Python: why does my list change when I'm not actually changing it? -

newbie question, please gentle:

list = [1, 2, 3, 4, 5] list2 = list  def fxn(list,list2):     number in list:         print(number)         print(list)         list2.remove(number)         print("after remove list  ", list, " , list 2  ", list2)     return list, list2  list, list2 = fxn(list, list2) print("after fxn list  ", list) print("after fxn list2  ", list2) 

this results in:

1 [1, 2, 3, 4, 5] after remove list   [2, 3, 4, 5]  , list 2   [2, 3, 4, 5] 3 [2, 3, 4, 5] after remove list   [2, 4, 5]  , list 2   [2, 4, 5] 5 [2, 4, 5] after remove list   [2, 4]  , list 2   [2, 4] after fxn list   [2, 4] after fxn list2   [2, 4] 

i don't understand why list changing when doing list2.remove(), not list.remove(). i'm not sure search terms use figure out.

that's because both list , list2 referring same list after did assignment list2=list.

try see if referring same objects or different:

id(list) id(list2) 

an example:

>>> list = [1, 2, 3, 4, 5] >>> list2 = list >>> id(list) 140496700844944 >>> id(list2) 140496700844944 >>> list.remove(3) >>> list [1, 2, 4, 5] >>> list2 [1, 2, 4, 5] 

if want create duplicate copy of list such list2 doesn't refer original list copy of list, use slice operator:

list2 = list[:] 

an example:

>>> list [1, 2, 4, 5] >>> list2 [1, 2, 4, 5] >>> list = [1, 2, 3, 4, 5] >>> list2 = list[:] >>> id(list) 140496701034792 >>> id(list2) 140496701034864 >>> list.remove(3) >>> list [1, 2, 4, 5] >>> list2 [1, 2, 3, 4, 5] 

also, don't use list variable name, because originally, list refers type list, defining own list variable, hiding original list refers type list. example:

>>> list <type 'list'> >>> type(list) <type 'type'> >>> list = [1, 2, 3, 4, 5] >>> list [1, 2, 3, 4, 5] >>> type(list) <type 'list'>