Python: list of lists

Question

Running the code

listoflists = [] list = [] for i in range(0,10):     list.append(i)     if len(list)>3:         list.remove(list[0])         listoflists.append((list, list[0])) print listoflists 

returns

[([7, 8, 9], 0), ([7, 8, 9], 0), ([7, 8, 9], 0), ([7, 8, 9], 1), ([7, 8, 9], 2), ([7, 8, 9], 3), ([7, 8, 9], 4), ([7, 8, 9], 5), ([7, 8, 9], 6), ([7, 8, 9], 7)] 

so somehow the first argument of each tuple (list) is being updated each time in the list of lists, but the second argument list[0] is not. Can someone explain what's going on here and suggest a way to fix this? I'd like to output

[([0],0), ([0,1],0), ... 

Answer 1

Lists are a mutable type - in order to create a copy (rather than just passing the same list around), you need to do so explicitly:

listoflists.append((list[:], list[0])) 

However, list is already the name of a Python built-in - it'd be better not to use that name for your variable. Here's a version that doesn't use list as a variable name, and makes a copy:

listoflists = [] a_list = [] for i in range(0,10):     a_list.append(i)     if len(a_list)>3:         a_list.remove(a_list[0])         listoflists.append((list(a_list), a_list[0])) print listoflists 

---

Note that I demonstrated two different ways to make a copy of a list above: [:] and list().

The first, [:], is creating a slice (normally often used for getting just part of a list), which happens to contain the entire list, and thus is effectively a copy of the list.

The second, list(), is using the actual list type constructor to create a new list which has contents equal to the first list. (I didn't use it in the first example because you were overwriting that name in your code - which is a good example of why you don't want to do that!)