如何将长度较短的列表平均分配到另一个列表？

Question

我有两份名单。第一个名为location，长度可以从1到无穷大不等。

location = ['loc1,'loc2','loc3','loc4,'loc5','loc6','loc7,'loc8','loc9','loc10,'loc11','loc12','loc13,'loc14','loc15','loc16,'loc17','loc18','loc19,'loc20','loc21','loc22,'loc23','loc24','loc25','loc26','loc27',] <- Please note that length could from 1 to infinite number

第二个列表名为审计师。它的长度通常大于位置列表长度。如果不是第一位或最后一位审核员，我想要将所有审计师平均分配到其他地点。

auditor = ['aone','atwo','athree','afour','afive','asix','aseven','aeight','anine','aten','aeleven','atwelve','athirteen','afourteen','afitheen','asixteen,'aseventeen''] <- Please note that length could from 1 to infinite number

下面的代码对大多数情况都很有用，但是当位置为28，审计师为17时，代码就失败了。

df2['location'] = location
df2['auditor'] = [auditor[int(df2)] for df2 in np.arange(0, len(auditor), (len(auditor)/len(df2)))]

我想要的输出是获得尽可能多的甚至被划分的列表，并且它必须在每一种情况下工作，只要位置大于审计师。

My_desired_output = 'aone'，'aone‘'atwo'，'atwo'，'aten'，'afour'，’a5‘，’a5‘，'asix'，'aseven'，'aseven'，’a8‘，'anine'，'aten'，’a11‘，’a11‘，'atwelve'，’a13‘，’a13‘，'afour'，‘A15’，‘16’，‘17’]

Answer 1

您可以考虑使用在chunked中找到的more-itertool函数

from more_itertools import chunked
from math import ceil
from typing import List, Any, Tuple, Iterator


def distribute_evenly(items: List[Any], cells: List[Any]) -> Iterator[Tuple[Any, List[Any]]]:
    if len(items) <= len(cells):
        item_chunks = [[item] for item in items] + [[]] * (len(cells) - len(items))
    else:
        chunk_size = int(ceil(len(items) / len(cells)))
        item_chunks = chunked(auditor, chunk_size)
    
    return zip(cells, item_chunks)


location = ["loc1", "loc2", "loc3"]
auditor = ["aud1", "aud2", "aud3", "aud4", "aud5", "aud6", "aud7", "aud8"]


auditor_per_location = list(distribute_evenly(auditor, location))

# auditor_per_location is now [('loc1', ['aud1', 'aud2', 'aud3']), ('loc2', ['aud4', 'aud5', 'aud6']), ('loc3', ['aud7', 'aud8'])]

祝好运!