In principle you could use:
np.equal(a,a).sum(0)
but, for unknown reason, np.equal operates only on "normal" arrays. maybe you
can transform the array to arrays of numbers, for example by hash.
Nadav
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מאת: numpy-discussion-boun...@scipy.org בשם josef.p...@gmail.com
נשל
I know David Cournapeau has done some work on using gcov for coverage
with Numpy.
Unaware of this, (doh! -- I should have Googled first), I wrote a small
C code-coverage tool built on top of valgrind's callgrind tool, so it
basically only works on x86/AMD64 unixy platforms, but unlike gcov it
On Mon, Oct 26, 2009 at 2:12 PM, Christopher Barker
wrote:
> Alan G Isaac wrote:
>> On 10/26/2009 4:04 AM, Nils Wagner wrote:
>>> how can I obtain the multiplicity of an entry in a list
>>> a = ['abc','def','abc','ghij']
>>
>> That's a Python question, not a NumPy question.
>
> but we can make it
Alan G Isaac wrote:
> On 10/26/2009 4:04 AM, Nils Wagner wrote:
>> how can I obtain the multiplicity of an entry in a list
>> a = ['abc','def','abc','ghij']
>
> That's a Python question, not a NumPy question.
but we can make it a numpy question!
In [15]: a = np.array(['abc','def','abc','ghij'])
Hi Sturla,
Thanks for the overview.
On Mon, Oct 26, 2009 at 5:24 PM, Sturla Molden wrote:
> Ralf Gommers skrev:
> >
> > If anyone with knowledge of the differences between the C and Fortran
> > versions could add a few notes at the above link, that would be great.
> >
> The most notable differe
Ralf Gommers skrev:
>
> If anyone with knowledge of the differences between the C and Fortran
> versions could add a few notes at the above link, that would be great.
>
The most notable difference (from a user perspective) is that the
Fortran version has more transforms, such as discrete sine and
Hi Everyone,
Is Numpy supposed to behave this like this when converting an array of
numbers to an array of strings with astype?
print(arange(20).astype(np.str))
['0' '1' '2' '3' '4' '5' '6' '7' '8' '9' '1' '1' '1' '1' '1' '1' '1'
'1' '1' '1']
When I do the following it works fine,
print(arange(
On Mon, Oct 26, 2009 at 12:04 AM, Ralf Gommers
wrote:
>
>
> On Sun, Oct 25, 2009 at 11:51 PM, Charles R Harris <
> charlesr.har...@gmail.com> wrote:
>
>>
>>
>> On Sun, Oct 25, 2009 at 4:21 PM, Ralf Gommers <
>> ralf.gomm...@googlemail.com> wrote:
>>
>>> Hi all,
>>>
>>> Can anyone tell me if fftpac
Mon, 26 Oct 2009 08:15:51 -0400, Neal Becker wrote:
> This link:
>
> http://docs.scipy.org/doc/scipy/reference/generated/
scipy.stats.var.html#scipy.stats.var
>
> gives 500 internal server error
Now that's strange. It's a static page.
--
Pauli Virtanen
___
On 10/26/2009 4:04 AM, Nils Wagner wrote:
> how can I obtain the multiplicity of an entry in a list
> a = ['abc','def','abc','ghij']
That's a Python question, not a NumPy question.
So comp.lang.python would be a better forum.
But here's a simplest solution::
a = ['abc','def','abc','ghij']
for it
This link:
http://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.var.html#scipy.stats.var
gives 500 internal server error
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On Mon, Oct 26, 2009 at 2:16 AM, David Goldsmith wrote:
> Technically, after "Needs Review," it's supposed to go through "Needs Work
> (Reviewed)" The "by the book" way to do it would be to:
>
> 0 & 1) Provide comments in the Discussion section and change status to
> "Needs Work (Reviewed)" (in e
Hi all,
how can I obtain the multiplicity of an entry in a list
a = ['abc','def','abc','ghij']
The multiplicity of 'abc' is 2.
'def' is 1.
'ghij' is 1.
Nils
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