On Sun, Feb 5, 2012 at 10:41 AM, Paolo p.zaff...@yahoo.it wrote:
I solved using 'rb' instead of 'r' option in the open file task.
that would do it, if it's binary data, but you might as well so it
right:
matrix=.join(f.readlines())
readlines is giving you a list of the data, as separated
I have two large matrices, say, ABC and DEF, each with a shape of 7000 by
4500. I have another list, say, elem, containing 850 values from ABC. I am
interested in finding out the corresponding values in DEF where ABC has
elem and store them *separately*. The code that I am using is:
for i in
basic difference between the commands:
import numpy as np
from numpy import *
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David: from 9-10 minutes to about 2-3 seconds, it's amazing!
Thanks,
Naresh
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The namespace is different. If you want to use numpy.sin(), for
example, you would use:
import numpy as np
np.sin(angle)
or
from numpy import *
sin(angle)
I generally prefer the first option because then I don't need to worry
about multiple imports writing on top of each other (i.e., having
Hello,
Is there a way to specify a format for the datetime64 constructor? The
constructor doesn't have a doc. I have dates in a file with the format
MM/dd/YY. datetime64 used to be able to parse these in 1.6.1 but the dev
version throws an error.
Cheers,
John
Short answer: Create 16 view arrays, each with a stride of 4 in both
dimensions. Test them against the conditions and combine the tests with an |=
operator. Thus you replace the nested loop with one that has only 16
iterations. Or reshape to 3 dimensions, the last with length 4, and you can do
On Mon, Feb 6, 2012 at 2:57 PM, Sturla Molden stu...@molden.no wrote:
Short answer: Create 16 view arrays, each with a stride of 4 in both
dimensions. Test them against the conditions and combine the tests with an
|= operator. Thus you replace the nested loop with one that has only 16
Something like this:
m,n = data.shape
x = data.reshape((m,n//4,4))
z = (x[0::4,...] = t1) (x[0::4,...] = t1)
z |= (x[1::4,...] = t1) (x[1::4,...] = t1)
z |= (x[2::4,...] = t1) (x[2::4,...] = t1)
z |= (x[3::4,...] = t1) (x[3::4,...] = t1)
found = np.any(z, axis=2)
Sturla
Sendt fra min iPad
The last t1 on each lineis of course t2. Sorry for the typo. Hard to code on an
ipad ;-)
Sturla
Sendt fra min iPad
Den 6. feb. 2012 kl. 22:12 skrev Sturla Molden stu...@molden.no:
Something like this:
m,n = data.shape
x = data.reshape((m,n//4,4))
z = (x[0::4,...] = t1) (x[0::4,...]
Hey John,
NumPy doesn't provide this, because it's already provided by the
datetime.date.strftime function in Python:
http://docs.python.org/library/datetime.html#datetime.date.strftime
One reason this format isn't supported automatically is that parsing
MM/dd/YY is inherently ambiguous, and
# Make a 4D view of this data, such that b[i,j]
# is a 2D block with shape (4,4) (e.g. b[0,0] is
# the same as a[:4, :4]).
b = as_strided(a, shape=(a.shape[0]/4, a.shape[1]/4, 4, 4),
strides=(4*a.strides[0], 4*a.strides[1], a.strides[0],
a.strides[1]))
Yes :-) Being
On Sat, Feb 4, 2012 at 3:55 PM, Ralf Gommers
ralf.gomm...@googlemail.com wrote:
On Wed, Dec 14, 2011 at 6:50 PM, Ralf Gommers ralf.gomm...@googlemail.com
wrote:
On Wed, Dec 14, 2011 at 3:04 PM, David Cournapeau courn...@gmail.com
wrote:
On Tue, Dec 13, 2011 at 3:43 PM, Ralf Gommers
That makes sense.
I figured that ambiguity was the reason it was removed.
Thank you for the explanation. I'm a big fan of your work.
John
On Mon, Feb 6, 2012 at 1:18 PM, Mark Wiebe mwwi...@gmail.com wrote:
Hey John,
NumPy doesn't provide this, because it's already provided by the
On Mon, Feb 6, 2012 at 1:17 AM, Wes McKinney wesmck...@gmail.com wrote:
Whenever I get motivated enough I'm going to make a pull request on
NumPy with something like khash.h and start fixing all the O(N log N)
algorithms floating around that ought to be O(N). NumPy should really
have a match
Hi,
On Mon, Feb 6, 2012 at 9:16 PM, Moroney, Catherine M (388D)
catherine.m.moro...@jpl.nasa.gov wrote:
Hello,
I have to write a code to downsample an array in a specific way, and I am
hoping that
somebody can tell me how to do this without the nested do-loops. Here is
the problem
2012/2/6 Stéfan van der Walt ste...@sun.ac.za
Hi all,
I noticed the following docstring on ``np.polynomial.polyval``:
In [116]: np.polynomial.polyval?
File: /home/stefan/src/numpy/numpy/lib/utils.py
Definition: np.polynomial.polyval(*args, **kwds)
Docstring:
`polyval` is
Hi,
Sorry for my latest post, hands way too quick ;(
On Mon, Feb 6, 2012 at 9:16 PM, Moroney, Catherine M (388D)
catherine.m.moro...@jpl.nasa.gov wrote:
Hello,
I have to write a code to downsample an array in a specific way, and I am
hoping that
somebody can tell me how to do this without
irfftn is an optimization for real input and does not take complex
input. You have to use numpy.fft.ifftn instead:
import numpy
a_shape = (63, 4, 98)
a = numpy.complex128(numpy.random.rand(*a_shape)+\
... 1j*numpy.random.rand(*a_shape))
axes = [0, 2]
numpy.fft.ifftn(a, axes=axes)
Or
On 6 February 2012 21:41, Ralf Gommers ralf.gomm...@googlemail.com wrote:
On Mon, Feb 6, 2012 at 8:17 AM, Scott Sinclair scott.sinclair...@gmail.com
wrote:
On 5 February 2012 13:07, Ralf Gommers ralf.gomm...@googlemail.com
wrote:
Does it need to be a new repo, or would permissions on
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