Question

I meet this problem when implementing my own swap() method with Python

def swap(a,b):
    temp=a
    a=b
    b=temp

list_=[5,4,6,3,7]
swap(list_[4],list_[2])

I expexted list_ to be updated with swap() call, since list_[4] and list_[2] are to be assigned new values during function call. However, list_ remains unchanged:

list_
   [5, 4, 6, 3, 7]

I misunderstand why swap function call is dealing with a copy. I don’t want to add a list argument to my swap function nor return the list during swap() call since I want the method to be adaptable to other datastructures, like in

 swap(mat[0][1],mat[2,3])

Answer 1

You are off on how Python works:

list_=[5,4,6,3,7]
swap(list_[4],list_[2]) # this is absolutely the same as
swap(7,6)

Python fundamental concept is the use of names and values. When you write down a name in code it stands for the value attached to it during runtime. In this case list_[4] is a name that stands for the value 7.

When you want to change something you must use one of it’s names. Here you want to change the list_, so you have to do this:

def swap(data, i1, i2):
    data[i1], data[i2] = data[i2], data[i1]

swap(list_, 4,2) # swaps list index 4 and 2