Original Question I am currently engaged in teaching my brother to program. He is a total beginner, but very smart. (And he actually wants to learn). I've noticed that some of our sessions have gotten bogged down in minor details, and I don't feel I've been very organized. (But the answers to this post have helped a lot.) What can I do better to teach him effectively? Is ther
原始问题 我目前正在教我的兄弟编程。 他是一个初学者，但非常聪明。 （他实际上想学习）。 我注意到，我们的一些会议已经陷入了一些小细节，我不觉得自己一直很有组织。 （但这篇文章的答案有很大帮助。） 我能更好地教他有效地做什么？ 有没有一种逻辑顺序可以用来贯穿概念概念？ 到了晚些时候我应该避免复杂吗？ 我们正在使用的语言是Python，但欢迎任何语言的建议。 如何帮助 如果你有好的，请在你的回答中
Python中的抽象类和接口有什么区别？ What you'll see sometimes is the following: class Abstract1( object ): """Some description that tells you it's abstract, often listing the methods you're expected to supply.""" def aMethod( self ): raise NotImplementedError( "Should have implemented this" ) Because Python doesn't have (and doesn't need) a formal Interface contr
Python中的抽象类和接口有什么区别？ 有时你会看到以下内容： class Abstract1( object ): """Some description that tells you it's abstract, often listing the methods you're expected to supply.""" def aMethod( self ): raise NotImplementedError( "Should have implemented this" ) 因为Python没有（也不需要）正式的Interface接口，所以抽象和接口之间的Java风格的区别并不存在。 如果有人经
I am curious about why we need @staticmethod decorator to declare method as static , actually I was reading about static method in python, and I came to know that static method can be callable without instantiating class. So I tried two examples below, but both does same. so why we need @Staticmethoad ? class StatMethoad(): def stat(): print("without Decorator") class StatMethaod_with_d
我很好奇为什么我们需要@staticmethod装饰器将方法声明为static ，实际上我正在阅读关于python中的静态方法，并且我开始知道静态方法可以在不实例化类的情况下进行调用。 所以我尝试了两个例子，但都是一样的。 那么为什么我们需要@Staticmethoad ？ class StatMethoad(): def stat(): print("without Decorator") class StatMethaod_with_decorator(): @staticmethod def stat(): print("With Decorator")
The question just about says it. I have an abstract class that calls a staticmethod in a helper function, and I want subclasses to simply define the staticmethod and run with it. Maybe I could use something along the lines of getattr? Should I use a @classmethod instead? Something like this: class A(object): def func(self): self.f() class B(A): @staticmethod def f():
这个问题就是这样说的。 我有一个抽象类在辅助函数中调用静态方法，我希望子类只需定义静态方法并使用它运行。 也许我可以沿着getattr的方式使用某些东西？ 我应该使用@classmethod吗？ 像这样的东西： class A(object): def func(self): self.f() class B(A): @staticmethod def f(): print "I'm B" 测试： >>> a=x.A() >>> a.func() Traceback (most recent call last
I'm looking to call a method in python from a different class like so: class foo(): def bar(name): return 'hello %s' % name def hello(name): a = foo().bar(name) return a Where hello('world') would return 'Hello World'. I'm aware I've done something wrong here, does anyone know what it is? I think it might be the way I'm handling the classes
我正在寻找从不同的类调用python的方法，如下所示： class foo(): def bar(name): return 'hello %s' % name def hello(name): a = foo().bar(name) return a 你好（'世界'）会返回'Hello World'。 我知道我在这里做错了什么，有人知道它是什么吗？ 我认为这可能是我处理课程的方式，但我还没有弄明白。 在Python中，非静态方法明确地将self作为第一个参数。 foo.bar()或者需要是
For lemmatization spacy has a lists of words: adjectives, adverbs, verbs... and also lists for exceptions: adverbs_irreg... for the regular ones there is a set of rules Let's take as example the word "wider" As it is an adjective the rule for lemmatization should be take from this list: ADJECTIVE_RULES = [ ["er", ""], ["est", ""], ["er", "e"], ["est", "e"] ] A
为了词法化，spacy有一个单词列表：形容词，副词，动词......还有例外情况列表：adverbs_irreg ...对于常规情况，有一组规则 我们以“更宽”这个词为例， 因为它是一个形容词，所以应该从这个列表中获取词形化的规则： ADJECTIVE_RULES = [ ["er", ""], ["est", ""], ["er", "e"], ["est", "e"] ] 据我了解，这个过程将是这样的： 1）获取单词的POS标签以知道它是否是名词，动词... 2）如果单词在不规
I want to access a static variable from a static method: #!/usr/bin/env python class Messenger: name = "world" @staticmethod def get_msg(grrrr): return "hello " + grrrr.name print Messenger.get_msg(Messenger) How to do it without passing grrrr to a method? Is this the true OOP?.. Anything like name or self.name seems not working: NameError: global name 'name' is not def
我想从一个静态方法访问一个静态变量： #!/usr/bin/env python class Messenger: name = "world" @staticmethod def get_msg(grrrr): return "hello " + grrrr.name print Messenger.get_msg(Messenger) 如何做到这一点，而不grrrr传递给方法？ 这是真正的面向对象吗？ 任何像name或self.name似乎不工作： NameError: global name 'name' is not defined 和 NameError: global name 'self' is not d
Maybe it's just late, but I cannot figure out why this isn't working. When I have a post_save signal call a generic function, it works, but when I have a post_save signal call a method from a model, nothing happens. Here is code that works: class Revision(models.Model): # Model junk... def send_email(sender, instance, created, **kwargs): if created: print "DO STUFF" s
也许这只是晚了，但我不明白为什么这不起作用。 当我有一个post_save信号调用一个通用函数时，它可以工作，但是当我有一个post_save信号从模型中调用一个方法时，什么都不会发生。 这是可以工作的代码： class Revision(models.Model): # Model junk... def send_email(sender, instance, created, **kwargs): if created: print "DO STUFF" signals.post_save.connect(send_email, sender=Revision) 但是
This is what my code looks like class InviteManager(): ALREADY_INVITED_MESSAGE = "You are already on our invite list" INVITE_MESSAGE = "Thank you! we will be in touch soon" @staticmethod @missing_input_not_allowed def invite(email): try: db.session.add(Invite(email)) db.session.commit() except IntegrityError: return ALREADY
这是我的代码看起来像 class InviteManager(): ALREADY_INVITED_MESSAGE = "You are already on our invite list" INVITE_MESSAGE = "Thank you! we will be in touch soon" @staticmethod @missing_input_not_allowed def invite(email): try: db.session.add(Invite(email)) db.session.commit() except IntegrityError: return ALREADY_INVITED_ME
In Python, if some methods of a class need a helper function, but the helper function itself doesn't use anything in the class, should I put the helper function inside or outside the class? I tried putting it inside but PyLint was complaining that this function could have been put outside. @Karl: The class is a software upgrader and the helper function creates a new folder if the folder
在Python中，如果类的某些方法需要辅助函数，但辅助函数本身不使用类中的任何内容，我应该将辅助函数放入类还是外部？ 我试图把它放进去，但PyLint抱怨说这个函数可能已经放到了外面。 @Karl： 该类是软件升级程序，如果该文件夹尚不存在，辅助函数将创建一个新文件夹。 现在，这个类在一个模块中几乎只有这个类的代码。 其他课程可能会在稍后添加。 当我决定把助手功能放在哪里时，我问的问题是，“这仅仅是为了这个班