Question

Instead of appending a value by list.append() to the list , Why can;t I assign a value to it like this?

In [24]: def a():
   ....:     a1=[5]
   ....:     print a1[0]
   ....:     

In [25]: a()
5

In [28]: def a():
   ....:     a1=[5]
   ....:     a1[1]=12
   ....:     print a1[1]
   ....:     

In [29]: a()
---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
/home/dubizzle/webapps/django/dubizzle/<ipython-input-29-72f2e37b262f> in <module>()
----> 1 a()

/home/dubizzle/webapps/django/dubizzle/<ipython-input-28-681d86164e67> in a()
      1 def a():
      2     a1=[5]
----> 3     a1[1]=12
      4     print a1[1]
      5 

IndexError: list assignment index out of range

Answer 1

a=[5] creates a list with size 1. The following is valid:

a[0] = 1

but you cannot alter the size of the list by just assigning a value to a given, non-existing position.

You can extend the list, though, by using append or +=:

a.append(2)
a += [2]

You can also pre-allocate a list with a given size:

a = [0] * 4

That would give you a list with four elements (all zeros). After that you could change any element in the range 0-3.

As a reference, here you have some documentation on lists from python.org.