Question

拥有list个dict个对象，其中50％将拥有密钥name，50％将拥有密钥full_name，我希望按字母顺序排序在任何一个键上。

某种类似的功能：

if 'name' in a:
    a_name = a['name']
elif 'full_name' in a:
    a_name = a['full_name']
# repeat with a b in the compare, sort based on value

这里有更有效的方式吗？

Answer 1

我认为它应该有效：

sorts = sorted(data, key= lambda x: x.get('name') or x.get('full_name'))

Answer 2

连接两个字符串，用空字符串替换缺少的字符串：

class PhoneResource(ModelResource):
    class Meta:
        queryset = Phone.objects.all()
        allowed_methods = ['get']
        resource_name = 'phone'

class CustomerResource(ModelResource):
    phone = fields.ManyToManyField(PhoneResource, "phone")

    class Meta:
        queryset = Customer.objects.all()
        allowed_methods = ['get', 'patch', 'put']
        resource_name = 'customer'
        authentication = Authentication()
        authorization = Authorization()

    def prepend_urls(self):
        return [
            url(r'^(?P<resource_name>%s)/(?P<pk>\w[\w/-]*)/phone%s$' %
                (self._meta.resource_name, trailing_slash()),
                self.wrap_view('get_customer_phone'), name='customer_phone'),
        ]

    def customer_phone(self, request, **kwargs):
        # My Question is what goes in this function 
        # I want to get only the phones for the given customer, and exclude other phones that does not belong to them

Answer 3

您是否尝试过基于lambda的功能方法：

testList = sorted(testList, key=lambda name: a_name['name'] if 'name'   in a_name else a_name['full_name'])

python：按一个键或另一个键排序dicts列表

3 个答案: