CODEWITHSUNDEEP
python
user3210005
user3210005

Reputation: 43

issue with unchangable input variable in python sort

I am writing a series of python methods with different sorts. What I have is user inputted numbers which I turn into a list. I then run my bubble_sort method on the input list and I then print the result. I have my initial input list saved so that I can re-sort using selection_sort but when I print out my original list I instead am printing the bubble_sorted list. I am new to python so I am not sure if I am missing a fundamental concept about variables in the language. Here is my code

def listify(i):
    temp_list = []
    input_list_local = []
    input_list_local = list(i)
    for char in input_list_local:
        if char.isdigit():
            temp_list.append(char)
    return temp_list


def bubble_sort(input):
    for i in range(len(input)-1):
        for j in range(len(input)-1):
            if(input[j] > input[j+1]):
                tmp = input[j]
                input[j] = input[j+1]
                input[j+1] = tmp
    return input

def selection_sort(input):
    pass


input = raw_input("enter random numbers here seperated by spaces-> ")
print("you entered "+input)
input_list = listify(input)
print(input_list)
pass_list = input_list
print(bubble_sort(pass_list))
print(input_list) #should print original input list.  Instead prints sorted list

Upvotes: 2

Views: 44

Answers (1)

Simeon Visser
Simeon Visser

Reputation: 122446

You are modifying the same list: pass_list = input_list means that pass_list is just a different name for the same list object. So after calling bubble_sort(pass_list) you've modified pass_list but also input_list because they're the same object (in memory). Hence you're seeing the sorted list when you're printing input_list.

You can fix it by doing:

pass_list = input_list[:]

This makes a copy of input_list by using Python's slicing notation. You can then safely sort pass_list and it won't affect the original input_list.

Upvotes: 4

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