[BangPypers] [New bie question ] Clarification on "Multiply" operator applied to list(data structure)

[Senthil Kumaran]

On Wed, Jun 10, 2009 at 04:57:39PM +0530, Aman Aggarwal wrote:
> 
> I am reading "How to think like a computer scientist" which is an
> introductory test in Python.
> 
> I wanna clarify the behaviour of multiply operator(*) when applied to
> list(data structure).
> 
> 
> I wonder why
> 
> [[0] * columns] * rows
> 
> causes "...each row is an alias of the other rows..."
> 
> but not
> 
> [[0] * columns]
> 
> i.e. why each [0] in a row is not an alias of other row element.


It is bit tricky in that code.. 
You have to understand that in the second option, the matrix is formed
row-by-row, wherein a new row (list) is created as [0] * columns.


This is what the second version essentially is doing:

>>> rows = 2
>>> cols = 2
>>> mat = []
>>> mat = [[0] * cols] # first row
>>> mat = mat + [[0] * cols] # second row
>>> mat
[[0, 0], [0, 0]]
>>> mat[0][0] = 'Spam'
>>> mat
[['Spam', 0], [0, 0]]
>>> 

This is the what the first version of the code was doing.

>>> rows = 2
>>> cols = 2
>>> mat = []
>>> mat = [[0] * cols] # first row
>>> mat = mat * 2 # second row
>>> mat
[[0, 0], [0, 0]]
>>> mat[0][0] = 'Spam'
>>> mat
[['Spam', 0], ['Spam', 0]]
>>> 

Does this clarify your doubt?

-- 
Senthil