python-nonlocal | 易学教程

Python 3 changing variable in function from another function

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问题 I would like to access the testing variable in main from testadder, such that it will add 1 to testing after testadder has been called in main. For some reason I can add 1 to a list this way, but not variables. The nonlocal declaration doesn't work since the functions aren't nestled. Is there a way to work around this? def testadder(test, testing): test.append(1) testing += 1 def main(): test = [] testing = 1 testadder(test, testing) print(test, testing) main() 回答1: Lists are mutable, but

nested function change variable in an outside function not working

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问题 def some_func(a): def access_a(): print(a) access_a() outputs the value of a . However, if I want to change a in the nested function like this: def some_func(a): def change_a(): a += 1 print(a) change_a() it raises UnboundLocalError exception. I know a is a nonlocal variable, but why can I access it without declaring nonlocal a ? 回答1: Python scoping rules 101: a name bound in a function body is considered local unless explicitely declared global (Python 2.x and 3.x) or nonlocal (Python 3.x

Where is nonlocals()?

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问题 How do I obtain the non-local variables for the current scope? The functions vars , locals , and globals exist, but is there a function to get the nonlocals ? Why aren't the nonlocals listed when calling vars ? Update My issue is that there's no way to enumerate the variables available in the current scope, as neither vars or globals includes the non-locals AFAICT. I frequently use vars in code such as the following: '{meh[0]}/{meh[3]} {a}{b}{c}'.format(**vars()) Which fails if any of these

Where is nonlocals()?

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Understanding nonlocal in Python 3

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问题 I am trying to understand Python 3 variable scoping and nonlocal . Consider the following function (it is just an example): def build_property(something): def deco(func): def getter(self): return getattr(self, something) def setter(self, value): setattr(self, something, value) return property(getter, setter) return deco This works fine without nonlocal . But if now I want to conditionally create getters and setters depending on something I need nonlocal. def build_property(something): def

Python, how can I change value of a variable in the parent scope?

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问题 for example: assginment statement will declare a new local variable. foo = 'global' def func1(): foo = 'func1' def func2(): foo = 'local variable in func2' use global declaration will use the foo in global: def func2(): global foo foo = 'global changed in func2' #changed the foo value in global scope how can I change the variable foo in func1 scope? Thanks for any help. Edit: Thank you Brandon Craig Rhodes, I finally understand your meaning. if there are more than 3 scopes nested, I can store

How to check if a variable with a given name is nonlocal?

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问题 Given a stack frame and a variable name, how do I tell if that variable is nonlocal? Example: import inspect def is_nonlocal(frame, varname): # How do I implement this? return varname not in frame.f_locals # This does NOT work def f(): x = 1 def g(): nonlocal x x += 1 assert is_nonlocal(inspect.currentframe(), 'x') g() assert not is_nonlocal(inspect.currentframe(), 'x') f() 回答1: Check the frame's code object's co_freevars , which is a tuple of the names of closure variables the code object

How to check if a variable with a given name is nonlocal?

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Given a stack frame and a variable name, how do I tell if that variable is nonlocal? Example: import inspect def is_nonlocal(frame, varname): # How do I implement this? return varname not in frame.f_locals # This does NOT work def f(): x = 1 def g(): nonlocal x x += 1 assert is_nonlocal(inspect.currentframe(), 'x') g() assert not is_nonlocal(inspect.currentframe(), 'x') f() Check the frame's code object's co_freevars , which is a tuple of the names of closure variables the code object uses: def is_nonlocal(frame, varname): return varname in frame.f_code.co_freevars Note that this is

Where is nonlocals()?

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How do I obtain the non-local variables for the current scope? The functions vars , locals , and globals exist, but is there a function to get the nonlocals ? Why aren't the nonlocals listed when calling vars ? Update My issue is that there's no way to enumerate the variables available in the current scope, as neither vars or globals includes the non-locals AFAICT. I frequently use vars in code such as the following: '{meh[0]}/{meh[3]} {a}{b}{c}'.format(**vars()) Which fails if any of these variables are in the scope of a containing function. From within running code, you can easily get the

Python, how can I change value of a variable in the parent scope?

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for example: assginment statement will declare a new local variable. foo = 'global' def func1(): foo = 'func1' def func2(): foo = 'local variable in func2' use global declaration will use the foo in global: def func2(): global foo foo = 'global changed in func2' #changed the foo value in global scope how can I change the variable foo in func1 scope? Thanks for any help. Edit: Thank you Brandon Craig Rhodes, I finally understand your meaning. if there are more than 3 scopes nested, I can store the variable in a list. foo = ['global', 'function1', 'function2'] def func1(): foo[1] = 'func1' def