Assign global variable to a function's local variable

Question

I want to loop over some global variable i and each time compile a new function that uses the instantaneous value of i. I can then use each of these functions in future, independent of the current value of i.

The problem is I can't find a good way to make the global variable i stay local within the namespace of the function. For example:

i = 0

def f():
    return i

i = 1

print(f()) #Prints 1, when I want 0

I found one solution but it seems very clumsy:

i = 0

def g(x):
    def h():
        return x
    return h

f = g(i)
i = 1

print(f()) #Prints 0 as I wanted.

Is there a better way to do this, maybe some kind of decorator?

Something doesn't sound quite right. It might require redesigning the solution. — CristiFati
– CristiFati, Commented Jun 6, 2018 at 16:37

Jacques Gaudin · Accepted Answer · 2018-06-06 16:44:47Z

1

Python functions are objects, you can set an attribute on them:

i = 0

def f():
    return f.i

f.i = i

i = 1

print(f()) # Prints 0 as you wanted

or:

for i in range(5):
    f.i = i
    print(f())

Note that an argument would do the same.

edited Jun 6, 2018 at 16:44

answered Jun 6, 2018 at 16:36

Jacques Gaudin

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3 Comments

henneray Over a year ago

That's a nice idea, but it doesn't quite work in my context. If I were to now append f to some list each time within a loop and then execute any member of the list at a later point, I just get the latest definition of f again.

Jacques Gaudin Over a year ago

Take no offence but it sounds like there is a flaw in the design. It sounds very complicated!

Jacques Gaudin Over a year ago

Just as tip: putting functions in a list is rare in Python. One of the beauty of this language is mapping.

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