On Monday, 20 April 2020 at 02:50:29 UTC, 9il wrote:
On Monday, 20 April 2020 at 02:42:33 UTC, 9il wrote:
On Sunday, 19 April 2020 at 20:29:54 UTC, p.shkadzko wrote:
On Sunday, 19 April 2020 at 20:06:23 UTC, jmh530 wrote:
[...]
Thanks. I somehow missed the whole point of "a *
a.transposed"
On Monday, 20 April 2020 at 19:06:53 UTC, p.shkadzko wrote:
[snip]
It is. I was trying to calculate the covariance matrix of some
dataset X which would be XX^T.
Incorrect. The covariance matrix is calculated with matrix
multiplication, not element-wise multiplication. For instance, I
often
On Sunday, 19 April 2020 at 21:27:43 UTC, jmh530 wrote:
On Sunday, 19 April 2020 at 20:29:54 UTC, p.shkadzko wrote:
[snip]
Thanks. I somehow missed the whole point of "a * a.transposed"
not working because "a.transposed" is not allocated.
a.transposed is just a view of the original matrix. E
On Monday, 20 April 2020 at 02:42:33 UTC, 9il wrote:
On Sunday, 19 April 2020 at 20:29:54 UTC, p.shkadzko wrote:
On Sunday, 19 April 2020 at 20:06:23 UTC, jmh530 wrote:
[...]
Thanks. I somehow missed the whole point of "a * a.transposed"
not working because "a.transposed" is not allocated.
On Sunday, 19 April 2020 at 20:29:54 UTC, p.shkadzko wrote:
On Sunday, 19 April 2020 at 20:06:23 UTC, jmh530 wrote:
On Sunday, 19 April 2020 at 19:20:28 UTC, p.shkadzko wrote:
[...]
Ah, you're right. I use it in other places where it hasn't
been an issue.
I can do it with an allocation (be
On Sunday, 19 April 2020 at 20:29:54 UTC, p.shkadzko wrote:
[snip]
Thanks. I somehow missed the whole point of "a * a.transposed"
not working because "a.transposed" is not allocated.
a.transposed is just a view of the original matrix. Even when I
tried to do a raw for loop I ran into issues
On Sunday, 19 April 2020 at 20:06:23 UTC, jmh530 wrote:
On Sunday, 19 April 2020 at 19:20:28 UTC, p.shkadzko wrote:
[snip]
well no, "assumeContiguous" reverts the results of the
"transposed" and it's "a * a".
I would expect it to stay transposed as NumPy does "assert
np.all(np.ascontiguous(a.T
On Sunday, 19 April 2020 at 19:20:28 UTC, p.shkadzko wrote:
[snip]
well no, "assumeContiguous" reverts the results of the
"transposed" and it's "a * a".
I would expect it to stay transposed as NumPy does "assert
np.all(np.ascontiguous(a.T) == a.T)".
Ah, you're right. I use it in other places
On Sunday, 19 April 2020 at 19:13:14 UTC, p.shkadzko wrote:
On Sunday, 19 April 2020 at 18:59:00 UTC, jmh530 wrote:
On Sunday, 19 April 2020 at 17:55:06 UTC, p.shkadzko wrote:
snip
So, lubeck mtimes is equivalent to NumPy
"a.dot(a.transpose())".
There are elementwise operation on two matric
On Sunday, 19 April 2020 at 18:59:00 UTC, jmh530 wrote:
On Sunday, 19 April 2020 at 17:55:06 UTC, p.shkadzko wrote:
snip
So, lubeck mtimes is equivalent to NumPy
"a.dot(a.transpose())".
There are elementwise operation on two matrices of the same
size and then there is matrix multiplication.
On Sunday, 19 April 2020 at 17:55:06 UTC, p.shkadzko wrote:
snip
So, lubeck mtimes is equivalent to NumPy "a.dot(a.transpose())".
There are elementwise operation on two matrices of the same size
and then there is matrix multiplication. Two different things.
You had initially said using an mx
On Sunday, 19 April 2020 at 17:22:12 UTC, jmh530 wrote:
On Sunday, 19 April 2020 at 17:07:36 UTC, p.shkadzko wrote:
I'd like to calculate XX^T where X is some [m x n] matrix.
// create a 3 x 3 matrix
Slice!(double*, 2LU) a = [2.1, 1.0, 3.2, 4.5, 2.4, 3.3, 1.5,
0, 2.1].sliced(3, 3);
auto b = a
On Sunday, 19 April 2020 at 17:07:36 UTC, p.shkadzko wrote:
I'd like to calculate XX^T where X is some [m x n] matrix.
// create a 3 x 3 matrix
Slice!(double*, 2LU) a = [2.1, 1.0, 3.2, 4.5, 2.4, 3.3, 1.5, 0,
2.1].sliced(3, 3);
auto b = a * a.transposed; // error
Looks like it is not possible
I'd like to calculate XX^T where X is some [m x n] matrix.
// create a 3 x 3 matrix
Slice!(double*, 2LU) a = [2.1, 1.0, 3.2, 4.5, 2.4, 3.3, 1.5, 0,
2.1].sliced(3, 3);
auto b = a * a.transposed; // error
Looks like it is not possible due to "incompatible types for (a)
* (transposed(a)): Slice!
14 matches
Mail list logo