如何将列表值映射到字典中并将列表值与字典值进行比较？

Question

我们有病人名单，诊断和住院时间。我们还有一本包含诊断和平均住院时间的字典。生成一个输出列表，列出患者，以及患者的停留时间是否“太长”、“太短”、“恰到好处”

avg_los = {
    "Hemolytic jaundice and perinatal jaundice" : 2,
    "Medical examination/evaluation" : 3.2,
    "Liveborn" : 3.2,
    "Trauma to perineum and vulva" : 2.1,
    "Normal pregnancy and/or delivery" : 2,
    "Umbilical cord complication" : 2.1,
    "Forceps delivery" : 2.2,
    "Administrative/social admission" : 4.2,
    "Prolonged pregnancy" : 2.4,
    "Other complications of pregnancy" : 2.5
}

#List
patients = [
    ['Boal', 'Medical examination/evaluation', 1.1],
    ['Boal', 'Other complications of pregnancy', 3.3],
    ['Jones', 'Liveborn', 3.2],
    ['Ashbury', 'Forceps delivery', 2.0]
]

如何将列表中的第三个值与avg_los的相应字典值中的值进行比较？

例如：

Boal接受了1.1天的医疗检查/评估。如果我将它与医疗检查/评估的avg_los进行比较，我得到的值是3.2。3.2大于1.1，所以我想输出“太长”。如果字典值小于列表值，则输出“太小”

如何使用for循环在python中对此进行编码？

Answer 1

使用子列表中的第二个值来查找字典中的参考值。然后比较这两个值。对于患者索引patient_num ...

procedure   = patients[patient_num][1]
patient_los = patients[patient_num][2]
reference   = avg_los[procedure]
if patient_los > reference:
    print("Too long")

明白了?如果遇到问题，可以使用更多的print语句来跟踪程序正在执行的操作。一旦你有了这个想法，你就可以组合几行代码来缩短代码。

Answer 2

avg_los = {
    "Hemolytic jaundice and perinatal jaundice" : 2,
    "Medical examination/evaluation" : 3.2,
    "Liveborn" : 3.2,
    "Trauma to perineum and vulva" : 2.1,
    "Normal pregnancy and/or delivery" : 2,
    "Umbilical cord complication" : 2.1,
    "Forceps delivery" : 2.2,
    "Administrative/social admission" : 4.2,
    "Prolonged pregnancy" : 2.4,
    "Other complications of pregnancy" : 2.5
}

#List
patients = [
    ['Boal', 'Medical examination/evaluation', 1.1],
    ['Boal', 'Other complications of pregnancy', 3.3],
    ['Jones', 'Liveborn', 3.2],
    ['Ashbury', 'Forceps delivery', 2.0]
]

for patient in patients:
    diagnosis = patient[1]
    time_frame = avg_los.get(diagnosis)

    if patient[2] > time_frame :
        message = 'too long'
    elif patient[2] < time_frame :
        message = 'too short'
    elif patient[2] == time_frame :
        message = 'exact'
    print(message)

Answer 3

这将是你的问题的解决方案。

for patient in patients:
    diagnosis = patient[1]
    length_of_stay = patient[2]

    average_length_of_stay = avg_los.get(diagnosis)
    if average_length_of_stay is not None:
        if average_length_of_stay < length_of_stay:
            print("too long")
        elif average_length_of_stay == length_of_stay:
            print("just right")
        else:
            print("too short")
    else:
        print("Diagnosis is not found")