Thank all for the very helpful replies. The goal of the matrix
multiply exercise was just to help my son and I learn Python better.
I now understand *why* my initialization of [c] was wrong and I am
continuing to check out numpy and scipy.
Regards,
David
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On 12 Jul, 14:59, sturlamolden wrote:
> ma = np.matrix(a)
> mb = np.matrix(b)
> a*b
ma*mb
Sorry for the typo.
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On 12 Jul, 07:39, David wrote:
> Should the following line work for defining a matrix with zeros?
>
> c= [[0]*col]*row
No. The rows will be aliased.
This will work:
c = [[0]*col for i in range(row)]
Note that Python lists are not ment to be used as matrices. We have
NumPy or the array module f
David writes:
> Should the following line work for defining a matrix with zeros?
>
> c= [[0]*col]*row
No. Python lists are not matrixes and are not arrays.
If you want good implementations of arrays and matrices, use NumPy
http://numpy.scipy.org/>.
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On Mon, Jul 11, 2011 at 10:39 PM, David wrote:
> Should the following line work for defining a matrix with zeros?
>
> c= [[0]*col]*row
>
> where "col" is the number of columns in the matrix and "row" is of
> course the number of rows.
Nope. See the FAQ:
http://docs.python.org/faq/programming.html
Should the following line work for defining a matrix with zeros?
c= [[0]*col]*row
where "col" is the number of columns in the matrix and "row" is of
course the number of rows.
If this a valid way of initializing a matrix in Python 3.2.1, then it
appears to me that a bug surfaces in Python when p
