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python - 试图在条件不佳的情况下使用Python为医生和医院创建匹配算法

首先，我想指出的是，我在python方面是个新手，所以如果我的措辞或问题本身听起来很愚蠢，请容忍我，只是想学习。
我创建了两本字典，一本是给医院排名的医生，一本是给医生排名的医院。
医生到医院：

{'Doctor_5': {1.0: 'Hospital_7', 2.0: 'Hospital_6', 3.0: 'Hospital_8', 4.0: 'Hospital_5', 5.0: 'Hospital_9', 6.0: 'Hospital_4', 7.0: 'Hospital_10', 8.0: 'Hospital_3', 9.0: 'Hospital_2', 10.0: 'Hospital_1'}, 'Doctor_9': {1.0: 'Hospital_9', 2.0: 'Hospital_8', 3.0: 'Hospital_10', 4.0: 'Hospital_7', 5.0: 'Hospital_6', 6.0: 'Hospital_5', 7.0: 'Hospital_4', 8.0: 'Hospital_3', 9.0: 'Hospital_2', 10.0: 'Hospital_1'}, 'Doctor_8': {1.0: 'Hospital_8', 2.0: 'Hospital_9', 3.0: 'Hospital_10', 4.0: 'Hospital_7', 5.0: 'Hospital_6', 6.0: 'Hospital_5', 7.0: 'Hospital_4', 8.0: 'Hospital_3', 9.0: 'Hospital_2', 10.0: 'Hospital_1'}, 'Doctor_1': {1.0: 'Hospital_1', 2.0: 'Hospital_2', 3.0: 'Hospital_3', 4.0: 'Hospital_4', 5.0: 'Hospital_5', 6.0: 'Hospital_6', 7.0: 'Hospital_7', 8.0: 'Hospital_8', 9.0: 'Hospital_9', 10.0: 'Hospital_10'}, 'Doctor_4': {1.0: 'Hospital_5', 2.0: 'Hospital_6', 3.0: 'Hospital_4', 4.0: 'Hospital_7', 5.0: 'Hospital_3', 6.0: 'Hospital_2', 7.0: 'Hospital_1', 8.0: 'Hospital_8', 9.0: 'Hospital_9', 10.0: 'Hospital_10'}, 'Doctor_3': {1.0: 'Hospital_4', 2.0: 'Hospital_3', 3.0: 'Hospital_2', 4.0: 'Hospital_1', 5.0: 'Hospital_5', 6.0: 'Hospital_6', 7.0: 'Hospital_7', 8.0: 'Hospital_8', 9.0: 'Hospital_9', 10.0: 'Hospital_10'}, 'Doctor_6': {1.0: 'Hospital_7', 2.0: 'Hospital_8', 3.0: 'Hospital_6', 4.0: 'Hospital_9', 5.0: 'Hospital_5', 6.0: 'Hospital_10', 7.0: 'Hospital_4', 8.0: 'Hospital_3', 9.0: 'Hospital_2', 10.0: 'Hospital_1'}, 'Doctor_7': {1.0: 'Hospital_8', 2.0: 'Hospital_7', 3.0: 'Hospital_9', 4.0: 'Hospital_6', 5.0: 'Hospital_5', 6.0: 'Hospital_4', 7.0: 'Hospital_3', 8.0: 'Hospital_2', 9.0: 'Hospital_1'}, 'Doctor_10': {1.0: 'Hospital_10', 2.0: 'Hospital_9', 3.0: 'Hospital_8', 4.0: 'Hospital_7', 5.0: 'Hospital_6', 6.0: 'Hospital_5', 7.0: 'Hospital_4', 8.0: 'Hospital_3', 9.0: 'Hospital_2', 10.0: 'Hospital_1'}, 'Doctor_2': {1.0: 'Hospital_3', 2.0: 'Hospital_2', 3.0: 'Hospital_4', 4.0: 'Hospital_1', 5.0: 'Hospital_5', 6.0: 'Hospital_6', 7.0: 'Hospital_7', 8.0: 'Hospital_8', 9.0: 'Hospital_9', 10.0: 'Hospital_10'}}

医院对医生：
{'Hospital_2': {1.0: 'Doctor_1', 2.0: 'Doctor_2', 3.0: 'Doctor_3', 4.0: 'Doctor_4', 5.0: 'Doctor_5', 6.0: 'Doctor_6', 7.0: 'Doctor_7', 8.0: 'Doctor_8', 9.0: 'Doctor_9', 10.0: 'Doctor_10'}, 'Hospital_1': {1.0: 'Doctor_1', 2.0: 'Doctor_2', 3.0: 'Doctor_3', 4.0: 'Doctor_4', 5.0: 'Doctor_5', 6.0: 'Doctor_6', 7.0: 'Doctor_7', 8.0: 'Doctor_8', 9.0: 'Doctor_9', 10.0: 'Doctor_10'}, 'Hospital_8': {1.0: 'Doctor_8', 2.0: 'Doctor_9', 3.0: 'Doctor_7', 4.0: 'Doctor_6', 5.0: 'Doctor_10', 6.0: 'Doctor_4', 7.0: 'Doctor_3', 8.0: 'Doctor_2', 9.0: 'Doctor_1'}, 'Hospital_7': {1.0: 'Doctor_5', 2.0: 'Doctor_6', 3.0: 'Doctor_7', 4.0: 'Doctor_8', 5.0: 'Doctor_3', 6.0: 'Doctor_2', 7.0: 'Doctor_9', 8.0: 'Doctor_1', 9.0: 'Doctor_10'}, 'Hospital_10': {1.0: 'Doctor_10', 2.0: 'Doctor_9', 3.0: 'Doctor_8', 4.0: 'Doctor_7', 5.0: 'Doctor_6', 6.0: 'Doctor_5', 7.0: 'Doctor_4', 8.0: 'Doctor_3', 9.0: 'Doctor_2', 10.0: 'Doctor_1'}, 'Hospital_3': {1.0: 'Doctor_1', 2.0: 'Doctor_2', 3.0: 'Doctor_3', 4.0: 'Doctor_4', 5.0: 'Doctor_5', 6.0: 'Doctor_6', 7.0: 'Doctor_7', 8.0: 'Doctor_8', 9.0: 'Doctor_9', 10.0: 'Doctor_10'}, 'Hospital_9': {1.0: 'Doctor_9', 2.0: 'Doctor_10', 3.0: 'Doctor_8', 4.0: 'Doctor_7', 5.0: 'Doctor_6', 6.0: 'Doctor_5', 7.0: 'Doctor_4', 8.0: 'Doctor_3', 9.0: 'Doctor_2', 10.0: 'Doctor_1'}, 'Hospital_5': {1.0: 'Doctor_4', 2.0: 'Doctor_3', 3.0: 'Doctor_2', 4.0: 'Doctor_1', 5.0: 'Doctor_5', 6.0: 'Doctor_6', 7.0: 'Doctor_7', 8.0: 'Doctor_8', 9.0: 'Doctor_9', 10.0: 'Doctor_10'}, 'Hospital_4': {1.0: 'Doctor_3', 2.0: 'Doctor_2', 3.0: 'Doctor_4', 4.0: 'Doctor_1', 5.0: 'Doctor_5', 6.0: 'Doctor_6', 7.0: 'Doctor_7', 8.0: 'Doctor_8', 9.0: 'Doctor_9', 10.0: 'Doctor_10'}, 'Hospital_6': {1.0: 'Doctor_4', 2.0: 'Doctor_5', 3.0: 'Doctor_6', 4.0: 'Doctor_3', 5.0: 'Doctor_2', 6.0: 'Doctor_1', 7.0: 'Doctor_8', 8.0: 'Doctor_9', 9.0: 'Doctor_10'}}

我还创建了一个医生列表，用于创建由算法生成的数据帧的索引：
['Doctor_5', 'Doctor_9', 'Doctor_8', 'Doctor_1', 'Doctor_4', 'Doctor_3', 'Doctor_6', 'Doctor_7', 'Doctor_10', 'Doctor_2']

现在我试图创建一个基于Gale-Shapley算法（不需要知道细节）的医生分配算法。
这就是我到目前为止所提出的，我在最后以一个数据框的形式呈现它，因为我发现这更容易解释：
Matches = {}
Matches['Doctors'] = (doctors)
First_round = []
for Doctor_ in ranking_by_doctors:
    First_round.append(ranking_by_doctors[Doctor_].get(1.0))
Matches['First round'] = (First_round)
Matches
Matches_round_1 = pd.DataFrame.from_dict(Matches)
Matches_round_1.set_index('Doctors', inplace=True)
Matches_round_1

这是我的结果：python - 试图在条件不佳的情况下使用Python为医生和医院创建匹配算法-LMLPHP
如你所见，我把每个医生中最喜欢的医院指派给那个医生。但我需要帮助来了解我的工作条件。就目前而言，我的结果存在重复：DOCTROR7和DOCTORY8都与医院A8相匹配，类似的医院7与两个不同的医生相匹配。不过，我想在我的功能中添加一个条件，在这种情况下，医院最喜欢哪个医生，匹配那个医生，而让另一个医生不匹配。之后，我想重新开始整个过程，为无与伦比的医生，导致新的结果在第二轮。在第二轮比赛中，应该有可能已经匹配的医院会因为医生接近他们，而不是他们的第一场比赛而打破他们的匹配。
然而，在花了几个小时调整我的功能，浏览谷歌和列表理解指南寻求帮助之后，我仍然没有找到解决方案。这就是为什么我转向stackoverflow寻求帮助，也许你们中的一个以前见过这个问题，可以帮助我解决它。如果你需要更多的信息，请告诉我！
非常感谢你提前，

最佳答案

我把第一轮换成了字典，使它更直截了当一点。我也会建议一些其他的结构改变，但更多的是在那之后。一种解决方法是简单地在if检查中pu，看看是否指定了特定的医院，然后如果医生有更好的分数，就交换他：

def hospital_ranking_of_doctor(hospital, doctor):
    return next(v for k, v in ranking_by_hospital[hospital].items() if v == doctor)

Matches = {}
Matches['Doctors'] = (doctors)
First_round = {}

# We loop through all the doctors one by one
for Doctor_ in ranking_by_doctors:

    # Then we find which hospital the doctor prefers
    favored_hospital = ranking_by_doctors[Doctor_].get(1.0)

    # We test if that hospital has already had a doctor assigned
    if favored_hospital not in First_round:
        # If it has not, then we just assign the current doctor
        First_round[favored_hospital] = Doctor_
    else:
        # If the hosptial was already assigned a doctor we compare
        # that doctor with the new one and set the new one only if
        # the hospital prefers him.
        previously_assigned_doctor = First_round[favored_hospital]
        if hospital_ranking_of_doctor(favored_hospital, previously_assigned_doctor) > hospital_ranking_of_doctor(favored_hospital, Doctor_):
            First_round[favored_hospital] = Doctor_

print(First_round)

现在对于结构的改变，我认为你也应该改变排名结构，从一个从数字到医院/医生的字典，或者只是一个偏好的有序列表，或者把它翻过来，关键是医院/医生，优先级/分数是值。这将使检查医生的评分更自然，就像：
ranking_by_hospital['Hospital_2']['Doctor_2'] > ranking_by_hospital['Hospital_2']['Doctor_3']

因此您可以将排名函数更新为：
def hospital_ranking_of_doctor(hospital, doctor):
    return ranking_by_hospital[hospital][doctor]

或者直接将这些代码位内联。

10-07 19:04
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