Question

正如您所说的，这是一种作业或“家庭作业”问题，我仅在开始时予以澄清。我是Python的新手，我真的对这种问题感到困惑。

指出的问题是：护理医院希望了解最多患者数所拜访的医学专业。假设将患者的患者ID以及患者所拜访的医学专业知识存储在列表中。医学专业的详细信息存储在字典中，如下所示： { “ P”：“儿科”， “ O”：“骨科”， “ E”：“ ENT }

编写一个函数以查找最多患者数所访问的医学专业，并返回该专业的名称。

我尝试的代码：

// I think UNIX open() will also work.
FILE *file = fopen("your_image.png", "r");

样本输入：[101，P，102，O，302，P，305，P]

预期产量：儿科

我得到的输出是：P

Answer 1

这应该做到：

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，您需要按专长计算每个患者：

# 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

在此之后，counts = {'E': 2, 'P': 3}因为有3位患者访问了'P'，有2位患者访问了'E'。然后将这些值用作max中的键：

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

这将向'P'返回最常访问的专长，然后在'P'字典中返回medical_speciality的值，

return medical_speciality[most_visited_speciality]

在这种情况下：'Pediatrics'。

进一步

max的文档。
有关get dict方法的文档。

Answer 2

如果您受到要求的限制，“将患者的病历和患者所拜访的医学专业知识存储在列表中” ，请使用以下命令优化和统一的方法：

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))

输出：

Pediatrics

P.S。养成“良好做法”的习惯。

Answer 3

你快到了。

在您的for循环中，words包含字符串P或E。现在，您只需要使用它来调用字典中的键：

示例：当word为'P'时，要获取值，您将使用medical_speciality['P']来获取值Pediatrics。因此，我们将其包含在您的函数中。

接下来，max在您在这里考虑时不起作用。您需要一种方法来计算“ P”或“ E”出现的次数，然后您真的只想要该最大值。

我也会移动您的部分

speciality=max(speciality_list)
return speciality`

在for循环之外，您需要该完整列表的最大值，因为它是每次迭代后当前正在执行的max和return所在的地方。

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

Answer 4

def max_visited_speciality(patient_medical_speciality_list,medical_speciality):
    # write your logic here
    a=len(patient_medical_speciality_list)
    b=patient_medical_speciality_list[1:a:2]
    c=[]
    d=[]
    for key in medical_speciality:
        c.append(key)
    #print(c)
    for i in c:
        d.append(b.count(i))
    #print(d)
    x=d.index(max(d))
    #print(x)
    return medical_speciality[c[x]]



#return speciality


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)

如何同时从列表和字典生成输出，并将它们的结果输出与字典值匹配？

4 个答案: