[BangPypers] [New bie question ] Clarification on "Multiply" operator applied to list(data structure)
Shivaraj M S
shivraj.ms at gmail.com
Wed Jun 10 15:09:16 CEST 2009
The first one is a case of reference/alias and second one is not.
case 1:
>>> m = [[0]*2]*4
>>> m[1][1]=7
>>> m
[[0, 7], [0, 7], [0, 7], [0, 7]]
case 2:
>>> m=[]
>>> for r in range(4):
... m+=[[0]*2]
...
>>> m
[[0, 0], [0, 0], [0, 0], [0, 0]]
>>> m[1][1]=7
>>> m
[[0, 0], [0, 7], [0, 0], [0, 0]]
We can make the second case look like first just by creating a reference to
[[0]*2] which is what was done implicitly when multiplication operator was
used.
case 2 - modified for reference:
>>> m=[]
>>> mi = [[0]*2]
>>> for r in range(4):
... m+=mi
...
>>> m
[[0, 0], [0, 0], [0, 0], [0, 0]]
>>> mi[0][1]=7
>>> m
[[0, 7], [0, 7], [0, 7], [0, 7]]
>>> m[1][1]=1
>>> m
[[0, 1], [0, 1], [0, 1], [0, 1]]
>>> mi
[[0, 1]]
Aman Aggarwal-4 wrote:
>
> Hello there
>
> 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).
>
> Consider the function make_matrix
>
> def make_matrix(rows, columns):
> """
> >>> make_matrix(4, 2)
> [[0, 0], [0, 0], [0, 0], [0, 0]]
> >>> m = make_matrix(4, 2)
> >>> m[1][1] = 7
> >>> m
> [[0, 0], [0, 7], [0, 0], [0, 0]]
> """
> return [[0] * columns] * rows
>
>
>
>
> The actual output is
>
> [[0, 7], [0, 7], [0, 7], [0, 7]]
>
>
>
>
> The correct version of make_matrix is :
>
>
>
> def make_matrix(rows, columns):
> """
> >>> make_matrix(3, 5)
> [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
> >>> make_matrix(4, 2)
> [[0, 0], [0, 0], [0, 0], [0, 0]]
> >>> m = make_matrix(4, 2)
> >>> m[1][1] = 7
> >>> m
> [[0, 0], [0, 7], [0, 0], [0, 0]]
> """
> matrix = []
> for row in range(rows):
> matrix += [[0] * columns]
> return matrix
>
>
> The reason why first version of make_matrix fails ( as explained in
> the book at 9.8 ) is that
>
> "...each row is an alias of the other rows..."
>
> 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.
>
>
>
>
>
> /*
> Everything worth doing is worth doing in excess
> */
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>
>
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