I like to programming in Python language to solve daily problems in system administration contexts, and I am happy for that, but lately I am learning Pharo Smalltalk because I am fascinating to the different approach of this language in object programming. I like very much the idea of Smalltalk about the virtual machine and system image concept which hold the environment all in one file. Does
我喜欢用Python语言编程来解决系统管理环境中的日常问题，我为此感到高兴，但最近我正在学习Pharo Smalltalk，因为我对这种语言在对象编程中的不同方法非常着迷。 我非常喜欢Smalltalk关于将环境全部保存在一个文件中的虚拟机和系统映像概念的想法。 它在Python中是否存在类似的东西？ AFAIK，没有。 Python没有什么像Smalltalk的图像。 如果你有足够的时间:)你当然可以使用Glamour和Petitparser在Pharo中构建一个Python
Is it possible to declare more than one variable using a with statement in Python? Something like: from __future__ import with_statement with open("out.txt","wt"), open("in.txt") as file_out, file_in: for line in file_in: file_out.write(line) ... or is cleaning up two resources at the same time the problem? It is possible in Python 3 since v3.1 and Python 2.7. The new with syn
是否有可能在Python中使用with语句声明多个变量？ 就像是： from __future__ import with_statement with open("out.txt","wt"), open("in.txt") as file_out, file_in: for line in file_in: file_out.write(line) ...或者在同一时间清理两个资源的问题？ 从v3.1和Python 2.7开始，它可能在Python 3中。 新with语法支持多个上下文管理器： with A() as a, B() as b, C() as c: doSomething(a,b,c) 不
What I understand from reading the documentation is that Python has a separate namespace for functions, and if I want to use a global variable in that function, I need to use global . I'm using Python 2.7 and I tried this little test >>> sub = ['0', '0', '0', '0'] >>> def getJoin(): ... return '.'.join(sub) ... >>> getJoin() '0.0.0.0' It seems things are wor
我从阅读文档不解的是，Python有功能的单独的命名空间，如果我想在函数中使用全局变量，我需要使用global 。 我正在使用Python 2.7，并且我尝试了这个小测试 >>> sub = ['0', '0', '0', '0'] >>> def getJoin(): ... return '.'.join(sub) ... >>> getJoin() '0.0.0.0' 即使没有global ，看起来事情仍然正常。 我能够访问全局变量没有任何问题。 我错过了什么？ 另外，以下是Python文档
I'm trying to make a build retrieval form, and seem to have issues with the buttons... I'm a novice at Python/tkinter GUI programming (and GUI programming in general) and borrowed the skeleton of a Hello World app, and sorta built off that. In the code below, I've set the "command" option of my Browse button to call my class's internal get_dir() function when it's
我试图制作一个构建检索表单，并且似乎遇到按钮问题...我是Python / tkinter GUI编程（通常是GUI编程）的新手，并且借用了Hello World应用程序的框架，和sorta建立起来的。 在下面的代码中，我设置了Browse按钮的“command”选项，当它被点击时调用我的类的内部get_dir（）函数。 但是，只要我尝试运行该应用程序，就会调用get_dir（）函数，并提示我选择一个目录。 任何想法为什么会发生这种情况，以及我能做些什么来使其行为
In Python, suppose I have a function f that I want to pass around with some secondary arguments (assume for simplicity that it's just the first argument that remains variable). What are the differences between doing it these two ways (if any)? # Assume secondary_args and secondary_kwargs have been defined import functools g1 = functools.partial(f, *secondary_args, **secondary_kwargs) g2
在Python中，假设我有一个函数f ，我想用一些辅助参数传递（为简单起见，假设它只是第一个参数保持可变）。 做这两种方式（如果有的话）有什么区别？ # Assume secondary_args and secondary_kwargs have been defined import functools g1 = functools.partial(f, *secondary_args, **secondary_kwargs) g2 = lambda x: f(x, *secondary_args, **secondary_kwargs) 在partial文档页面中，例如，有这样的引用： 类中定义
Recently I started playing around with Python and I came around something peculiar in the way closures work. Consider the following code: adders=[0,1,2,3] for i in [0,1,2,3]: adders[i]=lambda a: i+a print adders[1](3) It builds a simple array of functions that take a single input and return that input added by a number. The functions are constructed in for loop where the iterator i runs
最近我开始使用Python进行游戏，并且在关闭工作方式中遇到了一些特殊问题。 考虑下面的代码： adders=[0,1,2,3] for i in [0,1,2,3]: adders[i]=lambda a: i+a print adders[1](3) 它构建了一个简单的函数数组，它接受单个输入并返回由数字添加的输入。 这些函数在for循环中构建，其中迭代器i从0运行到3 。 对于这些数字中的每一个，都会创建一个lambda函数，它捕获i并将其添加到函数的输入中。 最后一行以3作为参数
I am trying to connect 9 different buttons to one handler using a lambda function and some PyQt5 QPushButtons in python 3.6. If I assign them individually using ints all works fine. However if I try to use a list and a loop they are all assigned to a button number of 10. I can't understand why, since I would have thought that my assignment was to the integer value and that my variable was o
我试图用一个lambda函数和python 3.6中的一些PyQt5 QPushButtons将9个不同的按钮连接到一个处理程序。 如果我使用整数分配它们，所有工作都正常。 但是，如果我尝试使用一个列表和一个循环，他们都被分配到一个按钮数为10.我不明白为什么，因为我会认为我的任务是整数值，我的变量超出范围。 很显然，这里有一些我不明白的事情。 任何人都可以解释这段代码的行为吗？ self.buttonList = [ self.sq1Button,
In the function main I have def main(): #... def button_callback(button,a, byMouse,label): #... date = date - oneday while(date.isoweekday()>5): date = date -oneday #... oneday = datetime.timedelta(1) date = datetime.date.today() python complains: local variable 'date' referenced before assignment, which is expected. In
在我有的功能主 def main(): #... def button_callback(button,a, byMouse,label): #... date = date - oneday while(date.isoweekday()>5): date = date -oneday #... oneday = datetime.timedelta(1) date = datetime.date.today() python抱怨：在赋值之前引用了局部变量'date'，这是预期的。 在main（）的其他部分，我注意不要指定而是修改，
What I essentially need is a mutable integer which can accomplish the equivalent of the below without having to resort to using single element lists. Below is a contrived code sample which is representative of the essence of my actual use case [1] >>> a = [5] >>> b = [77] >>> def swap(num1, num2): ... temp = num1[0] ... num1[0] = num2[0] ... num
我基本需要的是一个可变整数，它可以完成下面的等价操作，而不必求助于使用单个元素列表。 下面是一个人为的代码示例 ，代表了我的实际使用案例的本质[1] >>> a = [5] >>> b = [77] >>> def swap(num1, num2): ... temp = num1[0] ... num1[0] = num2[0] ... num2[0] = temp >>> swap(a, b) >>> a [77] >>> b [5] [1]我的实际用例
Possible Duplicate: UnboundLocalError in Python I have this code: import re def doReplace(toReplace): i = 1 def chapterReplacer(_): result = 'Chapter %i' % i i += 1 return result return re.sub('Chapter [a-zA-Z]+', chapterReplacer, test) test = 'Chapter one Chapter Two Chapter three' print doReplace(test) when I run it, I get the following error: Trace
可能重复： Python中的UnboundLocalError 我有这样的代码： import re def doReplace(toReplace): i = 1 def chapterReplacer(_): result = 'Chapter %i' % i i += 1 return result return re.sub('Chapter [a-zA-Z]+', chapterReplacer, test) test = 'Chapter one Chapter Two Chapter three' print doReplace(test) 当我运行它时，出现以下错误： Traceback (most recent call