如何将长度较短的列表平均分配给另一个列表？

Question

我有两张单子。第一个名为location，其长度可以从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员。它的长度通常大于位置列表长度。如果不是第一个或最后一个审计员被分配到其他位置，我想将所有审计员平均分配到各个位置

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

下面的代码在大多数情况下运行良好，但当位置为28，审计员为17时，代码失败

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

我想要的输出是得到最有可能的甚至分割的列表，并且它必须在任何情况下都能工作，只要位置大于审核员

我的期望输出= “aone”， “aone” “atwo”， “atwo”， “athree”， “athree”， “阿福”， “阿福”， “五”， “五”， "asix",， "asix",， “阿塞文”， “阿塞文”， “好的”， “好的”， “阿宁”， “阿宁”， “阿滕”， “阿滕”， “aeleven”， “aeleven”， “atwelve”， “阿瑟滕”， "昨天",， “阿菲菲菲”， “Asixten”， “Aseventen”]

Answer 1

可以考虑使用在{{CD2>}中发现的^ {CD1>}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'])]

祝你好运