This came up in [a matplotlib
issue](https://github.com/matplotlib/matplotlib/issues/5221):
>>> np.histogram(np.arange(10), range=(0.0, np.inf))
(array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0]),
array([ nan, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf]))
>>> np.histogram(np.arange(10), ran
I tried cleaning the git dir, and trying again. It still didn't work giving me
the report:
==
ERROR: test_scripts.test_f2py
--
Traceback (most recent call last)
Thanks Yu,
There was nothing in my PYTHONPATH at first, and adding my numpy directory
('/Users/lzkelley/Programs/public/numpy') didn't help (same error). In both
cases, adding 'print(np)' yields:
> On Oct 18, 2015, at 7:22 PM, Feng Yu wrote:
>
> Hi Luke
Thanks for the help Nathaniel --- but building via `./runtests.py` is failing
in the same way. Hopefully Numpy-discussion can help me out.
I'm able to build using `python setup.py build_ext --inplace` but both trying
to run `python setup.py install` or `./runtests.py` leads to the following
er
On Wed, Dec 24, 2014 at 7:56 AM, Sturla Molden
wrote:
> On 24/12/14 14:34, Sturla Molden wrote:
> > I would rather have SciPy implement this with the overlap-and-add method
> > rather than padding the FFT. Overlap-and-add is more memory efficient
> > for large n:
>
> (eh, the list should be)
>
>
0=No
1=Yes
Some typical values are:
arr=numpy.array([51199,37013,36885,36889,34841,2062,34837,2061,35033,349
61,2185,37013,36885,2185,4109,4233], dtype=numpy.uint16)
How would I extract groups of/individual bit values from such an array?
Regards
Luke Pinner
How about the masked aray filled method?
eg numpy.savetxt(somefilename, somemaskedarray.filled(somevalue))
Luke
-Original Message-
From: Guillaume Chérel
Sent: Monday, 6 September 2010 10:36 PM
To: Discussion of Numerical Python
Subject: [Numpy-discussion] saving masked array to a text
Simple as that...
Thanks!
-Original Message-
From: numpy-discussion-boun...@scipy.org
[mailto:numpy-discussion-boun...@scipy.org]
On Behalf Of Geordie McBain
Convert days to np.array, then you can index it with your indexmin:
days = [10, 20, 31, 41, 51, 59, 69, 79, 90, 100]
mydays
What I am currently doing is below,
#which is _rather_ slow on large arrays
#as I'm calling np.where len(days) times
daymin=np.zeros(indexmin.shape).astype(np.int)
for i,d in enumerate(days):
daymin[np.where(indexmin==i)]=d
Any better suggestions?
Regards
Luke Pinner
--
If you have
/
ideas discussed, or simply the presentation and discussion of the
algorithms/ideas themselves?
~Luke
On May 31, 10:34 am, Travis Oliphant <[EMAIL PROTECTED]> wrote:
> Christopher Barker wrote:
> >Anne Archibald wrote:
>
> >>I implemented the Kuiper statistic and would be
that if instead of ceil, round was used,
then it eliminated my problem, but I don't know if this would have
other undesirable effects in other situations.
I guess I could use range, but it is just a bit more tedious to code.
Thanks,
~Luke
___
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