I just learned python @ decorator, it's cool, but soon I found my modified code coming out weird problems. def with_wrapper(param1): def dummy_wrapper(fn): print param1 param1 = 'new' fn(param1) return dummy_wrapper def dummy(): @with_wrapper('param1') def implementation(param2): print param2 dummy() I debug it, it throws out exception at pr
我刚刚学会了python @ decorator，它很酷，但很快我发现我的修改代码出现了奇怪的问题。 def with_wrapper(param1): def dummy_wrapper(fn): print param1 param1 = 'new' fn(param1) return dummy_wrapper def dummy(): @with_wrapper('param1') def implementation(param2): print param2 dummy() 我调试它，它抛出了在打印参数1异常 UnboundLocalError: local variable
I'm wondering if it's possible for a closure in Python to manipulate variables in its namespace. You might call this side-effects because the state is being changed outside the closure itself. I'd like to do something like this def closureMaker(): x = 0 def closure(): x+=1 print x return closure a = closureMaker() a() 1 a() 2 Obviously what I hope to do is more comp
我想知道是否有可能使用Python中的闭包操作其名称空间中的变量。 您可能会称这种副作用是因为状态正在关闭本身之外进行更改。 我想要做这样的事情 def closureMaker(): x = 0 def closure(): x+=1 print x return closure a = closureMaker() a() 1 a() 2 显然我希望做的更复杂，但这个例子说明了我在说什么。 你不能在Python 2.x中完全做到这一点，但你可以使用一个技巧来获得相同的效果：使用一个可变对
This bit of Python does not work: def make_incrementer(start): def closure(): # I know I could write 'x = start' and use x - that's not my point though (: while True: yield start start += 1 return closure x = make_incrementer(100) iter = x() print iter.next() # Exception: UnboundLocalError: local variable 'start' referenced before assignment I
这一点Python不起作用： def make_incrementer(start): def closure(): # I know I could write 'x = start' and use x - that's not my point though (: while True: yield start start += 1 return closure x = make_incrementer(100) iter = x() print iter.next() # Exception: UnboundLocalError: local variable 'start' referenced before assignment 我知道如何
I have an application that loads a DLL with a class that handles jobs in a queue. In order to keep it thread safe, a mutex is locked whenever the queue is modified. When the application exits and the destructor is called, the mutex is locked in order to clear the queue. However, when I load this DLL in Python, create an instance of the object, and call exit() (in Python) an exception is throw
我有一个应用程序加载了一个处理队列中作业的类。 为了保持线程安全，只要队列被修改，就会锁定互斥锁。 当应用程序退出并且调用析构函数时，互斥锁被锁定以清除队列。 然而，当我在Python中加载这个DLL时，创建一个对象的实例，并调用exit() （在Python中），当互斥锁试图锁定时会引发异常： Microsoft Visual Studio C运行时库已在python.exe中检测到致命错误。 我已经简化了析构函数，直到在本地创建一个互斥锁并尝试
This question already has an answer here: What's the best way to find the inverse of datetime.isocalendar()? 7 answers EDIT Found question with solution: What's the best way to find the inverse of datetime.isocalendar()? My solution had mistakes. On googling saw previous question which upon testing below worked. (See top answered question in above link for definition of iso_to_gr
这个问题在这里已经有了答案： 查找datetime.isocalendar（）的逆函数的最佳方法是什么？ 7个答案 编辑发现问题的解决方案：什么是最好的方式来找到datetime.isocalendar（）的逆？ 我的解决方案有错误。 在谷歌上看到以前的问题，经过测试下面的工作。 （有关iso_to_gregorian定义，请参见上面的回答中iso_to_gregorian回答的问题）。 （基本上找到iso年开始日期，然后使用timedelta从日和周编号中查找当前日期。 for
I am attempting to use the hatching feature in matplotlib, which works fine when displaying to screen. However when I save the figure to pdf format, the hatch marks are not rendered: import matplotlib import matplotlib.pyplot as plt import numpy as np x = np.linspace(0,2*np.pi,100) plt.figure() plt.fill(x,np.sin(x),color='blue',alpha=0.5,hatch='/') plt.show() plt.savefig('./test.pdf',format='
我试图在matplotlib中使用阴影功能，在显示到屏幕时工作正常。 但是，当我将图保存为pdf格式时，不会呈现阴影标记： import matplotlib import matplotlib.pyplot as plt import numpy as np x = np.linspace(0,2*np.pi,100) plt.figure() plt.fill(x,np.sin(x),color='blue',alpha=0.5,hatch='/') plt.show() plt.savefig('./test.pdf',format='pdf') 我在OS X 10.6.6的pylab中使用matplotlib 1.0.1。 这可能是与后端渲染
I'm working on a GUI application in WxPython, and I am not sure how I can ensure that only one copy of my application is running at any given time on the machine. Due to the nature of the application, running more than once doesn't make any sense, and will fail quickly. Under Win32, I can simply make a named mutex and check that at startup. Unfortunately, I don't know of any facili
我正在WxPython中开发一个GUI应用程序，并且我不确定如何确保在计算机上的任何给定时间只运行我的应用程序的一个副本。 由于应用程序的性质，多次运行没有任何意义，并且会很快失败。 在Win32下，我可以简单地创建一个有名的互斥锁并在启动时检查它。 不幸的是，我不知道Linux中的任何工具都可以做到这一点。 我正在寻找应用程序意外崩溃时会自动发布的内容。 我不想让我的用户不得不手动删除锁定文件，因为我崩溃了。
The local/global/free variable definitions from python doc: If a name is bound in a block, it is a local variable of that block, unless declared as nonlocal. If a name is bound at the module level, it is a global variable . (The variables of the module code block are local and global.) If a variable is used in a code block but not defined there, it is a free variable . Code 1: >>>
来自python doc的local / global / free变量定义： 如果名称被绑定在一个块中，那么它是该块的局部变量 ，除非声明为非局部变量 。 如果名称在模块级绑定，它是一个全局变量 。 （模块代码块的变量是局部变量和全局变量。）如果在代码块中使用了一个变量，但在那里未定义，则它是一个自由变量 。 代码1： >>> x = 0 >>> def foo(): ... print(x) ... print(locals()) ... >>> foo() 0 {}
From Programming Language Pragmatics by Scott, Figure 3.14 Deep binding in Python. At right is a conceptual view of the run-time stack. Referencing environments captured in closures are shown as dashed boxes and arrows. When B is called via formal parameter P, two instances of I exist. Because the closure for P was created in the initial invocation of A , B's static link (solid arrow) p
从编程语言语用学作者Scott， 图3.14 Python中的深度绑定。 右边是运行时栈的概念视图。 在闭包中捕获的参考环境显示为虚线框和箭头。 当通过形式参数P调用B时，存在两个I的实例。 因为P的闭包是在A的初始调用中创建的，所以 B的静态链接（实线箭头）指向该早期调用的框架。 B在其打印语句中使用该调用的I实例，并且输出为1。 问题在于正在运行的程序可能有多个在递归子例程中声明的对象实例。 使用静态范围的语言的闭
I was wondering about the scope of the variable a in the following Python snippet, # ============================ def get_plotter(): def get_a(): return a a = 3.14 return get_a # ============================ if __name__ == '__main__': f = get_plotter() print f() # ============================ The output is 3.14 , but looking at the code I would have expected a to go
我想知道下面的Python代码片段中的变量a的范围， # ============================ def get_plotter(): def get_a(): return a a = 3.14 return get_a # ============================ if __name__ == '__main__': f = get_plotter() print f() # ============================ 输出为3.14 ，但看代码我本来期望a尽快走出去的范围为get_plotter()终止，有效地留下一个不确定的东西。 这仅仅是打