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To: python-list@python.org
From: Peter Otten <__peter__@web.de>
Subject: Re: Passing a list into  a list .append() method
Date: Tue, 09 Sep 2014 09:13:16 +0200
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JBB wrote:

> I have a list with a fixed number of elements which I need to grow; ie.
> add rows of a fixed number of elements, some of which will be blank.
> 
> e.g. [['a','b','c','d'], ['A','B','C','D'], ['', 'aa', 'inky', ''], ['',
> 'bb', 'binky', ''], ... ]
> 
> This is a reduced representation of a larger list-of-lists problem that
> had me running in circles today.
> 
> I think I figured out _how_ to get what I want but I am looking to
> understand why one approach works and another doesn't.

The actual problem is that you are appending the same list multiple times. 
In nuce:

>>> inner = [1, 2, 3]
>>> outer = [inner, inner, inner]

Remember that outer[0] is inner just like the two other items of the outer 
list:

>>> outer[0] is inner
True
>>> outer[1] is inner
True
>>> outer[2] is inner
True

So

>>> outer[0][0] = 42

effectively changes inner

>>> inner
[42, 2, 3]

which is reflected in the following output:

>>> outer
[[42, 2, 3], [42, 2, 3], [42, 2, 3]]

With list(inner) you create a (shallow) copy of inner

>>> inner = [1, 2, 3]
>>> outer = [list(inner), list(inner), list(inner)]
>>> outer[0] is inner
False

and a subsequent assignment changes only that specific copy of inner:

>>> outer[0][0] = 42
>>> outer
[[42, 2, 3], [1, 2, 3], [1, 2, 3]]

So the key to the solution is that you create a new list on every iteration 
of the loop. With that in mind I'd write

for a, b, in zip(qq, rr):
    proc_file.append(["", a, b, ""])

or alternatively if the actual inner list is more complex

template = ["", "", "", ""]
for p in zip(qq, rr):
    inner = list(template)
    inner[1:3] = p
    proc_file.append(inner)

> 1) What does NOT work as desired:
>     proc_file.append((blank_r))  # Add a row of blanks
>     proc_file[i+2][1] = j[0]
>     proc_file[i+2][2] = j[1]

> 2) What works as desired:

>     proc_file.append(list(blank_r))  # Change it to list(blank_r) and it
>     works 
>     proc_file[i+2][1] = j[0]
>     proc_file[i+2][2] = j[1]
>     print len(proc_file), blank_r, proc_file