Hi,
here is something I am thinking about for some time and I am wondering whether
there is a better solution
within numpy.
The task is:
I have an array (30+ entries) with arrays each with length == 3, that is
initially empty like this:
n = 100 # for test otherwise ~30
a1 =
On Tue, Feb 8, 2011 at 09:24, EMMEL Thomas thomas.em...@3ds.com wrote:
Hi,
here is something I am thinking about for some time and I am wondering
whether there is a better solution
within numpy.
The task is:
I have an array (30+ entries) with arrays each with length == 3, that is
Tue, 08 Feb 2011 15:24:10 +, EMMEL Thomas wrote:
[clip]
n = 100 # for test otherwise ~30
a1 = reshape(zeros(3*n).astype(float), (n,3))
a2 = zeros(n).astype(int)
for indices, data in [...]:
#data = array((1.,2.,3.))
#indices = (1,5,60)
for index in indices:
I have an array of (say, row) vectors,
v = [ [ a1, a2, a3 ],
[ b1, b2, b3 ],
[ c1, c2, c3 ],
...
]
What is the optimal way to compute the norm of each vector,
norm(v)**2 = [
[ a1**2 + a2**2 + a3**2 ],
[ b1**2 + b2**2 + b3**2 ],
...
]
It
On Tue, Feb 8, 2011 at 10:44, Ben Gamari bgamari.f...@gmail.com wrote:
I have an array of (say, row) vectors,
v = [ [ a1, a2, a3 ],
[ b1, b2, b3 ],
[ c1, c2, c3 ],
...
]
What is the optimal way to compute the norm of each vector,
norm(v)**2 = [
[ a1**2 +
On Tue, 8 Feb 2011 10:46:34 -0600, Robert Kern robert.k...@gmail.com wrote:
(v*v).sum(axis=1)[:,np.newaxis]
You can leave off the newaxis bit if you don't really need a column vector.
Fair enough, I unfortunately neglected to mention that I ultimately want
to normalize these vectors, hence
On Tue, Feb 8, 2011 at 11:55, Ben Gamari bgamari.f...@gmail.com wrote:
On Tue, 8 Feb 2011 10:46:34 -0600, Robert Kern robert.k...@gmail.com wrote:
(v*v).sum(axis=1)[:,np.newaxis]
You can leave off the newaxis bit if you don't really need a column vector.
Fair enough, I unfortunately
On 08/02/2011 16:44, Ben Gamari wrote:
I have an array of (say, row) vectors,
v = [ [ a1, a2, a3 ],
[ b1, b2, b3 ],
[ c1, c2, c3 ],
...
]
What is the optimal way to compute the norm of each vector,
norm(v)**2 = [
[ a1**2 + a2**2 + a3**2 ],
On Tue, 08 Feb 2011 18:06:48 +, Andrew Jaffe a.h.ja...@gmail.com wrote:
For this shape=(N,3) vector, this is not what you mean: as Robert Kern
also has it you want axis=1, which produces a shape=(N,) (or the
[:,newaxis] version which produces shape=(N,1).
But what is the point of the
I don't know who handles these permissions, I didn't see a way to do it
myself.
Chuck
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/numpy-discussion
Tue, 08 Feb 2011 11:35:12 -0700, Charles R Harris wrote:
Permission to close ticket for Mark Wiebe
I don't know who handles these permissions, I didn't see a way to do it
myself.
Granted. (You'd need a shell account to do these changes.)
--
Pauli Virtanen
Hi,
I have an NxM array, which I am indexing with a 1-d, length N boolean
array. For example, with a 3x5 array:
In [1]: import numpy
In [2]: data = numpy.arange(15)
In [3]: data.shape = 3, 5
Now, I want to select rows 0 and 2, so I can do:
In [4]: mask = numpy.array([True, False, True])
In
12 matches
Mail list logo
