I am using a list on which some functions works in my program. This is a shared list actually and all of my functions can edit it. Is it really necessary to define it as "global" in all the functions?
I mean putting the global keyword behind it in each function that uses it, or defining it outside of all the functions is enough without using the global word behind its definition?
This question is related to
python
list
global-variables
When you assign a variable (x = ...), you are creating a variable in the current scope (e.g. local to the current function). If it happens to shadow a variable fron an outer (e.g. global) scope, well too bad - Python doesn't care (and that's a good thing). So you can't do this:
x = 0
def f():
x = 1
f()
print x #=>0
and expect 1. Instead, you need do declare that you intend to use the global x:
x = 0
def f():
global x
x = 1
f()
print x #=>1
But note that assignment of a variable is very different from method calls. You can always call methods on anything in scope - e.g. on variables that come from an outer (e.g. the global) scope because nothing local shadows them.
Also very important: Member assignment (x.name = ...), item assignment (collection[key] = ...), slice assignment (sliceable[start:end] = ...) and propably more are all method calls as well! And therefore you don't need global to change a global's members or call it methods (even when they mutate the object).
No, you can specify the list as a keyword argument to your function.
alist = []
def fn(alist=alist):
alist.append(1)
fn()
print alist # [1]
I'd say it's bad practice though. Kind of too hackish. If you really need to use a globally available singleton-like data structure, I'd use the module level variable approach, i.e. put 'alist' in a module and then in your other modules import that variable:
In file foomodule.py:
alist = []
In file barmodule.py:
import foomodule
def fn():
foomodule.alist.append(1)
print foomodule.alist # [1]
Yes, you need to use global foo if you are going to write to it.
foo = []
def bar():
global foo
...
foo = [1]
Source: Stackoverflow.com