First look at my Python code,
s = [[1, 2], [3, 6], [9, 6]]
t = s
i = s[0]
del i
del t[0]
There is three references for [2, 1]; t[0], s[0] and "i", so if we try and delete of them it shouldn't effect on [1,2] because [1,2] still has two another reference so Python garbage-collection system doesn't remove [1,2] do to the having no reference. So why deleting "i" doesn't effect on [1,2] but deleting t[0] does.
You mistakenly understand the function of the del statement. For variable names:
del variable_name
The del statement deletes the binding of the name to the object from the global variable.
For the combination of container and key:
del container[key]
It is responsible for deleting the elements in the container according to the key.
Use some simple symbols to explain what you are doing. Here I use a pair of parentheses to represent an object, and a word ref or a variable name represents a reference. After you create these objects, it will look like this:
t s
↓ ↓
(list [ref, ref])
i ┌───┘ └───────┐
↓ ↓ ↓
(list [ref, ref]) (list [ref, ref])
┌───┘ ┌─┘ ┌──────┘┌───┘
↓ ↓ ↓ ↓
(int 1) (int 2) (int 3) (int 6)
If you execute del i, the final result is as follows:
t s
↓ ↓
(list [ref, ref])
┌───┘ └───────┐
↓ ↓
(list [ref, ref]) (list [ref, ref])
┌───┘ ┌─┘ ┌──────┘┌───┘
↓ ↓ ↓ ↓
(int 1) (int 2) (int 3) (int 6)
If you execute del t[0], the final result is as follows:
t s
↓ ↓
(list [ref])
i └────────────┐
↓ ↓
(list [ref, ref]) (list [ref, ref])
┌───┘ ┌─┘ ┌──────┘┌───┘
↓ ↓ ↓ ↓
(int 1) (int 2) (int 3) (int 6)