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BigCodeBench / BigCodeBench v0.1.0_hf 22951ecf-9652-56bc-b5c5-e04916ddc8b4

Problem

Answer published by the source. Consult the official source to check your work against its answer.

complete prompt

from random import shuffle
import pandas as pd
import numpy as np

# Constants



def task_func(l, n_groups = 5):
    """
    Given a list `l`, this function shuffles the list, constructs a dataframe using the shuffled list,
    and then for each row in the dataframe, moves the first n_groups elements to the end of the same row.

    Parameters:
    - l (list): A list of elements.
    - n_groups (int): number of groups. Default value is 5.

    Returns:
    - DataFrame: A modified DataFrame constructed from the shuffled list.

    Requirements:
    - pandas
    - numpy
    - random

    Example:
    >>> df = task_func(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'])
    >>> df.shape == (5, 10)
    True
    >>> set(df.iloc[0]) == set(['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'])
    True
    """

instruct prompt

Given a list `l`, this function shuffles the list, constructs a dataframe using the shuffled list, and then for each row in the dataframe, moves the first n_groups elements to the end of the same row.
The function should output with:
    DataFrame: A modified DataFrame constructed from the shuffled list.
You should write self-contained code starting with:

Code

from random import shuffle
import pandas as pd
import numpy as np
# Constants
def task_func(l, n_groups = 5):

code prompt

Code

from random import shuffle
import pandas as pd
import numpy as np
# Constants
def task_func(l, n_groups = 5):

entry point

task_func

libs

  • pandas
  • numpy
  • random

Discussion

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Source and history

task_id
22951ecf-9652-56bc-b5c5-e04916ddc8b4
task_key
default--v0~2e1~2e0~5fhf--22951ecf-9652-56bc-b5c5-e04916ddc8b4
task_revision_id
1
config / split
default / v0.1.0_hf

Official source

initial import