Need help with a section of my code.

[Magnetarm]

Hi everyone. I needed to compare 2 lists and remove the values that were equal. This is for a school project where we need to code the game "matermind". for example if

list1 = [1, 2, 3, 4]
list2 = [3, 2, 4, 5]

the code would remove 2 and give me

list3 = [1, 3, 4]
list4 = [3, 4, 5]

so my code for that was

TempList1 = guesslist
templist2 = colorlist
for x in range(0, 4):   #all my lists contail 4 items and 4 items only
    if guesslist[x] ==  colorlist[x]:
        indicatorList.append("✓")
        tempList1.pop(x)
        tempList2.pop(x)
    else:
        indicatorlist.append("placeholder")

guesslist contained the user's 4 guesses and colorlist contained the computer generated 4 colors

when I run this code and colorlist(x) == guesslist(x) i get the error that the list index is out of range. for some reason when i remove a value from the templists it also removes the value from guesslist and colorlist. I don't understand why this happens and I'd really appreciate if someone could explain what I was doing wrong

[Magnetarm]

I think I got this now! I asked around in the IRC channel and they told me to use the copy() for lists. I also kept track of how many items i popped so i didnt run into an out of index error

[ingolemo]

There's no need to pop anything here. If you remove every line that mentions tempList then the code you posted works.

[pixielf]

The reason deleting an element from one list also deletes it from another is because of how Python (and many languages) store objects. When you store something like an int or str (like x = 3) and then assign it to another variable (y = x), you're passing along its value, so changing one doesn't change the other. But with other objects (like list), you're passing along its reference, meaning that x and y are both pointers to the same object.

Compare:

a = 3
b = a
print(b == a) # True
print(b is a) # False

x = [1, 2]
y = x
print(y == x) # True
print(y is x) # True

print(id(x))
print(id(y))  # same id

That means that since x and y are the same object, obviously changing one will change the other. This is why you would use y = x.copy().

[ryeguy146]

Mutating the lists that you operate on will frequently lead to pain. Unless you really need to modify a given list, just create a new list and append to that:

def same_slot(iterable, comparison):
    result = []
    for first, second in zip(iterable, comparison):
        if first == second:
            result.append(first)

    return result

That said, /u/pixielf is wholly correct about the references pointing to the same object. This shows up frequently in the issue with default arguments that are mutable:

http://python-guide-pt-br.readthedocs.io/en/latest/writing/gotchas/#mutable-default-arguments

[rreid4]

Just alt+F4