On Jul 19, 2011, at 3:15 PM, Chad Netzer wrote:
> %python
import timeit
import numpy as np
>
t=timeit.Timer('k = m - 0.5', setup='import numpy as np;m =
np.ones([8092,8092],float); k = np.zeros(m.size, m.dtype)')
np.mean(t.repeat(repeat=10, number=1))
> 0.5855752944946288
On Mar 8, 2011, at 11:56 AM, Josh Hykes wrote:
> At some point in my code, I need to do a cell-wise multiplication of the
> properties with a state variable. The ideal method would (1) be fast (no
> Python loops) and (2) not waste memory constructing an entire property map.
> My best attempt usi
On Nov 18, 2010, at 6:49 AM, Venkat wrote:
> I am trying to reshape my text data which is in one single column (10,000
> rows).
> I want the data to be in 100x100 array form.
If all you want to do is converting the actual files, and you are using a
unix-ish operating system, you don't even need
Is there a way to execute tests even if they are marked as KNOWNFAIL? For
example, the module scipy.sparse.linalg has a few tests that are marked as such
on my architecture (OS-X 64 bit). Running
scipy.sparse.linalg(extra_argv=["--no-knownfail"])
changes the test result to ERROR, but it looks
On Oct 19, 2010, at 6:09 PM, Dewald Pieterse wrote:
> for xiter in range(xindex):
> for yiter in range(yindex):
> if edges[xiter,yiter,:] == [255,0,0]:
> groenpixelarea = groenpixelarea + 1
> if edges[xiter,yiter,:] == [0,255,0]:
>
According to the docstring, ndarray.copy should accept a keyword argument
"order". This doesn't seem to work for me:
>>> np.array([[1,2],[3,4]]).copy(order='C')
Traceback (most recent call last):
File "", line 1, in
TypeError: copy() takes no keyword arguments
Calling ndarray.copy with a posi
Hello,
I am trying to use a C library function from Python using numpy and
ctypes. Assume the function has the signature
void foo(int* bar)
so it usually takes an integer array as a parameter. In python, I define
foo.argtypes=[ndpointer(dtype="intc",flags="C_CONTIGUOUS")]
and can then call the
Hi Joe,
> Travis has freed his original book and large parts of it (e.g.,
> the C API docs) are now being incorporated into the
> actively-maintained manuals at docs.scipy.org. Please go there for
> the latest docs. You'll find that the fft section gives the 1/n
> formula when discussing ifft.
On Thu, Mar 26, 2009 at 7:02 PM, Gideon Simpson
wrote:
> I thought it was the same as the MATLAB format:
>
> http://www.mathworks.com/access/helpdesk/help/techdoc/index.html?/access/helpdesk/help/techdoc/ref/fft.html&http://www.google.com/search
> ?client=safari&rls=en-us&q=MATLAB+fft&ie=UTF-8&oe=
Hi Michael,
> this documentation is saying that the difference between the equations
> for the fft and ifft is a factor of 1/n (not the numpy implementations).
> if you do
>
> output = numpy.ifft( numpy.fft( input ) )
>
> and you get output = input, then the normalizations are appropriately
> wei
Hello,
I just started to use python and numpy for some numerical analysis. I
have a question about the definition of the inverse Fourier transform.
The user gives the formula (p.180)
x[m] = Sum_k X[k] exp(j 2pi k m / n)
where X[k] are the Fourier coefficients, and n is the length of the arrays
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