Python - Function parameter value changing

Question

I'm very new to Python and am having trouble trying to write a replace function. The aim is to display a list, and then display the same list with a character being changed.

The code being used to call the function is:

str_list = ['s', 'h', 'r', 'u', 'g', 'g', 'e', 'd']

print("\nreplace Test")
new_str = list_function.replace(str_list, 'g', 'z')
print(list_function.to_string(str_list))
print(list_function.to_string(new_str))

new_str = list_function.replace(str_list, 'a', 'y')
print(list_function.to_string(new_str))

The code defining the function is:

def replace(my_list, old_value, new_value):
    new_list = my_list
    for k in range(0, length(new_list)):
        if new_list[k] == old_value:
            new_list[k] = new_value
    return my_list

However when I run the program, the output is:

replace Test
s, h, r, u, z, z, e, d
s, h, r, u, z, z, e, d
s, h, r, u, z, z, e, d

I'd like only the second list to be altered, and I ran the debugger and found out that str_list is being altered as well as new_list, but I just cannot work out what I'm doing wrong. Any help would be greatly appreciated.

thefourtheye · Accepted Answer · 2013-11-05 13:00:47Z

3

when you do

new_list = my_list

you are not copying the list, but assigning the reference to a new variable. What you need is

new_list = my_list[:]

my_list[:] actually creates a new copy in the memory. With that change, the program produces

replace Test
s, h, r, u, g, g, e, d
s, h, r, u, g, g, e, d
s, h, r, u, g, g, e, d

You can confirm whether the new copy is getting created or not using id in CPython.

new_list = my_list
print id(new_list), id(my_list)

will print the same address twice, which means that they both are pointing to the same element.

new_list = my_list[:]
print id(new_list), id(my_list)

will print two different addresses.

edited Nov 5, 2013 at 13:00

answered Nov 5, 2013 at 12:55

thefourtheye

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1 Comment

Dylsbo Over a year ago

Thank you very much, I've been spending hours trying to fix this and it was something so small.

score 0 · Accepted Answer · 2013-11-05 13:08:44Z

0

The best way to understand the difference is with some examples.

a = []
b = []

id(a)==id(b)
#False

a = []
b = a

id(a)==id(b)
#True

The id function simply returns the object’s unique id or object’s memory address.

Now to create a list with the same elements of a but with different id you can do the following:

a = [1,2,3,4,5]
#Both works
b = list(a) 
b = a[:]

id(a)==id(b)
#False

list() works with a generator, [:] doesn't. list() will construct a new list based on the sequence given. Most of the times list() can be replace by [:].

edited Nov 5, 2013 at 13:08

answered Nov 5, 2013 at 13:02

user1301404

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