Python初学者类变量Error [英] Python beginner Class variable Error

问题描述

这是我的第一个问题，很抱歉...
我是一个初学者python和编码，一般来说，我想创建一个名为Map的类，它将有以下类变量： / p>

this is my first question, so sorry... I'm a beginner with python and coding in general, and i wanted to create a class called 'Map' that would have the following class variables:

class Map:
    height = 11  
    width = 21

    top = [['#']*width]
    middle = [['#']+[' ']*(width-2)+['#'] for i in range(height-2)]
    field = top + middle + top

b = Map()

Shell:
>>> middle = [['#']+[' ']*(width-2)+['#'] for i in range(height-2)]
    NameError: name 'width' is not defined

如果我把变量放在类外面，它工作。
我做错了什么？

If i put the variables outside of the class it works. What am I doing wrong??

感谢您的帮助。

推荐答案

从文档：

名称引用对象。名称是通过名称绑定操作引入的。

Names refer to objects. Names are introduced by name binding operations. Each occurrence of a name in the program text refers to the binding of that name established in the innermost function block containing the use.

一个块是一段Python程序文本作为一个单元执行。以下是块：模块，函数体和类定义。交互式输入的每个命令都是一个块。脚本文件（作为解释器的标准输入或作为解释器的命令行参数指定的文件）是代码块。脚本命令（在解释器命令行中使用-c选项指定的命令）是一个代码块。传递给内置函数eval（）和exec（）的字符串参数是一个代码块。

A block is a piece of Python program text that is executed as a unit. The following are blocks: a module, a function body, and a class definition. Each command typed interactively is a block. A script file (a file given as standard input to the interpreter or specified as a command line argument to the interpreter) is a code block. A script command (a command specified on the interpreter command line with the ‘-c‘ option) is a code block. The string argument passed to the built-in functions eval() and exec() is a code block.

代码块在执行框架中执行。框架包含一些管理信息（用于调试），并确定在代码块执行完成后继续执行的位置和方式。

A code block is executed in an execution frame. A frame contains some administrative information (used for debugging) and determines where and how execution continues after the code block’s execution has completed.

范围定义名称的可见性块。如果在块中定义了局部变量，则其范围包括该块。如果定义出现在功能块中，则作用域扩展到定义的块中包含的任何块，除非包含的块引入了对名称的不同绑定。 在类块中定义的名称范围仅限于类块;它不扩展到方法的代码块 - 这包括了理解和生成器表达式，因为它们是使用函数范围实现的。这意味着以下操作将失败：

A scope defines the visibility of a name within a block. If a local variable is defined in a block, its scope includes that block. If the definition occurs in a function block, the scope extends to any blocks contained within the defining one, unless a contained block introduces a different binding for the name. The scope of names defined in a class block is limited to the class block; it does not extend to the code blocks of methods – this includes comprehensions and generator expressions since they are implemented using a function scope. This means that the following will fail:

class A:
    a = 42
    b = list(a + i for i in range(10))

list comps in python3有你自己的范围，而不是python2，你的代码将工作原样。

list comps in python3 have their own scope, as opposed to python2 where your code would work as is.

如果你使用python2下面的例子，你可以看到变量如何泄露的范围列表comp可能会导致一些问题：

If you take the following example using python2, you can see how variables leaking the scope of the list comp could cause some problems:

class A:
    a = 42
    b = [a for a in range(10)]

a = A()

print(a.a)
9

您有几个选项，您可以使用 __ init __ 创建实例属性：

You have a few options, you could use __init__ creating instance attributes:

class Map:
    def __init__(self):
        self.height = 11
        self.width = 21
        self.top = [['#']*self.width]
        self.middle = [['#']+[' ']*(self.width-2)+['#'] for i in range(self.height-2)]
        self.field = self.top + self.middle + self.top
m=Map()
print(m.field)

使用方法：

class Map:
    @staticmethod
    def meth():
        height = 11
        width = 21
        top = [['#']*width]
        middle = [['#']+[' ']*(width-2)+['#'] for i in range(height-2)]
        field = top + middle + top
        return field

b = Map()


print(b.meth())

你选择的是什么取决于你想做什么。

What you choose is really dependant on what you want to do.

这篇关于Python初学者类变量Error的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！