试图用 Python 为医生和医院创建匹配算法，但遇到了条件

Question

首先，我想指出我在python方面是一个新手，所以如果我的措辞或问题本身听起来很愚蠢，请多多包涵，我只是在努力学习。

我创建了两个字典，一个为医院分配排名的医生，另一个为医生分配排名的医院。 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'}}

医院到医生：

{'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，因为它用于创建由算法产生的 DataFrame 的索引：

['Doctor_5', 'Doctor_9', 'Doctor_8', 'Doctor_1', 'Doctor_4', 'Doctor_3', 'Doctor_6', 'Doctor_7', 'Doctor_10', 'Doctor_2']

现在我正在尝试创建一个算法，根据 Gale-Shapley 算法将医生分配到医院（无需了解细节）。

这是我到目前为止的想法，最后我将它呈现在一个 DataFrame 中，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：

如您所见，我将每位医生最喜欢的医院分配给该医生。但我需要有关我的职能条件的帮助。截至目前，我的结果中存在重复项：医生_7 和医生_8 都与医院_8 匹配，类似地，医院_7 与两个不同的医生匹配。但是，我想在我的函数中添加一个条件，在这种情况下检查医院最喜欢哪个医生，匹配那个医生，而让其他医生不匹配。之后，我想为未匹配的医生重新开始整个过程​​，在第 2 轮中产生新的结果。对于第二轮，已经匹配的医院应该有可能打破他们的匹配，因为医生接近他们，他们更喜欢他们的第一次匹配。

但是，在调整我的功能几个小时并探索谷歌和列表理解指南寻求帮助之后，我仍然没有设法找到解决方案。这就是为什么我转向stackoverflow寻求帮助的原因，也许你们中的一个人以前见过这个问题并且可以帮助我解决它。如果您需要上述信息以外的更多信息，请告诉我！

非常感谢您，

Answer 1

我将 First_round 更改为字典以使其更直接。我还建议进行一些其他结构更改，但之后会更多。一个解决方案是简单地输入 if 检查以查看是否已分配特定医院，然后如果他的分数更高则交换医生：

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)

现在对于结构更改，我认为您还应该将排名结构从字典范围从数字到医院/医生更改为只是一个有序的偏好列表或翻转它，因此关键是医院/医生优先级/分数是价值。这将使检查医生的分数更自然，例如：

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]

或者只是内联那些代码位。