In django.utils.tree.py: def _new_instance(cls, children=None, connector=None, negated=False): obj = Node(children, connector, negated) obj.__class__ = cls return obj _new_instance = classmethod(_new_instance) I don't know what classmethod does in this code sample. Can someone explain what it does and how to use it? classmethod is a descriptor, wrapping a func
在django.utils.tree.py中： def _new_instance(cls, children=None, connector=None, negated=False): obj = Node(children, connector, negated) obj.__class__ = cls return obj _new_instance = classmethod(_new_instance) 我不知道这个代码示例中使用了什么classmethod 。 有人可以解释它做了什么以及如何使用它？ classmethod是一个描述符，包装一个函数，并且你可以在一个类或者（等价
This question already has an answer here: What is the difference between @staticmethod and @classmethod in Python? 24 answers If both are working it means that inside make_attr you are not using self . Making it a regular non-static method only makes sense if the code could logically depend on the instance and only incidentally doesn't depend on it in the current implementation (but fo
这个问题在这里已经有了答案： Python中@staticmethod和@classmethod有什么区别？ 24个答案 如果两者都工作，这意味着make_attr里面没有使用self 。 使它成为一个常规的非静态方法只有在代码能够在逻辑上依赖于实例并且在当前实现中不依赖它的情况下才有意义（但是例如，它可以依赖于派生自该类的类中的实例）。 在功能方面，@staticmethod并不重要。 它的价值在于语义 - 你告诉自己或其他编码人员，即使这个函数属于
Possible Duplicate: What is the difference between @staticmethod and @classmethod in Python? I am learning OOP in python and came to know about these two methods It seems that the difference in terms of syntax is that class methods are implicitly passed the class they belong to as their first parameter class Circle: all_circles = [] # class variable @staticmethod def total_area():
可能重复： Python中@staticmethod和@classmethod有什么区别？ 我正在学习python中的OOP，并开始了解这两种方法 看起来，语法方面的差异在于类方法隐式地传递了它们所属的类作为它们的第一个参数 class Circle: all_circles = [] # class variable @staticmethod def total_area(): for c in Circle.all_circles: # hardcode class name # do somethig @classmethod def total_area(cls):
Possible Duplicate: What is the difference between @staticmethod and @classmethod in Python? I have a few questions about staticmethods in classes. I will start by giving an example. Example one: class Static: def __init__(self, first, last): self.first = first self.last = last self.age = randint(0, 50) def printName(self): return self.first + self.
可能重复： Python中@staticmethod和@classmethod有什么区别？ 我有几个关于类中staticmethods的问题。 我将首先举一个例子。 示例一： class Static: def __init__(self, first, last): self.first = first self.last = last self.age = randint(0, 50) def printName(self): return self.first + self.last @staticmethod def printInfo(): return "Hello %s,
This question already has an answer here: What is the difference between @staticmethod and @classmethod in Python? 24 answers There's little difference in your example, but suppose you created a subclass of Foo and called the create_new method on the subclass... class Bar(Foo): pass obj = Bar.create_new() ...then this base class would cause a new Bar object to be created... class
这个问题在这里已经有了答案： Python中@staticmethod和@classmethod有什么区别？ 24个答案 你的例子几乎没有什么区别，但是假设你创建了Foo的子类并且在子类上调用了create_new方法。 class Bar(Foo): pass obj = Bar.create_new() ...然后这个基类会导致创建一个新的Bar对象... class Foo: @classmethod def create_new(cls): return cls() ...而这个基类会导致创建一个新的Foo对象...... class
Does anyone know of a Python 2.7 compliant Perforce API build? On their FTP site they have only up to Python 2.6. It explains in the README how to build it for other versions of Python, however I was running into a lot of issues because I was using VS2010 (I even tried MinGW). Any help would be appreciated. The official P4Python build for Python 2.7 is in the works, but you can use the bin
有谁知道Python 2.7兼容的Perforce API构建？ 在他们的FTP站点上，他们只能使用Python 2.6。 它在自述文件中解释了如何为其他版本的Python构建它，但是由于我使用VS2010（我甚至尝试过MinGW），我遇到了很多问题。 任何帮助，将不胜感激。 Python 2.7的官方P4Python构建正在开发中，但您可以使用来自// guest / sven_erik_knop / P4Pythonlib / bin / ...中的Perforce Public Server（public.perforce.com：1666）的二进
CPython stores unicode strings as either utf-16 or utf-32 internally depending on compile options. In utf-16 builds of Python string slicing, iteration, and len seem to work on code units, not code points, so that multibyte characters behave strangely. Eg, on CPython 2.6 with sys.maxunicode = 65535: >>> char = u'U0001D49E' >>> len(char) 2 >>> char[0:1] u'uu835' >
根据编译选项，CPython将Unicode字符串存储为内部的utf-16或utf-32。 在utf-16版本的Python字符串切片中，迭代和len似乎在代码单元上工作，而不是代码点，所以多字节字符的行为很奇怪。 例如，在CPython 2.6上使用sys.maxunicode = 65535： >>> char = u'U0001D49E' >>> len(char) 2 >>> char[0:1] u'uu835' >>> char[1:2] u'udc9e' 根据Python文档， sys.maxunicode是“为Unicode字符
Is there any reason for a class declaration to inherit from object ? I just found some code that does this and I can't find a good reason why. class MyClass(object): # class code follows... Is there any reason for a class declaration to inherit from object ? tl;dr: In Python 3, apart from compatibility between Python 2 and 3, no reason. In Python 2, many reasons. Python 2.x story
是否有任何理由让类声明继承object ？ 我刚刚发现了一些代码，我找不到一个很好的理由。 class MyClass(object): # class code follows... 是否有任何理由让类声明继承object ？ tl; dr：在Python 3中，除了Python 2和3之间的兼容性外，没有理由。 在Python 2中，有很多原因。 Python 2.x故事： 在Python 2.x中（从2.2开始），根据是否存在object作为基类，有两种类型的类型： “经典”风格类：它们没有object作
What is this language construct called? In Python I can say: def a(b,c): return b+c a(*[4,5]) and get 9. Likewise in Ruby: def a(b,c) b+c end a(*[4,5]) What is this called, when one passes a single array to a function which otherwise requires multiple arguments? What is the name of the * operator? What other languages support this cool feature? The Python docs call this Unpacking Argu
这种语言结构称为什么？ 在Python中，我可以说： def a(b,c): return b+c a(*[4,5]) 并得到9.同样在Ruby中： def a(b,c) b+c end a(*[4,5]) 当一个数组传递给一个需要多个参数的函数时，这叫做什么？ *运算符的名称是什么？ 其他什么语言支持这个很酷的功能？ Python文档称这个解包参数列表。 这是一个非常方便的功能。 在Python中，您还可以使用双星号（**）将字典（散列）解包到关键字参数中。 他们也反过来工
This question already has an answer here: What does ** (double star/asterisk) and * (star/asterisk) do for parameters? 15 answers Now that you've edited your question, it's clear what's happening. You're method has a the decorator @classmethod. According to the docs: A class method receives the class as implicit first argument, just like an instance method receives the in
这个问题在这里已经有了答案： **（双星/星号）和*（星号/星号）对参数做什么？ 15个答案 现在你已经编辑了你的问题，很清楚发生了什么事情。 你的方法有一个装饰器@classmethod。 根据文档： 类方法将类作为隐式第一个参数接收，就像实例方法接收实例一样。 为了声明一个类方法，使用这个习惯用法： class C: @classmethod def f(cls, arg1, arg2, ...): ... 这（有点）等同于： class MyClass: def f(