尝试为 Python 中的医生和医院创建匹配算法，卡在条件下

Question

首先，我想指出，我在python中是一个菜鸟，所以如果我的措辞或问题本身听起来很愚蠢，请耐心等待，只是想学习。

我创建了两本词典，一本是给医院排名的医生，另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生成的 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种算法，根据 Gale-Shapley 算法（无需了解具体细节）将医生分配给医院。

这是我到目前为止的想法，最后我将它呈现在 DataFrame 中，因为我发现它更容易解释：

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，我正在将每个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：doctor_7 和 doctor_8 都与 hospital_8 匹配，类似地 hospital_7 与两个不同的医生匹配。但是，我想在我的功能中添加一个条件，在这种情况下，检查医院最喜欢哪个医生，匹配那个医生，让另一个医生不匹配。之后，我想为未匹配的医生重新开始整个过程，从而在第 2 轮中得出新的结果。对于第二轮，已经匹配的医院应该有可能打破他们的匹配，因为医生接近他们时他们更喜欢他们的第一场匹配。

然而，在调整我的函数几个小时并探索 google 和列出理解指南以寻求帮助之后，我仍然没有找到解决方案。这就是为什么我转向 Whosebug 寻求帮助，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。如果您还需要以上提供的信息，请告诉我！

非常感谢您，

Answer 1

我将 First_round 更改为字典，使其更直接一些。我还会建议一些其他的结构更改，但之后会更多。一个解决方案是简单地 pu in 一个 if 检查是否已经分配了特定的医院，然后如果他有更好的分数交换医生：

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，我认为您也应该将排名结构从一个范围从数字到 hospitals/doctors 的字典更改为只是一个有序的偏好列表或翻转它，所以关键是hospital/doctor 和 priority/score 是值。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]

或者简单地内联那些代码位。

尝试为 Python 中的医生和医院创建匹配算法，卡在条件下

Trying to create a matching algorithm for doctors and hospitals in Python, stuck on conditions

python

dictionary

list-comprehension

matching

conditional-statements