is it possible to add a local varible to a function, just before calling it ? if yes how ?
EDIT:REASON
i noticed that all my views in django are using
render_to_response(template_name,locals())
now i created a middleware and i wanted to add one more local variable using the
def process_view():
method of it .so that i don't have to modify the views .
The local scope for a function does not exist until the function is called, so it's not possible to do this. You could do this for a closure, but the next person to have to maintain the code would hunt you down and kill you.
Although I also think it is pretty useless, I thought that you may enclose the function in either a 'with' statement or another function, like the code below. Of course, this approach can be accomplished directly within the function of interest. In fact, you are adding the local variable 'during' the function declaration. See if this fits your needs!
#!/usr/bin/python
def my_funct(_local):
"""My function of interest
"""
print "Local argument was %s" % str(_local)
return "Finished"
def localize(fct, local_var):
"""
"""
return fct(_local = local_var)
## Use function to 'localize' variable
localize(my_funct, local_var="LOCAL_VARIABLE")
## Same effect without supplementary function :
my_funct(_local="LOCAL_VARIABLE")
try:
print local_var
except:
print "No such global variable"
Just some thoughts :)
Cheers
So if you’re one of those lazy
programmers and you like keeping code
particularly concise, you can take
advantage of a built-in Python
function called locals(). It returns a
dictionary mapping all local variable
names to their values, where “local”
means all variables that have been
defined within the current scope.
source
It is a trick in order to not have to explicitly list all of the variables you need to pass in to the function. In this case, you need to explicitly state a variable to pass in. Therefore, you should not be using locals() in the calls you are making in your middle-ware, as the trick was not designed to be used like that.
i mangaged to do that using decorators.
Related
I am looking for a way in python to stop certain parts of the code inside a function but only when the output of the function is assigned to a variable. If the the function is run without any assignment then it should run all the inside of it.
Something like this:
def function():
print('a')
return ('a')
function()
A=function()
The first time that I call function() it should display a on the screen, while the second time nothing should print and only store value returned into A.
I have not tried anything since I am kind of new to Python, but I was imagining it would be something like the if __name__=='__main__': way of checking if a script is being used as a module or run directly.
I don't think such a behavior could be achieved in python, because within the scope of the function call, there is no indication what your will do with the returned value.
You will have to give an argument to the function that tells it to skip/stop with a default value to ease the call.
def call_and_skip(skip_instructions=False):
if not skip_instructions:
call_stuff_or_not()
call_everytime()
call_and_skip()
# will not skip inside instruction
a_variable = call_and_skip(skip_instructions=True)
# will skip inside instructions
As already mentionned in comments, what you're asking for is not technically possible - a function has (and cannot have) any knowledge of what the calling code will do with the return value.
For a simple case like your example snippet, the obvious solution is to just remove the print call from within the function and leave it out to the caller, ie:
def fun():
return 'a'
print(fun())
Now I assume your real code is a bit more complex than this so such a simple solution would not work. If that's the case, the solution is to split the original function into many distinct one and let the caller choose which part it wants to call. If you have a complex state (local variables) that need to be shared between the different parts, you can wrap the whole thing into a class, turning the sub functions into methods and storing those variables as instance attributes.
I have a file
outSimulation.py this file has a function called
outlet1 and inside this function there is a variable called data
I want to access this data variable from another program called main.py
I have tried it with absolute imports
from outSimulation import outlet1
values = outlet1.data()
but it doesnt work. Any help is appreciated.
You cannot* access a variable which is local to a function from outside the function.
If you want to access the value of data from a function which is called by outlet1(), you should pass data to that function.
If you want to access the value of data after outlet1() finishes running, you should return data (possibly along with other things) from outlet1().
Alternatively, you can make data a module-level variable. Assuming you assign something to data within outlet1(), you would make it a module-level variable using the global keyword,
def outlet1():
global data
# code
data = ...
# more code
but this tends to be less clean than the other methods.
There are other solutions if outlet1() is defined within a class, but your question suggests that's not the case so I won't get into that.
*Well... uh, never mind. There are some wacky things you can do for debugging, but you're better off not considering those for your purposes.
Functions are not classes, and are specifically designed so that you CAN'T do what you're trying to do. That's the whole basis of functional programming and the concepts behind encapsulation. Even if your function was a class, your code still wouldn't work because you're trying to call the variable as a function (by adding () to data).
Either create data as a global variable that can be referenced inside the function or have the function return a tuple with multiple items, including the original return value and the data variable (messy). You could also rewrite the function in an object-oriented way if that approach suits your problem domain, in which case your code above would make more sense.
Let's say I have a code like this:
def read_from_file(filename):
list = []
for i in filename:
value = i[0]
list.append(value)
return list
def other_function(other_filename):
"""
That's where my question comes in. How can I get the list
from the other function if I do not know the value "filename" will get?
I would like to use the "list" in this function
"""
read_from_file("apples.txt")
other_function("pears.txt")
I'm aware that this code might not work or might not be perfect. But the only thing I need is the answer to my question in the code.
You have two general options. You can make your list a global variable that all functions can access (usually this is not the right way), or you can pass it to other_function (the right way). So
def other_function(other_filename, anylist):
pass # your code here
somelist = read_from_file("apples.txt")
other_function("pears.txt.", somelist)
You need to "catch" the value return from the first function, and then pass that to the second function.
file_name = read_from_file('apples.txt')
other_function(file_name)
You need to store the returned value in a variable before you can pass it onto another function.
a = read_from_file("apples.txt")
There are at least three reasonable ways to achieve this and two which a beginner will probably never need:
Store the returned value of read_from_file and give it as a parameter to other_function (so adjust the signature to other_function(other_filename, whatever_list))
Make whatever_list a global variable.
Use an object and store whatever_list as a property of that object
(Use nested functions)
(Search for the value via garbage collector gc ;-)
)
Nested functions
def foo():
bla = "OK..."
def bar():
print(bla)
bar()
foo()
Global variables
What are the rules for local and global variables in Python? (official docs)
Global and Local Variables
Very short example
Misc
You should not use list as a variable name as you're overriding a built-in function.
You should use a descriptive name for your variables. What is the content of the list?
Using global variables can sometimes be avoided in a good way by creating objects. While I'm not always a fan of OOP, it sometimes is just what you need. Just have a look of one of the plenty tutorials (e.g. here), get familiar with it, figure out if it fits for your task. (And don't use it all the time just because you can. Python is not Java.)
I am trying to access a particular variable that is defined in a function in python module. I was able to get to the variable name(using inspect.getmembers()) function, I was able to extract all variables that are defined in the function of interest in the (co_varnames) attribute. How do I get the value assigned to the variable ? Here is the sample code:-
File 1:- input.py
class foo(self):
def some_fun():
some_var='abc'
File 2:- sample.py
imported_module = importlib.import_module(input)
all_varnames = inspect.getmembers(imported_module)
-->> need value of all_varnames['some_var']
You don't in that context.
Local variables are only present during the execution of the function, that is between it has been called and returned. Your only possibility would be to do that in a function that the function calles (or if it's a generator during iteration).
Anyhow, if you do it in that way it will still be non-valid again after the function has returned.
Thanks for all the comments and suggestions. #Thomas Wagenaar.#skyking.
This is what I finally ended up doing that, I defined the variable like this inside the function:-
def testMethod(self):
self.myvar = {'a':1,'b':2}
Then using a inspect module and getattr(), I could instantiate the class, call this testMethod(which failed with exception), but got to a point where the variable was initialized and then just read the variable. I dont know if the solution is "correct" way of doing things in python, but this little hack worked for me.
I would like to write a function which receives a local namespace dictionary and update it. Something like this:
def UpdateLocals(local_dict):
d = {'a':10, 'b':20, 'c':30}
local_dict.update(d)
When I call this function from the interactive python shell it works all right, like this:
a = 1
UpdateLocals(locals())
# prints 20
print a
However, when I call UpdateLocals from inside a function, it doesn't do what I expect:
def TestUpdateLocals():
a = 1
UpdateLocals(locals())
print a
# prints 1
TestUpdateLocals()
How can I make the second case work like the first?
UPDATE:
Aswin's explanation makes sense and is very helpful to me. However I still want a mechanism to update the local variables. Before I figure out a less ugly approach, I'm going to do the following:
def LoadDictionary():
return {'a': 10, 'b': 20, 'c': 30}
def TestUpdateLocals():
a = 1
for name, value in LoadDictionary().iteritems():
exec('%s = value' % name)
Of course the construction of the string statements can be automated, and the details can be hidden from the user.
You have asked a very good question. In fact, the ability to update local variables is very important and crucial in saving and loading datasets for machine learning or in games. However, most developers of Python language have not come to a realization of its importance. They focus too much on conformity and optimization which is nevertheless important too.
Imagine you are developing a game or running a deep neural network (DNN), if all local variables are serializable, saving the entire game or DNN can be simply put into one line as print(locals()), and loading the entire game or DNN can be simply put into one line as locals().update(eval(sys.stdin.read())).
Currently, globals().update(...) takes immediate effect but locals().update(...) does not work because Python documentation says:
The default locals act as described for function locals() below:
modifications to the default locals dictionary should not be
attempted. Pass an explicit locals dictionary if you need to see
effects of the code on locals after function exec() returns.
Why they design Python in such way is because of optimization and conforming the exec statement into a function:
To modify the locals of a function on the fly is not possible without
several consequences: normally, function locals are not stored in a
dictionary, but an array, whose indices are determined at compile time
from the known locales. This collides at least with new locals added
by exec. The old exec statement circumvented this, because the
compiler knew that if an exec without globals/locals args occurred in
a function, that namespace would be "unoptimized", i.e. not using the
locals array. Since exec() is now a normal function, the compiler does
not know what "exec" may be bound to, and therefore can not treat is
specially.
Since global().update(...) works, the following piece of code will work in root namespace (i.e., outside any function) because locals() is the same as globals() in root namespace:
locals().update({'a':3, 'b':4})
print(a, b)
But this will not work inside a function.
However, as hacker-level Python programmers, we can use sys._getframe(1).f_locals instead of locals(). From what I have tested so far, on Python 3, the following piece of code always works:
def f1():
sys._getframe(1).f_locals.update({'a':3, 'b':4})
print(a, b)
f1()
However, sys._getframe(1).f_locals does not work in root namespace.
The locals are not updated here because, in the first case, the variable declared has a global scope. But when declared inside a function, the variable loses scope outside it.
Thus, the original value of the locals() is not changed in the UpdateLocals function.
PS: This might not be related to your question, but using camel case is not a good practice in Python. Try using the other method.
update_locals() instead of UpdateLocals()
Edit To answer the question in your comment:
There is something called a System Stack. The main job of this system stack during the execution of a code is to manage local variables, make sure the control returns to the correct statement after the completion of execution of the called function etc.,
So, everytime a function call is made, a new entry is created in that stack,
which contains the line number (or instruction number) to which the control has to return after the return statement, and a set of fresh local variables.
The local variables when the control is inside the function, will be taken from the stack entry. Thus, the set of locals in both the functions are not the same. The entry in the stack is popped when the control exits from the function. Thus, the changes you made inside the function are erased, unless and until those variables have a global scope.