tuples. It then creates a set of these normalized tuples, which automatically removes duplicates, and returns the number of unique elements in the set. The provided test cases should pass with this implementation. |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| kimi | Certainly! Let's write the function to extract the frequency of unique tuples in the given list order irrespective. ```python def extract_freq(tuples_list): # Create a set to store unique tuples unique_tuples = set() # Iterate through each tuple in the list for tup in tuples_list: # Add the tuple to the set unique_tuples.add(tup) # The number of unique tuples is the size of the set return len(unique_tuples) # Test cases assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)]) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)]) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)]) == 4 ``` This function will count the number of unique tuples in the list, regardless of the order of the elements within the tuples. The `assert` statements will check if the function returns the correct number of unique tuples for the given test cases. |
wrong_answer 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| wenxinyiyan | To solve the tasks given, let's write the Python functions as described. ### Task 1: Finding Similar Elements in Two Tuple Lists The provided code snippet is almost correct. However, it returns a single tuple within a tuple due to the extra parentheses around `res`. We can fix this by removing the extra parentheses. ```python def similar_elements(test_tup1, test_tup2): res = tuple(set(test_tup1) & set(test_tup2)) return res ``` ### Task 2: Identifying Non-Prime Numbers The provided code snippet correctly identifies non-prime numbers. It iterates from 2 to the square root of `n` and checks if `n` is divisible by any number in that range. If it finds a divisor, it sets `result` to `True` and returns it. No changes are needed for this function. ### Task 3: Finding the Largest Integers Using Heap Queue Algorithm The provided code snippet uses the `heapq.nlargest()` function to find the `n` largest numbers in the given list. This is a valid solution, but it's worth noting that `heapq.nlargest()` doesn't necessarily use a heap queue internally for small values of `n`. However, for large lists and relatively small values of `n`, it is an efficient approach. No changes are needed for this function. ### Task 4: Extracting the Frequency of Unique Tuples in a Given List (Order Irrespective) For this task, we need to write a function that counts the frequency of unique tuples in a list, ignoring the order of elements within the tuples. We can use a dictionary to keep track of the counts and convert the tuples to sorted tuples before using them as keys in the dictionary. Here's the code to achieve this: ```python def extract_freq(lst): freq_dict = {} for tup in lst: sorted_tup = tuple(sorted(tup)) freq_dict[sorted_tup] = freq_dict.get(sorted_tup, 0) + 1 return len(freq_dict) ``` This function iterates over each tuple in the list, sorts it, and then uses the sorted tuple as a key in the `freq_dict` dictionary. It increments the count for each unique sorted tuple encountered. Finally, it returns the length of the dictionary, which represents the number of unique tuples (ignoring order) in the given list. Note: The provided test cases check for the number of unique tuples, not the total frequency count. If you want the total frequency count, you can modify the function to return `sum(freq_dict.values())` instead of `len(freq_dict)`. However, based on the test cases, it seems like we only need the number of unique tuples. |
failed 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| zhipu | To achieve this task, we need to write a function that counts the frequency of unique tuples in a list, considering that tuples are unordered (i.e., (3, 4) is the same as (4, 3)). We can use a dictionary to keep track of the frequencies. Here's the implementation: ```python def extract_freq(tuples_list): freq_dict = {} for tup in tuples_list: # Normalize the tuple to ensure (3, 4) and (4, 3) are considered the same normalized_tup = tuple(sorted(tup)) if normalized_tup in freq_dict: freq_dict[normalized_tup] += 1 else: freq_dict[normalized_tup] = 1 return len(freq_dict) # Test cases assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)]) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)]) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)]) == 4 print("All tests passed!") ``` ### Explanation: 1. **Normalization**: Each tuple is sorted to ensure that (3, 4) and (4, 3) are treated as the same. 2. **Frequency Dictionary**: A dictionary `freq_dict` is used to keep track of the frequency of each normalized tuple. 3. **Count Unique Tuples**: The function returns the number of unique tuples by returning the length of the dictionary. This code should pass the provided test cases, ensuring that the function correctly counts the frequency of unique tuples irrespective of their order. |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| gpto1 | [BEGIN] 'def extract_freq(lst): sorted_tuples = [tuple(sorted(t)) for t in lst] unique_tuples = set(sorted_tuples) return len(unique_tuples)' [DONE] |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |
| deepseekv3 | ```python def extract_freq(tuples_list): # Convert each tuple to a frozenset to make them hashable and order-irrelevant unique_tuples = set(frozenset(t) for t in tuples_list) return len(unique_tuples) # Test cases assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)]) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)]) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)]) == 4 ``` ### Explanation: 1. **Conversion to Frozenset**: Each tuple in the list is converted to a `frozenset`. This is because sets are unordered collections, and converting tuples to sets ensures that the order of elements within the tuples does not matter. For example, `(3, 4)` and `(4, 3)` will be considered the same. 2. **Unique Tuples**: The `set` data structure is used to store these `frozenset`s, which automatically removes any duplicates. 3. **Count Unique Tuples**: The length of this set gives the number of unique tuples, considering the order of elements within the tuples irrelevant. This approach ensures that the function correctly counts the number of unique tuples, regardless of the order of elements within each tuple. |
pass 测试用例：assert extract_freq([(3, 4), (1, 2), (4, 3), (5, 6)] ) == 3 assert extract_freq([(4, 15), (2, 3), (5, 4), (6, 7)] ) == 4 assert extract_freq([(5, 16), (2, 3), (6, 5), (6, 9)] ) == 4 |