python - python 中用于实现 37% 规则的字典

Question

我正在尝试布莱恩·克里斯蒂安 (Brian Christian) 所著的《生活算法》一书中著名的 37% 规则。

37% 规则基本上是说，当您需要在有限的时间内筛选一系列选项时 - 无论是工作候选人、新公寓还是潜在的浪漫伴侣 - 做出决定的最佳时机是您已经看过其中 37% 的选项。

在选择过程中，您将收集足够的信息来做出明智的决定，但您不会浪费太多时间寻找不必要的选项。在 37% 的分数上，您处于一个很好的位置，可以从中挑选出最好的。

证明这一理论的常见思想实验是由非电脑数学专家在 20 世纪 60 年代开发的，称为“秘书问题”。

该计划正在运行，但我想开始考虑在 37% 的候选人之后选择候选人。由于我使用的是字典，因此我无法访问指定数量的候选者之后的元素。我怎样才能做到这一点？

import matplotlib.pyplot as plt
# for visualising scores
def initiate(candidates):
print("Total candidates are : ",len(candidates))
lookup_num=int(len(candidates) *0.37)
#finds 37% of the candidates
average=lookup(candidates)
# returns average of lookup phase
chosen=select_cad(candidates,lookup_num,average)
# selects a candidate based on lookUp average
print("The chosen candidate is : {} ".format(chosen))

def lookup(candidates):
    average_score=0
    for cands,score in candidates.items():
        average_score+=score
    average_score=int(average_score/len(candidates))
    print("The average score in lookup is : ",average_score)
    #return the average score to average local variable
    return average_score



def select_cad(candidates,lookup_num,average):
    for cands,score in candidates.items():
        if(score>average):
            return cands
        else:
            continue
        print("Something went wrong!")
        quit


candidates={"Husain":85, "Chirag":94 ,"Asim":70,"Ankit":65 ,"Saiteja":65 ,"Absar":75 ,"Premraj":70 ,"Sagar":75 ,"Himani":75 ,"Parth":76 ,"Sumedha":70 ,"Revati":65 ,"Sageer":65 ,"Noorjahan":60 ,"Muzammil":65 ,"Shifa":56 , "Dipti":65 , "Dheeraj":70 }
initiate(candidates)

plt.bar(range(len(candidates)), list(candidates.values()), align='center', color='green')
plt.xticks(range(len(candidates)), list(candidates.keys()))
plt.show()

即使在选择阶段，如何才能更灵活地更新平均分数？

Answer 1

刚刚阅读了这个“37% 规则”，所以我希望我理解正确。我会实现类似的东西:

import random
def rule_of_37(candidates):
    # first I date random 37% of the candidates
    who_i_date = random.sample(list(candidates), int(len(candidates)*.37))
    print("I meet those 37% of the candidates", who_i_date)
    # then I calculate their average score
    average = sum(candidates[name] for name in who_i_date) / len(who_i_date)
    print("The average score was", average)
    # then I settle with the next person that has a score higher than the average (obviously I cannot re-date candidates)
    # hopefully there is still someone with an higher score than average...
    try:
        who_i_marry = next(name 
                           for name, score in candidates.items()
                           if name not in who_i_date
                           and score > average)
        print("I end up with", who_i_marry, "who has a score of", candidates[who_i_marry])
    except StopIteration:
        print("I end up all alone, there was nobody left with an higher score than", average, "...")

candidates={"Husain":85, "Chirag":94 ,"Asim":70,"Ankit":65 ,"Saiteja":65 ,"Absar":75 ,"Premraj":70 ,"Sagar":75 ,"Himani":75 ,"Parth":76 ,"Sumedha":70 ,"Revati":65 ,"Sageer":65 ,"Noorjahan":60 ,"Muzammil":65 ,"Shifa":56 , "Dipti":65 , "Dheeraj":70 }

rule_of_37(candidates)

示例执行(您的示例可能会有所不同，因为前 37% 的候选者是随机挑选的):

I meet those 37% of the candidates ['Dipti', 'Chirag', 'Revati', 'Sumedha', 'Dhe eraj', 'Muzammil']

The average score was 71.5

I end up with Husain who has a score of 85

如果您想自己选择第一批候选人而不是依靠随机，您可以简单地将 who_i_date 替换为您预先选择的列表:

who_i_date = ['Husain', 'Chirag', 'Asim', 'Ankit', 'Saiteja', 'Absar']

但是其他 63% 将被任意排序，因此您可能不会总是选择相同的(除非您使用默认情况下保持字典顺序的 Python 3.6+)。如果您想按顺序确定剩余 63% 的日期，则必须迭代候选人姓名列表，而不是遍历将姓名映射到分数的字典。我把这个留给你。