This question already has answers here:
How do I forward-declare a function to avoid `NameError`s for functions defined later?
(17 answers)
Closed 3 years ago.
Simple python script is giving error, what is wrong?
var ="first variable"
myfun(var)
def myfun(var):
print(var)
Error -> NameError: name 'myfun' is not defined
This thing is quite obvious though. Python reads the code line by line and not like C.
Just switch your two blocks i.e. definition of function and calling it.
var ="first variable"
def myfun(var):
print(var)
myfun(var)
This should be good to go.
When python interpreter sees the statement myfun(var), the name myfun is not defined yet. You need to move this line after your function definition.
Related
This question already has answers here:
How to get local variables updated, when using the `exec` call?
(3 answers)
Closed 1 year ago.
Given a string (e.g. "lista=[1,2,3]") I would like to be able to use the variable lista.
exec() does the work outside a function, but when used inside a function the variables cannot be used in that same function. I guess it has something to do with local and global variables but I don't really understand the problem.
For example,
def funcion(texto):
exec(texto)
print(lista)
funcion("lista = [3,4,5]")
Gives the error: NameError: name 'lista' is not defined.
add globals
def funcion(texto):
exec(texto, globals())
print(lista)
funcion("lista = [3,4,5]")
This question already has answers here:
Given a function with closure, can I refer back to it's closure scope?
(1 answer)
What exactly is contained within a obj.__closure__?
(4 answers)
Closed 2 years ago.
I would like to know if there is any method to check whether two functions have the same arguments at runtime in python 3.
Basically, I have this function (func) that takes another function and an argument. I want to check the values assigned to args in the lambda function
func(another_func, args):
return lambda(x : another_func(x, args))
It is not feasible to run the code before and check the results because I am implementing a lazy framework. My main goal is to be able to understand what are the arguments of the function because there is one variable argument that I do not care but there is one static that is created before running the function.
##EDIT
I actually solved this problem using the inspect module (getclosure) for those who are interested!
I actually solved this problem using the inspect module (getclosure) for those who are interested!
Extension (Martijn Pieters):
I think you are referring to getclosurevars(). You can also just access function.closure, and access the value of each cell via its cell_contents attribute.
This question already has answers here:
Scope of variable within "with" statement?
(2 answers)
Closed 3 years ago.
I didn't find the answer for this question on stackoverflow, so I thought it might be helpful to ask it, and have it here -
I am declaring a new dictionary after I open a file, in the following way -
with open('some_file.txt','r') as f:
dict = json.loads(f.read()) #converts text to a dictionary
my question is - will I be able to reach dict content's even after the 'with' scope ends.
Thanks
Yes, in Python the scope of a variable ends only when the code block it's defined in ends, and the with statement is not a code block per the documentation:
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.
In python scope is defined by functions. There is no indentation scope (similar to "bracket" scope in other languages). The with part affects just the f object.
Yes you will, you will not be able to access f, everything else is fair game.
This question already has answers here:
Creating functions (or lambdas) in a loop (or comprehension)
(6 answers)
Closed 8 years ago.
when working with python, it bothered me that while obj.method() is perfectly fine, method(obj) isn't allowed. So I figured I'd try to write some code to fix that. I came up with the next:
def globalclassfuncs(defobj):
for i in inspect.getmembers(defobj, predicate=inspect.ismethod):
def scope():
var = i[0];
setattr(sys.modules[__name__], i[0], lambda obj, *args: getattr(obj, var)(*args));
scope();
However, there's something weird with this. When I remove def scope(): and scope(), so that it'll run without a function definition in the for loop, or when I change the getattr() function to use i[0] directly instead of through var, somehow all new defined functions point to the last defined function instead of the function they should point to. Why does this behaviour change so much on such small changes in the code?
Seems like a case of late binding closure
This question already has answers here:
Why doesn't exec work in a function with a subfunction?
(6 answers)
Closed 7 years ago.
I'm trying to make a "compiler" for my game (so that people could do intresting stuff but not inject code), for mainly declaritive "code" (It would look like this: {"player_location":"IceHall.A7", "print", "You are teleported somewhere", "tiles":{"FirePlace.B3":{'Type':"Corner", "Actions+":{....}}}}. This is how a action is represented; It is called when the player does it.
Anyways, it would have to be compiled into a function. When I tryed out something similar in the interactive interpreter (specifically:
def compile(code):
def act():
exec code
return act
). This (which is would be more or less what would be in the final, with the exception of "code" being constructed by me) raised a odd error:
File "", line 3 SyntaxError:
unqualified exec is not allowed in
function 'act' it is a nested
function.
How do I get around this?
The answer, as said in this question, is the lack of context. I wanted exec code in locals(), globals()