[BangPypers] [New bie question ] Clarification on "Multiply" operator applied to list(data structure)
Aman Aggarwal
aman.coe at gmail.com
Wed Jun 10 13:27:39 CEST 2009
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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