python - Appending to range function

Question

EDIT: Thank you for your kind replies on how to accomplish what I am trying to do, but the question is about WHY range().append() returns None if you try it in one step, and WHY it works if you two-step the procedure.

Im trying to create a numerical list but with a twist. I don't want a couple of numbers at the beggining of my list:

mlist = [0, 5, 6, 7, ...]

So i thought to do the following:

 mlist = range(5,n+1).append(0)

but silently fails because type(mlist) afterwards equals to NoneType ?! (related: type(range(5,10) evaluates to list Type)

If i try to do that in two steps eg:

>>> mlist = range(5,10)
#and then
>>> mlist.append(0)
>>> mlist
[5, 6, 7, 8, 9, 10, 0]

What's happening?

Answer 1

list.append() alters the list in place, and returns None. By assigning you assigned that return value to mlist, not the list you wanted to build. This is a standard Python idiom; methods that alter the mutable object never return the altered object, always None.

Separate the two steps:

mlist = range(5, n + 1)
mlist.append(0)

This adds [0] to the end; if you need the 0 at the start, use:

mlist = [0] + range(5, n + 1)

or you can use list.insert(), again as a separate call:

mlist = range(5, n + 1)
mlist.insert(0, 0)

but the latter has to shift up all elements one step and creating a new list by concatenation is the faster option for shorter lists, insertion wins on longer lists:

>>> from timeit import timeit
>>> def concat(n):
...     mlist = [0] + range(5, n)
... 
>>> def insert(n):
...     mlist = range(5, n)
...     mlist.insert(0, 0)
... 
>>> timeit('concat(10)', 'from __main__ import concat')
1.2668070793151855
>>> timeit('insert(10)', 'from __main__ import insert')
1.4820878505706787
>>> timeit('concat(1000)', 'from __main__ import concat')
23.53221583366394
>>> timeit('insert(1000)', 'from __main__ import insert')
15.84601092338562