I am trying to use defaultdict(list) as,
dict = defaultdict(list)
dict['A'][1] = [1]
or
dict['A'][1] = list([1])
I got an error,
IndexError: list assignment index out of range
if I do
dict['A'][1].append(1)
IndexError: list index out of range
I am wondering what is the issue here.
There are some issues here:
You're using 1 for the index value while the default list you have start at index 0.
When you append, you don't need to specify the index.
And finally, it's not a good idea to declare a variable with the same name as a built-in type (dict, in this case) since that would usually result in unexpected behavior later on when you would use the built-in type.
Revised, your code would be:
>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> d['A'].append(1)
>>> d['A']
[1]
Firstly, don't use dict as a name. Use something like d so you don't mask the built-in name dict.
Secondly, d['A'][1] = list([1]) tries to access the index 1 of an empty list (the value you initialize the defaultdict with) and assign to it, that's a no-no.
You are probably looking for:
d = defaultdict(list)
d['A'].append(1)
1 in d['A'][1], is a key and assign to it [1] as value, what should I do, so d['A'] is a nested dict.
defaultdict(dict)defaultdict with a list and not a dict?
[][1] = [1]. An empty list doesn't have a1index to assign to.list.appendis how it's done.