如何在Python中输出变化的参数？

Question

I am writing a function in Python.

So lets say I have 3 arguments in this function:

Function (arg1, arg2, arg3)
  arg1 = arg2 * arg3

When the function is called, it is as such:

function(arg1=var1, arg2=var2, arg3=var3)

Now the thing is, these variables, ( var1 , var2 , var3 ) vary in the case of this function being called. So they could be var4 , var5 , var6 , etc.

Given that these variables vary, what would be the syntax to make this function change a varying argument?

ie. arg1 = arg2 * arg3 changes arg1 . Because arg1 = var1 .

What I need is somehow for arg1 to stand in as an actual proxy to var1 so that not only is arg1 changed but var1 is changed also.

This is evidently not doable with simply with arg1=var1

I assume this could also be done with other function commands or structure. How does not really concern me I just need to output a change into a varying argument.

Edit This function is specifically designed to take an argument called, not necessarily knowing what it is, and then inside of the function, that argument which is converted into a formula variable such as x in y = mx + b, needs to be converted back into its variable target, all inside of the function.

The problem here is that I don't know how to target the original variable.

So if it were function(a=cats, b=dogs, c=mice)

a = c/2 - b*2

how do I then set cat=a when cat is variable?

I can't just say cat=a because maybe the input is function(a=bird, b=flys, c=cats)

Answer 1

def f(x, y):
    return x*y

a, b = 1, 2

c = f(a, b)

Instead of having your function try to alter its arguments directly, return any results the function produces. This turns out to be a much cleaner, more useful way to structure your code. For example, you can do

print f(1, f(2, 3))

instead of having to create temporary variables:

f(a, 2, 3)
f(b, 1, a)
print b

Instead of doing

f(arg1=var1, arg2=var2, arg3=var3)
f(arg1=var4, arg2=var5, arg3=var6)

do the following:

var1 = f(var2, var3)
var4 = f(var5, var6)

Answer 2

If it is not doable with arg1=var1 , it's because of this:

Python passes references-to-objects by value (like Java), and everything in Python is an object. This sounds simple, but then you will notice that some data types seem to exhibit pass-by-value characteristics, while others seem to act like pass-by-reference... what's the deal?

It is important to understand mutable and immutable objects. Some objects, like strings, tuples, and numbers, are immutable. Altering them inside a function/method will create a new instance and the original instance outside the function/method is not changed. Other objects, like lists and dictionaries are mutable, which means you can change the object in-place. Therefore, altering an object inside a function/method will also change the original object outside.

Why not just return arg1 and set val1 to it?

Answer 3

Python doesn't have a specific syntax regarding references (not that I know of at least). If you want your argument to be passed as reference to a function, it has to be mutable (eg. list, dictionary).

def foobar(a, b):
    a[0] = 42   # This will modify the list
    b[0] = '-'  # This will modify the value of b passed to the function

foobar(
    [4,8,15,16,23,42],  # This argument is passed by reference, mutable
    'asdf'              # This one is passed by value, immutable
)