如何从列表和字典中生成输出，并将它们的结果输出与字典值匹配？

Question

这是一个作业或“家庭作业”的问题，就像你们说的那样，我只是在一开始就澄清了。我是Python新手，我真的对这类问题感到困惑。

问题是:护理医院想知道病人人数最多的医疗专业。假设病人的身份与病人所访问的医学专业一起存储在一个列表中。医学专业的详细信息存储在字典中如下：{“P”：“儿科”，“O”：“骨科”，"E":"ENT }

编写一个函数，找出尽可能多的病人访问的医学专业，并返回专科的名称。

我尝试的代码：

def max_visited_speciality(patient_medical_speciality_list,medical_speciality):
    speciality_list=[]
    for words in patient_medical_speciality_list:
        if words in medical_speciality:
                speciality_list.append(words)
                speciality=max(speciality_list)
                return speciality

#provide different values in the list and test your program
patient_medical_speciality_list=[301,'P',302, 'P' ,305, 'P' ,401, 'E' ,656, 'E']
medical_speciality={"P":"Pediatrics","O":"Orthopedics","E":"ENT"}
speciality = max_visited_speciality(patient_medical_speciality_list,medical_speciality)
print(speciality)

样本输入: 101，P,102，O,302，P,305，P

预期产出:儿科

我得到的输出:P

Answer 1

这应该可以做到：

def max_visited_speciality(patient_medical_speciality_list, medical_speciality):

    # count each speciality patients
    counts = {}
    for _, speciality in zip(patient_medical_speciality_list[::2], patient_medical_speciality_list[1::2]):
        counts[speciality] = counts.get(speciality, 0) + 1

    # get most visited speciality by count of it's patients
    most_visited_speciality = max(medical_speciality, key=lambda e: counts.get(e, 0))

    # return value of most visited speciality
    return medical_speciality[most_visited_speciality]


# provide different values in the list and test your program
patient_medical_speciality_list = [301, 'P', 302, 'P', 305, 'P', 401, 'E', 656, 'E']
medical_speciality = {"P": "Pediatrics", "O": "Orthopedics", "E": "ENT"}
speciality = max_visited_speciality(patient_medical_speciality_list, medical_speciality)
print(speciality)

输出

Pediatrics

首先，你需要按专业来统计每一个病人：

# count each speciality patients
    counts = {}
    for _, speciality in zip(patient_medical_speciality_list[::2], patient_medical_speciality_list[1::2]):
        counts[speciality] = counts.get(speciality, 0) + 1

在counts = {'E': 2, 'P': 3}之后，因为有3例患者出现了'P‘，2例出现了'E’。然后在max中使用这些值作为键：

most_visited_speciality = max(medical_speciality, key=lambda e: counts.get(e, 0))

这将返回访问次数最多的专业'P'，然后在medical_speciality字典中返回'P'的值，

return medical_speciality[most_visited_speciality]

在本例中：'Pediatrics'。

进一步

1. 最大值的文档。
2. dict的到达方法的文档。

Answer 2

如果您受限于“将病人的id与病人所访问的医疗专业一起存储在list__中”，请使用以下优化和统一的方法：

from collections import Counter


class MedicalSpecialityError(Exception):
    pass


medical_speciality_map = {"P": "Pediatrics", "O": "Orthopedics", "E": "ENT"}
patient_medical_speciality_list = [301, 'P', 302, 'P', 305, 'P', 401, 'E', 656, 'E']


def max_visited_speciality(patient_medical_speciality_list: list):
    counts = Counter(s for s in patient_medical_speciality_list if str(s).isalpha())
    try:
        med_spec = medical_speciality_map[counts.most_common()[0][0]]
    except IndexError:
        raise MedicalSpecialityError('Bad "patient_medical_speciality_list"')
    except KeyError:
        raise MedicalSpecialityError('Unknown medical speciality key')

    return med_spec


print(max_visited_speciality(patient_medical_speciality_list))

产出：

Pediatrics

养成“良好实践”的习惯。

Answer 3

你快到了。

在for循环中，words保存字符串P或E。现在，您只需要使用它调用字典中的键：

示例:当word为'P‘时，要获得值，您可以使用medical_speciality['P']获取值Pediatrics。所以我们就把它包括在你的功能里。

接下来，max不像您在这里想的那样工作。你需要一种方法来计算“P”或“E”出现的次数，然后你真的只想要那个最大值。

我也会把你的那部分

speciality=max(speciality_list)
return speciality`

在for循环之外，当您想要整个列表的最大值时，当它当前在每次迭代之后执行max和return时，这是不需要的。

def max_visited_speciality(patient_medical_speciality_list,medical_speciality):
    speciality_list=[]
    for words in patient_medical_speciality_list:
        if words in medical_speciality:
                speciality_list.append(words)

    counts = dict(map(lambda x  : (x , speciality_list.count(x)) , speciality_list))
    most_visited_speciality = max(counts, key=lambda e: counts.get(e, 0))
    return medical_speciality[most_visited_speciality]

#provide different values in the list and test your program
patient_medical_speciality_list=[301,'P',302, 'P' ,305, 'P' ,401, 'E' ,656, 'E']
medical_speciality={"P":"Pediatrics","O":"Orthopedics","E":"ENT"}
speciality = max_visited_speciality(patient_medical_speciality_list,medical_speciality)
print(speciality)

输出：

>>> print(speciality)
>>> Pediatrics