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The dataset generation failed
Error code: DatasetGenerationError
Exception: CastError
Message: Couldn't cast
problem_key: string
rollout_index: int64
arm: string
steer: string
label: string
committed: bool
correct: bool
hit_token_cap: bool
finish_reason: string
output_tokens: int64
max_tokens_used: int64
completion: string
exec_correct: bool
exec_error: string
exec_extract: string
to
{'problem_key': Value('string'), 'rollout_index': Value('int64'), 'arm': Value('string'), 'steer': Value('string'), 'label': Value('string'), 'committed': Value('bool'), 'correct': Value('bool'), 'hit_token_cap': Value('bool'), 'finish_reason': Value('string'), 'output_tokens': Value('int64'), 'max_tokens_used': Value('int64'), 'completion': Value('string')}
because column names don't match
Traceback: Traceback (most recent call last):
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1827, in _prepare_split_single
for key, table in generator:
^^^^^^^^^
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 613, in wrapped
for item in generator(*args, **kwargs):
~~~~~~~~~^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/packaged_modules/json/json.py", line 343, in _generate_tables
self._cast_table(pa_table, json_field_paths=json_field_paths),
~~~~~~~~~~~~~~~~^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/packaged_modules/json/json.py", line 132, in _cast_table
pa_table = table_cast(pa_table, self.info.features.arrow_schema)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2378, in table_cast
return cast_table_to_schema(table, schema)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2306, in cast_table_to_schema
raise CastError(
...<3 lines>...
)
datasets.table.CastError: Couldn't cast
problem_key: string
rollout_index: int64
arm: string
steer: string
label: string
committed: bool
correct: bool
hit_token_cap: bool
finish_reason: string
output_tokens: int64
max_tokens_used: int64
completion: string
exec_correct: bool
exec_error: string
exec_extract: string
to
{'problem_key': Value('string'), 'rollout_index': Value('int64'), 'arm': Value('string'), 'steer': Value('string'), 'label': Value('string'), 'committed': Value('bool'), 'correct': Value('bool'), 'hit_token_cap': Value('bool'), 'finish_reason': Value('string'), 'output_tokens': Value('int64'), 'max_tokens_used': Value('int64'), 'completion': Value('string')}
because column names don't match
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1369, in compute_config_parquet_and_info_response
parquet_operations, partial, estimated_dataset_info = stream_convert_to_parquet(
~~~~~~~~~~~~~~~~~~~~~~~~~^
builder, max_dataset_size_bytes=max_dataset_size_bytes
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
)
^
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 948, in stream_convert_to_parquet
builder._prepare_split(split_generator=splits_generators[split], file_format="parquet")
~~~~~~~~~~~~~~~~~~~~~~^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1694, in _prepare_split
for job_id, done, content in self._prepare_split_single(
~~~~~~~~~~~~~~~~~~~~~~~~~~^
gen_kwargs=gen_kwargs, job_id=job_id, **_prepare_split_args
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
):
^
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1880, in _prepare_split_single
raise DatasetGenerationError("An error occurred while generating the dataset") from e
datasets.exceptions.DatasetGenerationError: An error occurred while generating the datasetNeed help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
problem_key string | rollout_index int64 | arm string | steer string | label string | committed bool | correct bool | hit_token_cap bool | finish_reason string | output_tokens int64 | max_tokens_used int64 | completion string |
|---|---|---|---|---|---|---|---|---|---|---|---|
HumanEval/0 | 0 | none | base | humaneval | false | false | false | stop | 1,179 | 31,744 | ```python
Okay, I need to write a Python function called has_close_elements that checks if there are any two numbers in a list that are closer to each other than a given threshold. Let's think about how to approach this.
First, the problem says that if any two numbers in the list have a difference less than the thre... |
HumanEval/0 | 1 | none | base | humaneval | false | false | false | stop | 1,315 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if there are any two numbers in a list that are closer than a given threshold. Let's think about how to approach this.
First, the problem says that if any two numbers in the list have an absolute difference less than the threshold, return... |
HumanEval/0 | 2 | none | base | humaneval | false | false | false | stop | 1,356 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if there are any two numbers in a list that are closer to each other than a given threshold. Let's see. The examples given are helpful.
First, the first example: [1.0, 2.0, 3.0] with threshold 0.5. The differences between consecutive ele... |
HumanEval/0 | 3 | none | base | humaneval | false | false | false | stop | 2,035 | 31,744 | The function has_close_elements takes a list of numbers and a threshold as input, and returns True if any two numbers in the list are closer to each other than the threshold, and False otherwise. The function can be implemented using a nested loop to compare each pair of numbers in the list and check if their differen... |
HumanEval/0 | 4 | none | base | humaneval | false | false | false | stop | 1,012 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if any two numbers in a list are closer to each other than a given threshold. Let's think about how to approach this.
First, the problem says that if there are two numbers in the list where the absolute difference between them is less tha... |
HumanEval/0 | 5 | none | base | humaneval | false | false | false | stop | 1,544 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if any two numbers in a list are within a certain threshold of each other. Let's see. The examples given: the first one returns False because the numbers are 1.0, 2.0, 3.0 with a threshold of 0.5. The differences between consecutive elemen... |
HumanEval/0 | 6 | none | base | humaneval | false | false | false | stop | 1,508 | 31,744 | ERROR: type should be string, got " https://drive.google.com/file/d/1Fq9lP3L2Y9yj9k4QF3XZB7b9YiKxR4hK/view?usp=sharing\n\nOkay, I need to write a Python function called has_close_elements that checks if any two numbers in a list are closer to each other than a given threshold. Let's think about how to approach this.\n\nFirst, the problem says that two numbers are considered close if their absolute difference is less than the threshold. So for every pair of numbers in the list, I need to check if the absolute difference between them is less than the threshold. If any such pair exists, return True; otherwise, return False.\n\nBut wait, checking all possible pairs might be inefficient, especially for large lists. Because the time complexity would be O(n²), which isn't great. But maybe the problem expects a straightforward approach. Let's see the sample inputs. The first example has three numbers, and the second has six. For small n, the O(n²) approach is acceptable. But if the list is very large, this could be a problem. However, the problem doesn't specify any constraints on the input size, so perhaps the intended solution is the straightforward one.\n\nSo the steps would be:\n\n1. Iterate through each pair of elements in the list.\n2. For each pair, compute the absolute difference.\n3. If any difference is less than the threshold, return True immediately.\n4. If after checking all pairs no such pair is found, return False.\n\nBut how to generate all pairs? One way is to use two nested loops. The outer loop runs from index i=0 to len(numbers)-1, and the inner loop runs from j=i+1 to len(numbers)-1. This way, each pair is considered once, avoiding duplicates like (i,j) and (j,i).\n\nWait, but in Python, for a list, you can use combinations from itertools to generate all unique pairs. But importing itertools might be necessary. However, the problem's code template doesn't have any imports except for List from typing. So maybe the solution should avoid using itertools to keep it simple.\n\nAlternatively, use two nested loops. Let's think about that.\n\nSo, for each i in range(len(numbers)), and for each j in range(i+1, len(numbers)), compute the absolute difference between numbers[i] and numbers[j]. If any of these is less than threshold, return True.\n\nYes, that's manageable.\n\nNow, let's test this logic with the sample inputs.\n\nFirst sample: [1.0, 2.0, 3.0], threshold 0.5. The differences are 1.0 (2-1), 2.0 (3-2), and 1.0 (3-1). All are >=0.5, so return False. Correct.\n\nSecond sample: [1.0, 2.8, 3.0, 4.0, 5.0, 2.0], threshold 0.3. Let's see. The numbers are 1, 2.8, 3,4,5,2. The pair 2.8 and 2.0 has a difference of 0.8, which is more than 0.3. Wait, but maybe another pair. Let's check all possible pairs. For example, 2.8 and 3.0: difference is 0.2, which is less than 0.3. So that pair would trigger a True. So the function should return True. Which matches the sample.\n\nSo the approach works.\n\nNow, code structure:\n\nThe function is given as:\n\ndef has_close_elements(numbers: List[float], threshold: float) -> bool:\n\nSo, inside the function, we need to loop through all pairs.\n\nInitialize a flag, maybe, but in Python, we can return as soon as a pair is found.\n\nSo:\n\nfor i in range(len(numbers)):\n for j in range(i+1, len(numbers)):\n if abs(numbers[i] - numbers[j]) < threshold:\n return True\nreturn False\n\nThat's the code. Let's test this with the samples.\n\nFirst sample: numbers = [1.0, 2.0, 3.0], threshold 0.5. The pairs are (0,1) difference 1.0, (0,2) 2.0, (1,2) 1.0. All >=0.5. So returns False.\n\nSecond sample: numbers = [1.0, 2.8, 3.0, 4.0, 5.0, 2.0], threshold 0.3. Let's see:\n\nCheck pairs:\n\n1.0 and 2.8: 1.8 >=0.3.\n\n1.0 and 3.0: 2.0 >=0.3.\n\n1.0 and 4.0: 3.0 >=0.3.\n\n1.0 and 5.0: 4.0 >=0.3.\n\n1.0 and 2.0: 1.0 >=0.3.\n\nThen 2.8 and 3.0: 0.2 <0.3 → return True. So the code returns True here, which is correct.\n\nSo the code works for the samples.\n\nBut what about edge cases? For example, when the list has only one element. Then, no pairs, returns False. If two elements, check their difference. If the list is empty, returns False.\n\nAnother edge case: threshold is zero. Then, any two equal elements would return True. But the problem says \"closer than threshold\", so if threshold is zero, then the difference must be less than zero, which is impossible. So in that case, the function would return False unless there are duplicates. Wait, no. Because if two elements are equal, their difference is zero, which is not less than zero. So even if threshold is zero, the function would return False unless there are elements with a difference less than zero, which can't happen. So in that case, the function would return False. But if the threshold is zero, the function would return True only if there are two elements with a difference less than zero, which is impossible. So the code is correct.\n\nAnother case: numbers with negative values. The absolute difference is handled correctly.\n\nSo the code seems correct.\n\nBut what about the time complexity? For a list of n elements, the number of pairs is n*(n-1)/2. For n=1e4, that's about 5e7 operations, which might be slow. But the problem doesn't mention performance constraints, so this approach is acceptable.\n\nTherefore, the code should be as written.```python\nfrom typing import List\n\ndef has_close_elements(numbers: List[float], threshold: float) -> bool:\n \"\"\" Check if in given list of numbers, are any two numbers closer to each other than\n given threshold.\n \"\"\"\n for i in range(len(numbers)):\n for j in range(i + 1, len(numbers)):\n if abs(numbers[i] - numbers[j]) < threshold:\n return True\n return False\n```" |
HumanEval/0 | 7 | none | base | humaneval | false | false | false | stop | 1,355 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if any two numbers in a list are closer to each other than a given threshold. Let's see.
The problem says that if there are any two elements in the list where their absolute difference is less than the threshold, then return True. Otherwi... |
HumanEval/0 | 8 | none | base | humaneval | false | false | false | stop | 1,342 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if any two numbers in a list are closer than a given threshold. Let's think about how to approach this.
First, the problem says that the function takes a list of floats and a threshold. The goal is to determine if there are at least two e... |
HumanEval/0 | 9 | none | base | humaneval | false | false | false | stop | 2,149 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if any two numbers in a list are closer than a given threshold. Let's see.
The problem says, given a list of numbers and a threshold, return True if there are at least two numbers whose absolute difference is less than the threshold. Othe... |
HumanEval/0 | 10 | none | base | humaneval | false | false | false | stop | 198 | 31,744 | ```python
def has_close_elements(numbers: List[float], threshold: float) -> bool:
""" Check if in given list of numbers, are any two numbers closer to each other than
given threshold.
>>> has_close_elements([1.0, 2.0, 3.0], 0.5)
False
>>> has_close_elements([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)
... |
HumanEval/0 | 11 | none | base | humaneval | false | false | false | stop | 951 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if there are any two numbers in a list that are closer to each other than a given threshold. Let's think about how to approach this.
First, the problem says that the function should return True if any two elements in the list have an abso... |
HumanEval/0 | 12 | none | base | humaneval | false | false | false | stop | 1,246 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if any two numbers in a list are closer to each other than a given threshold. Let's see. The examples given are like, for [1.0, 2.0, 3.0] with threshold 0.5, it returns False. But when the threshold is 0.3 and the list has 2.8 and 2.0, whi... |
HumanEval/0 | 13 | none | base | humaneval | false | false | false | stop | 1,130 | 31,744 | ```python
Okay, I need to write a Python function called has_close_elements that checks if there are any two numbers in a list that are closer than a given threshold. Let's think about how to approach this.
First, the problem says that if any two numbers in the list are within the threshold of each other, the functi... |
HumanEval/0 | 14 | none | base | humaneval | false | false | false | stop | 1,124 | 31,744 |
Okay, I need to write a Python function called has_close_elements that checks if there are any two numbers in a list that are closer than a given threshold. Let's think about how to approach this.
First, the problem says that any two numbers in the list must be checked. So for every pair of numbers in the list, I n... |
HumanEval/0 | 15 | none | base | humaneval | false | false | false | stop | 2,107 | 31,744 | ERROR: type should be string, got " https://drive.google.com/file/d/1nXa5W8L6aTQ6q0V5Z2pW5B1R1N1N1N1/view?usp=sharing\n\nOkay, I need to write a Python function called has_close_elements that checks if there are any two numbers in a list that are closer than a given threshold. Let's think about how to approach this.\n\nFirst, the problem says that any two numbers in the list should be checked to see if their absolute difference is less than the threshold. So, the straightforward way is to compare every possible pair of numbers in the list. But wait, comparing all pairs might be inefficient for large lists. However, since the problem doesn't specify any constraints on the list size, maybe the straightforward approach is acceptable here, especially since the examples given are small.\n\nSo, the steps would be:\n\n1. Iterate through each element in the list.\n2. For each element, compare it with every subsequent element (to avoid checking the same pair twice).\n3. Calculate the absolute difference between the two elements.\n4. If any difference is less than the threshold, return True immediately.\n5. If after checking all pairs no such pair is found, return False.\n\nWait, but comparing each element with every other element could be O(n^2) time complexity. For very large lists, this might be slow. But given that the problem doesn't mention handling large inputs, maybe this is acceptable. Let's proceed with this approach for now.\n\nLet me think about the example given. The first example is [1.0, 2.0, 3.0] with threshold 0.5. The differences are 1.0 and 1.0, both of which are larger than 0.5, so returns False. The second example has 2.8 and 2.0, which are 0.8 apart. Wait, no, 2.8 - 2.0 is 0.8, which is larger than 0.3. Wait, maybe I'm misunderstanding the example. Let me check the second example again.\n\nWait, the second example input is [1.0, 2.8, 3.0, 4.0, 5.0, 2.0], threshold 0.3. Let's see the pairs. The elements 2.8 and 2.0 are 0.8 apart. Hmm, but maybe there's another pair. Let's see: 2.8 and 3.0 are 0.2 apart. Oh right, 3.0 - 2.8 is 0.2, which is less than 0.3. So that's why the example returns True. So the function should find that pair.\n\nSo the approach of checking all pairs would work here. But for a list of n elements, this would be O(n^2) time. For very large n, this could be slow. But perhaps the problem expects this approach.\n\nAlternatively, if the list is sorted, we can check adjacent elements only. Because in a sorted list, the closest elements are adjacent. So sorting the list first would reduce the problem to O(n log n) time for sorting, and then O(n) time to check adjacent pairs. This is more efficient for large n.\n\nSo the steps would be:\n\n1. Sort the list.\n2. Iterate through the sorted list, comparing each element with the next one.\n3. If any adjacent pair has a difference less than the threshold, return True.\n4. If none are found, return False.\n\nThis approach is better for performance, especially with large lists. Let's see how that works with the examples.\n\nFirst example: [1.0, 2.0, 3.0] sorted is the same. Adjacent differences are 1.0 and 1.0, both >=0.5. So returns False. Correct.\n\nSecond example: sorted list would be [1.0, 2.0, 2.8, 3.0, 4.0, 5.0]. The adjacent differences are 1.0, 0.8, 0.2, 1.0, 1.0. The 0.2 is less than 0.3, so returns True. Correct.\n\nSo this approach works for the examples. And it's more efficient.\n\nSo the plan is to sort the list first, then check adjacent pairs.\n\nBut wait, the problem says \"any two numbers\" in the list, not just adjacent ones. But after sorting, the closest pair must be adjacent. So this approach is correct.\n\nSo the steps for the function:\n\n- Sort the input list.\n- Iterate from index 0 to len(numbers)-2:\n - Compare numbers[i] and numbers[i+1]\n - If the absolute difference is < threshold, return True\n- If all pairs are checked and none meet the condition, return False.\n\nBut wait, the original list could have duplicates. For example, if the list is [1.0, 1.0], the difference is 0.0, which is less than any positive threshold. So the function should return True. The sorted approach would catch that.\n\nSo this approach is correct.\n\nNow, how to implement this in Python.\n\nFirst, sort the list. Then loop through each pair.\n\nBut wait, the original list is a list of floats. So sorting is straightforward.\n\nSo code steps:\n\ndef has_close_elements(numbers: List[float], threshold: float) -> bool:\n # Sort the list\n sorted_numbers = sorted(numbers)\n # Iterate through adjacent pairs\n for i in range(len(sorted_numbers) - 1):\n if abs(sorted_numbers[i] - sorted_numbers[i+1]) < threshold:\n return True\n return False\n\nWait, but what if the list has only one element? Then len(sorted_numbers) -1 is 0, so the loop doesn't run, returns False. Which is correct, because there's no pair.\n\nWhat if the list is empty? Then the function returns False, which is correct.\n\nTesting the first example:\n\nnumbers = [1.0, 2.0, 3.0], threshold 0.5. Sorted is same. The differences are 1.0 and 1.0. So returns False. Correct.\n\nSecond example: [1.0, 2.8, 3.0, 4.0, 5.0, 2.0] sorted becomes [1.0, 2.0, 2.8, 3.0, 4.0, 5.0]. The adjacent differences are 1.0, 0.8, 0.2, 1.0, 1.0. The 0.2 is less than 0.3, so returns True. Correct.\n\nAnother test case: [1.0, 1.0], threshold 0.1. The difference is 0.0 < 0.1, returns True.\n\nAnother test case: [5.0, 5.0, 5.0], threshold 0.1. The adjacent differences are 0.0, so returns True.\n\nWhat about a list with two elements where the difference is exactly equal to the threshold? For example, [1.0, 2.0], threshold 1.0. The difference is 1.0, which is not less than the threshold, so returns False.\n\nSo the code seems to handle these cases.\n\nBut wait, the problem says \"closer to each other than given threshold\". So the condition is if the absolute difference is strictly less than the threshold. So the code uses <, which is correct.\n\nSo this code should work.\n\nBut let's check the sample inputs.\n\nSample 1:\n\nhas_close_elements([1.0, 2.0, 3.0], 0.5) → returns False. Correct.\n\nSample 2:\n\nhas_close_elements([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3) → returns True. Correct.\n\nSo the code works for the samples.\n\nTherefore, the correct implementation is to sort the list and check adjacent pairs.To solve this problem, we need to determine if there are any two numbers in a given list that are closer to each other than a specified threshold. The solution should efficiently check all possible pairs of numbers to find the closest pair, ensuring optimal performance even for larger lists.\n\n### Approach\nThe key insight here is that after sorting the list, the closest pair of numbers will always be adjacent. This allows us to reduce the problem from checking all possible pairs (O(n^2) complexity) to checking only adjacent pairs after sorting (O(n log n) complexity due to sorting, followed by O(n) complexity for checking adjacent pairs). This approach is both efficient and straightforward.\n\n### Solution Code\n```python\nfrom typing import List\n\n\ndef has_close_elements(numbers: List[float], threshold: float) -> bool:\n sorted_numbers = sorted(numbers)\n for i in range(len(sorted_numbers) - 1):\n if abs(sorted_numbers[i] - sorted_numbers[i + 1]) < threshold:\n return True\n return False\n```\n\n### Explanation\n1. **Sorting the List**: By sorting the list, we ensure that the closest numbers are adjacent. This step is crucial as it allows us to check only consecutive elements instead of all possible pairs.\n2. **Checking Adjacent Pairs**: After sorting, we iterate through the list and check the absolute difference between each consecutive pair of elements. If any pair has a difference less than the threshold, we immediately return `True`.\n3. **Efficiency**: Sorting the list takes O(n log n) time, and checking adjacent pairs takes O(n) time, making the overall complexity O(n log n), which is efficient for large lists.\n\nThis approach ensures that we efficiently determine the presence of any two numbers closer than the given threshold, leveraging sorting to minimize the number of comparisons needed." |
End of preview.
rollouts-olmo32b-screens
Model: allenai/Olmo-3-1125-32B (snapshot c2b61dae). Tokenizer: allenai/Olmo-3-1125-32B. Protocol: entropy screens: none + the model's top-20 beam nominees, first 30 MATH-train problems x 16 rollouts, budget 16,384, T 0.6, top-p 0.95, seed 20260819; code screens: MBPP beam nominees on HumanEval 164 x 16 (the 32B, on 2 x H200), budget 31,744, execution-graded copies included.
Rollouts generated on the CSAIL cluster for the reasoning-registers paper (Sophie Wang, MIT), uploaded 2026-09-13; sister repositories of the same organisation hold the cells run by Rulin Shao (see HF_MANIFEST.md there). Layout: each top-level folder is one local store with its date prefix dropped, keeping the store's own layout (rollouts_base_<arm>_p<start>.jsonl, one line per rollout with problem_key, rollout_index, arm, completion, correct, output_tokens, hit_token_cap; *_b2 = the second seeded batch; *_execgraded/ = the execution-graded copy for code). summaries/ holds the cell summaries the paper's tables read (scripts/eval/summarize_cell.py), tables/ the derived tables. Folders: entropy20_rlzero, entropy20_boxed, code_nominees, tables. Files are plain JSON Lines, not compressed.
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