| 模型 | 模型答案 | 测试结果 |
|---|---|---|
| 金声玉亮2.0 | ```python def encode_list(lst): if not lst: return [] encoded = [] current_element = lst[0] count = 1 for element in lst[1:]: if element == current_element: count += 1 else: encoded.append([count, current_element]) current_element = element count = 1 encoded.append([count, current_element]) return encoded ``` [DONE] |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| deepseekr1 | To solve this problem, we need to create a function that performs run-length encoding on a given list or string. Run-length encoding is a simple form of data compression where consecutive identical elements are represented as a single value and a count. The function should handle both lists of numbers and strings, returning the encoded result as a list of lists, where each sublist contains the count followed by the element. ### Approach 1. **Check for Empty Input**: If the input is empty (either an empty list or string), return an empty list. 2. **Initialize Variables**: Start with the first element of the input as the current value and initialize a count to 1. 3. **Iterate Through Elements**: Loop through the remaining elements of the input. For each element: - If it matches the current value, increment the count. - If it does not match, append the current count and value to the result list, then update the current value and reset the count to 1. 4. **Final Append**: After the loop, append the last accumulated count and value to the result list. 5. **Return Result**: Return the result list containing all the run-length encoded pairs. ### Solution Code ```python def encode_list(s): if not s: return [] result = [] current_value = s[0] count = 1 for elem in s[1:]: if elem == current_value: count += 1 else: result.append([count, current_value]) current_value = elem count = 1 result.append([count, current_value]) return result ``` ### Explanation - **Handling Empty Input**: The function first checks if the input is empty and returns an empty list immediately if true. - **Initialization**: The first element is stored as `current_value` with an initial count of 1. - **Loop Through Elements**: The loop starts from the second element. Each element is compared to `current_value`. If they are the same, the count is incremented. If different, the current count and value are added to the result, and the current value and count are updated. - **Final Append**: After processing all elements, the last accumulated count and value are added to the result to ensure the final group is included. This approach efficiently processes each element exactly once, resulting in a time complexity of O(n), where n is the length of the input. This ensures the solution is both optimal and straightforward. |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| gpto1 | [BEGIN] 'def encode_list(lst): result = [] if not lst: return result iterator = iter(lst) prev_element = next(iterator) count = 1 for element in iterator: if element == prev_element: count += 1 else: result.append([count, prev_element]) prev_element = element count = 1 result.append([count, prev_element]) return result' [DONE] |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| GPT-4o | To write a function that performs run-length encoding, we need to iterate through the input list or string, count consecutive identical elements, and create a list of lists where each sublist contains the count and the element itself. Here is the function to perform run-length encoding: ```python def encode_list(data): if not data: return [] encoded = [] count = 1 previous = data[0] for current in data[1:]: if current == previous: count += 1 else: encoded.append([count, previous]) previous = current count = 1 # Append the last group encoded.append([count, previous]) return encoded # Tests assert encode_list([1, 1, 2, 3, 4, 4.3, 5, 1]) == [[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically') == [[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python') == [[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] ``` This function should now pass all the provided test cases. |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| llama3.1-70B-Instruct | def encode_list(lst): result = [] current = [1, lst[0]] for i in range(1, len(lst)): if lst[i] == lst[i-1]: current[0] += 1 else: result.append(current) current = [1, lst[i]] result.append(current) return result |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| llama3.1-405B-Instruct | def encode_list(input_list): if not input_list: return [] encoded_list = [] count = 1 for i in range(1, len(input_list)): if input_list[i] == input_list[i-1]: count += 1 else: encoded_list.append([count, input_list[i-1]]) count = 1 encoded_list.append([count, input_list[-1]]) return encoded_list |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| DeepSeek-Coder-V2 | ```python def encode_list(lst): if not lst: return [] encoded = [] current_element = lst[0] count = 1 for element in lst[1:]: if element == current_element: count += 1 else: encoded.append([count, current_element]) current_element = element count = 1 encoded.append([count, current_element]) return encoded ``` [DONE] |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| Qwen2-72B-Instruct | 'def encode_list(input_list): if isinstance(input_list, str): input_list = list(input_list) res = [] count = 1 for i in range(1, len(input_list)): if input_list[i] == input_list[i-1]: count += 1 else: res.append([count, input_list[i-1]]) count = 1 res.append([count, input_list[-1]]) return res' [DONE] |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| baichuan4 | To reflect the run-length encoding from a list, you can use the following function: ```python def encode_list(input_list): if not input_list: return [] encoded_list = [] current_element = input_list[0] count = 1 for element in input_list[1:]: if element == current_element: count += 1 else: encoded_list.append([count, current_element]) current_element = element count = 1 encoded_list.append([count, current_element]) return encoded_list ``` This function iterates through the input list and counts consecutive occurrences of each element. When a new element is encountered, it appends the count and the element to the encoded list and resets the count. Finally, it appends the count and the last element to the encoded list. Here are the test cases: ```python assert encode_list([1,1,2,3,4,4.3,5,1]) == [[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically') == [[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python') == [[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] ``` |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| doubao | 'def encode_list(lst): result = [] current = None count = 0 for item in lst: if item == current: count += 1 else: if current is not None: result.append([count, current]) current = item count = 1 result.append([count, current]) return result' [DONE] |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| minimax | ```python def encode_list(lst): if not lst: return [] # Initialize the list to store the encoded pairs encoded = [] # Initialize the count and the previous element count = 1 prev = lst[0] # Iterate over the list starting from the second element for item in lst[1:]: if item == prev: # If the current item is the same as the previous, increment the count count += 1 else: # If the current item is different, append the count and previous item to the encoded list encoded.append([count, prev]) # Reset the count and update the previous item count = 1 prev = item # Append the last count and previous item to the encoded list encoded.append([count, prev]) return encoded # Test cases assert encode_list([1, 1, 2, 3, 4, 4.3, 5, 1]) == [[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list(list('automatically')) == [[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list(list('python')) == [[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] ``` This function `encode_list` takes a list `lst` and returns a list of lists, where each sublist contains a count and the corresponding element that was repeated consecutively in the input list. The test cases check that the function correctly encodes the input lists according to the run-length encoding scheme. |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| kimi | Here's the Python code for each of the tasks you've described: 1. Function to find similar elements from two tuple lists: ```python def similar_elements(test_tup1, test_tup2): res = tuple(sorted(set(test_tup1) & set(test_tup2))) return res ``` 2. Function to identify non-prime numbers: ```python import math def is_not_prime(n): if n <= 1: return True for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return True return False ``` 3. Function to find the largest integers from a given list using heap queue algorithm: ```python import heapq def heap_queue_largest(nums, n): largest_nums = heapq.nlargest(n, nums) return largest_nums ``` 4. Function to reflect the run-length encoding from a list: ```python def encode_list(input_list): if not isinstance(input_list, list): input_list = list(input_list) encoded = [] prev_item = None count = 1 for item in input_list: if item == prev_item: count += 1 else: if prev_item is not None: encoded.append([count, prev_item]) count = 1 prev_item = item encoded.append([count, prev_item]) return encoded ``` Each of these functions should pass the tests you've provided. |
failed 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| wenxinyiyan | Here are the solutions to the tasks you've described: 1. Finding similar elements from two tuple lists: ```python def similar_elements(test_tup1, test_tup2): res = tuple(set(test_tup1) & set(test_tup2)) # The solution you provided is almost correct, but it returns a tuple within a tuple. # To match the expected output, we should return `res` directly without the extra tuple. return res # Test cases assert similar_elements((3, 4, 5, 6),(5, 7, 4, 10)) == (4, 5) assert similar_elements((1, 2, 3, 4),(5, 4, 3, 7)) == (3, 4) assert similar_elements((11, 12, 14, 13),(17, 15, 14, 13)) == (13, 14) ``` 2. Identifying non-prime numbers: ```python import math def is_not_prime(n): if n < 2: # Numbers less than 2 are not prime return True for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return True return False # Test cases assert is_not_prime(2) == False assert is_not_prime(10) == True assert is_not_prime(35) == True ``` 3. Finding the largest integers from a given list using the heap queue algorithm: ```python import heapq as hq def heap_queue_largest(nums, n): largest_nums = hq.nlargest(n, nums) return largest_nums # Test cases assert heap_queue_largest([25, 35, 22, 85, 14, 65, 75, 22, 58], 3) == [85, 75, 65] assert heap_queue_largest([25, 35, 22, 85, 14, 65, 75, 22, 58], 2) == [85, 75] assert heap_queue_largest([25, 35, 22, 85, 14, 65, 75, 22, 58], 5) == [85, 75, 65, 58, 35] ``` 4. Reflecting the run-length encoding from a list: ```python def encode_list(lst): encoded = [] count = 1 for i in range(1, len(lst)): if lst[i] == lst[i-1]: count += 1 else: encoded.append([count, lst[i-1]]) count = 1 encoded.append([count, lst[-1]]) # Append the last element's count and value return encoded # Test cases assert encode_list([1,1,2,3,4,4.3,5,1]) == [[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically') == [[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python') == [[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] ``` Note: The `encode_list` function assumes that the input list is not empty. If you want to handle empty lists as well, you can add a check at the beginning of the function. |
failed 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| zhipu | To create a function that performs run-length encoding on a list, we need to iterate through the list and count consecutive occurrences of each element. The function should handle both lists of numbers and strings. Here's the implementation: ```python def encode_list(input_list): if not input_list: return [] encoded_list = [] current_element = input_list[0] count = 1 for element in input_list[1:]: if element == current_element: count += 1 else: encoded_list.append([count, current_element]) current_element = element count = 1 encoded_list.append([count, current_element]) return encoded_list # Test cases assert encode_list([1,1,2,3,4,4.3,5,1]) == [[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically') == [[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python') == [[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] print("All tests passed!") ``` ### Explanation: 1. **Initialization**: - `encoded_list` to store the result. - `current_element` to keep track of the current element being counted. - `count` to count the occurrences of the current element. 2. **Iteration**: - Loop through the input list starting from the second element. - If the current element is the same as the previous one, increment the count. - If it differs, append the count and the current element to `encoded_list`, reset the count, and update the current element. 3. **Final Append**: - After the loop, append the last counted element and its count to `encoded_list`. 4. **Return**: - Return the `encoded_list`. This function should pass the provided test cases and handle both lists of numbers and strings. |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| gpto1 | [BEGIN] 'def encode_list(lst): result = [] if not lst: return result iterator = iter(lst) prev_element = next(iterator) count = 1 for element in iterator: if element == prev_element: count += 1 else: result.append([count, prev_element]) prev_element = element count = 1 result.append([count, prev_element]) return result' [DONE] |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |
| deepseekv3 | To implement the run-length encoding function, we need to iterate through the list and count consecutive occurrences of each element. The function should then return a list of lists, where each sublist contains the count followed by the element. Here's the implementation: ```python def encode_list(elements): if not elements: return [] encoded = [] prev_element = elements[0] count = 1 for element in elements[1:]: if element == prev_element: count += 1 else: encoded.append([count, prev_element]) prev_element = element count = 1 encoded.append([count, prev_element]) return encoded # Test cases assert encode_list([1, 1, 2, 3, 4, 4.3, 5, 1]) == [[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically') == [[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python') == [[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] print("All test cases passed!") ``` ### Explanation: 1. **Initialization**: We start by checking if the input list is empty. If it is, we return an empty list. 2. **Iteration**: We iterate through the list starting from the second element. For each element, we compare it with the previous element. 3. **Counting**: If the current element is the same as the previous, we increment the count. Otherwise, we append the count and the previous element to the result list and reset the count. 4. **Final Append**: After the loop, we append the last counted element and its count to the result list. 5. **Return**: Finally, we return the encoded list. This function works for both lists of numbers and strings, as demonstrated by the test cases. |
pass 测试用例：assert encode_list([1,1,2,3,4,4.3,5,1])==[[2, 1], [1, 2], [1, 3], [1, 4], [1, 4.3], [1, 5], [1, 1]] assert encode_list('automatically')==[[1, 'a'], [1, 'u'], [1, 't'], [1, 'o'], [1, 'm'], [1, 'a'], [1, 't'], [1, 'i'], [1, 'c'], [1, 'a'], [2, 'l'], [1, 'y']] assert encode_list('python')==[[1, 'p'], [1, 'y'], [1, 't'], [1, 'h'], [1, 'o'], [1, 'n']] |