Question

我有一个if elif else块，它只为if语句生成输出，即使它应该返回其他条件。即使我改变条件的顺序，它总是返回第一个。我的代码如下所示：

age, education, stability ....是我定义的函数，我将它们添加到feat_list

中

class operable:
def __init__(self, f):
    self.f = f
def __call__(self, x):
    return self.f(x)

def op_to_function_op(op):
    def function_op(self, operand):
        def f(x):
            return op(self(x), operand(x))
        return operable(f)
    return function_op

for name, op in [(name, getattr(operator, name)) for name in dir(operator) 
    if "__" in name]:
    try:
        op(1,2)
    except TypeError:
        pass
    else:
        setattr(operable, name, op_to_function_op(op))

@operable
def age(ModelAdmin, request, queryset):
    age = 0
    if customer.Age == '60 +':
        age = 0
    elif customer.Age == '36 - 59':
        age = 1
    else:
        age = 2
    return age

def education(ModelAdmin, request, queryset):
    education = 0
    if customer.Education == 'Highschool and below':
        education = 0
    else:
        education = 1
    return education

def employment(ModelAdmin, request, queryset):
    employment = 0
    if customer.Employment == 'Student':
        employment = 0
    elif customer.Employment == 'Contract':
        employment = 1
    else:
        employment = 2
    return employment

def stability(ModelAdmin, request, queryset):
    stability = 0
    if customer.Employer_Stability == 'Unstable':
        stability = 0
    else:
        stability = 1
    return stability

def residential(ModelAdmin, request, queryset):
    residential = 0
    if customer.Residential_Status == 'Rented':
        residential = 0
    else:
        residential = 1
    return residential

def salary(ModelAdmin, request, queryset):
    salary = 0
    if customer.Salary <= 1000:
        salary = 0
    elif 1000 < customer.Salary <= 10001:
        salary = 1
    else:
        salary = 2
    return salary

def loyalty(ModelAdmin, request, queryset):
    loyalty = 0
    if customer.Customer_Loyalty <= 2:
        loyalty = 0
    else:
        loyalty = 1
   return loyalty

def balance(ModelAdmin, request, queryset):
    balance = 0
    if customer.Balance <= 2500:
        balance = 0
    elif 2500 < customer.Balance <= 10001:
        balance = 1
    else:
        balance = 2
    return balance

def feat_list():
    total = age + education + employment + stability + residential + salary 
        + loyalty + balance
return total

for customer in queryset:

    if feat_list() < 12:
        customer.Service_Level = Service.objects.get(service_name = 'Silver 
            Package')
        silver_customers.append(customer.Name)

    elif 11 < feat_list() <= 15:
        customer.Service_Level = Service.objects.get(service_name = 'Gold 
            Package')
        gold_customers.append(customer.Name)

    else:
         customer.Service_Level = Service.object.get(service_name = 
            "Platinum Package")
         platinum_customers.append(customer.name)
   customer.save()

如何更改这些语句以使其不使用逻辑？

Answer 1

如果你说feat_list() < 12 dand值是11.5，那么它是真的，并执行firt if，但下一个条件也是真的11 < feat_list() <= 15，因为11.5，是11到15之间。我的观点是......你在第一个条件中的第一个条件是，＃34;包含＆＃34;在elif的第二个条件中：

12介于11和15之间，因为11 < 12 <= 15，所以，第一个条件是错误的。

Answer 2

仅当if不为真时才会评估elif。

因此，如果您想检查11 < feat_list() <= 15:即使feat_list() < 12:为真，请使用if代替elif

Answer 3

在您的示例中，Silver和Gold包之间的比较范围重叠（11-12）。此外，您可能不希望白金套餐成为＆＃34;默认＆＃34;一个（没有检查，只要它是最高级别）。如果用以下样式（伪代码）重写比较，可能会更清楚代码在做什么：

flist = feat_list()
# You can check the returned value here, ie. print(flist), and the value
# won't change before you're finished with the comparison.

if flist > 15:
    #...Platinum processing...
elif flist > 11:
    #...Gold processing...
else:
    #...Silver processing...

python如果elif else只返回第一个输出而不使用逻辑

3 个答案: